Readme - Fizwebfizweb.elte.hu/download/Fizika-BSc/Kalkulus-2/Kalkulus2...Kalkulus II Balog Dániel 2010.02.16 Inpropius integrál Motiváló példák: 1. R1 0 1 x dx A rendes Riemann-integrál

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Balog Daacuteniel

2010 juacutenius 2

A jegyzet Dr Simon Peacuteter Kalkulus II(mf1n1a02)oacuteraacutei alapjaacuten keacuteszuumllt a20092010 taneacutev maacutesodik feacutel-eacuteveacuteben

A jegyzet letoumlltheto a honlapomroacutel httpbalogdaniwebeltehuKalkulusIIA jegyzet kinyomtathatoacute nyomtatott formaacuteban illetve e-mailbenIM programokkal ezen readme csatolaacute-saacuteval szabadon terjeszthetoMaacutes taacuterhelyen valoacute elhelyezeacuteshez e-mailben kell eacutertesiacuteteni Ciacutemem skippersdgmailcomA jegyzet nem moacutedosiacutethatoacute Amennyiben hibaacutet talaacutelsz bennekeacuterlek eacutertesiacutets a fenti ciacutemen

Kuumlloumln koumlszoumlnet

Dr Simon Peacuteternek amieacutert a jegyzetet szabadidejeacuteben aacutetneacutezte valamintKuumlszter Melindaacutenak a 20100330-as jegyzet begeacutepeleacuteseacuteeacutert

utoacuteirat

A jegyzet teljesen ingyen hasznaacutelhatoacute de a koumlszisoumlroumlket szivesen fogadom

1

Kalkulus II

Balog Daacuteniel

20100209

1 Integraacutelaacutes

11 Racionaacutelis toumlrtfuumlggveacutenyek

polinompolinom Ezeket mindig lehet integraacutelni az ilyen primitiacutev fuumlggveacutenye

meghataacuterozhatoacute

Ellenpeacuteldaint

eminusx2

Ez szimpatikusan neacutez ki meacutegsem lehet kiszaacute-molni

1int

1ax+b

plint

1

x= ln(x) + c

int

1

x+ 3= ln |x+ 3|+ c

int

1

2xminus 8=

1

2middot ln(2xminus 8) + c

2int

cx+d

ax+bpl

int

x+ 2

x+ 1=

int

x+ 1 + 1

x+ 1=

int

x+ 1

x+ 1+

1

x+ 1dx = x+ ln(x + 1) + c

int

3x+ 1

2x+ 3= 1

int

(2x+ 3) middot 32 minus 7

2

2x+ 3=

int

3

2minus

72

2x+ 3dx =

3

2xminus 7

2middot 12ln(2x+ 3) + c

3int

1(xminusa)(xminusb) pl

int

3

x2 + xminus 22 =

int

3

(xminus (minus2)) middot (xminus 1)=

int minus1

x+ 2+

1

xminus 1dx3 = minus ln(x+2)+ln(xminus1)+c = ln

xminus 1

x+ 2+c

4int

cx+d

(xminusa)(xminusb) plint

3 + x

2 + 3xminus 2x2= 4

int

3 + x

(2minus x)(2x+ 1)=

int

A

2minus x+

B

2x+ 1dx

1a 3

2azeacutert kell hogy a 3x stimmeljen a7

2hogy az 1 stimmeljen

2 minusbplusmn

radicb2minus4ac

2arArr x1 = minus2 x2 = 1

3(koumlzoumls nevezore hozaacutes)42 + 3xminus 2x2 = 0 x1 = 2 x2 = 1

2

Truumlkk a kibontottat beszorzomminus2-vel minus2(x minus 2)(x + 12 ) iacutegy ki

is joumln A szeacutetbontaacutes koacutedneveparciaacutelis toumlrtekre bontaacutes

A(2x+ 1) +B(2minus x) = 3 + x(2AminusB)x+A+ 2B = 3 + x

2AminusB = 1 A+ 2B = 3

A = 1 B = 1

=

int

1

2minus x+

1

2x+ 1= minus ln(2minus x) +

1

2ln(2x+ 1) + c

5int

1(xminusa)2 pl

int

1

x2= minus 1

x+ c

6int

cx+d

(xminusa)2 pl

int

2x+ 3

(xminus 1)2=

int

(xminus 1) middot 2 + 5

(xminus 1)2=

int

2

xminus 1+

5

(x minus 1)2dx = 2 ln(xminus 1)minus 5

xminus 1+ c

7int

1(xminusc)2+d

plint

1

x2 + 1= arctg(x) + c

int

1

x2 + 4=

int

1

4middot 1

(x2 )2 + 1

=1

4middot 2 middot arctg(x

2)

int

1

x2 + 3=

1

3middotint

1

( xradic3)2 + 1

=1

3middotradic3 middot arctg( xradic

3)

int

1

x2 minus 2x+ 2=

int

1

(xminus 1)2 + 1= arctg(xminus 1)

A megoldaacutes meneteA valoacutes gyoumlkoumlket kell ellenorizni Ha van 2 darabvaloacutes gyoumlk laacutesd 4 peacutelda ha 1 db van ld 5 ha nincs ld 7

8int

ax+b

(xminusc)2+d2 pl

int

2x

1 + x2Felismerheto hogy ez

f prime

ftipusuacute

int

2x+ 3

1 + x2=

int

2x

1 + x2+

3

1 + x2= ln(1 + x2) + 3 arctg(x) + c

int

x+ 3

1 + x2=

int

2x middot 12 + 3

1 + x2=

1

2ln(1 + x2) + 3 arctg(x)

1

12 Gyoumlkoumls fuumlggveacutenyek integraacutelaacutesaint radic

ax2 + bx+ c

Maacutes fokuacutera nincs aacuteltalaacutenos keacuteplet

1 a lt 0 plint

radic

1minus x2 = 5

intradic

1minus sin2(t) cos dt6 =

int

cos2(t)dt

Keacutet lehetseacuteges megoldaacutesi uacutet

bull cos(2t) = cos2(t)minus sin2(t)

bull cos2(t) = cos(t) middot sinprime(t)int

cos2(t) =

int

cos(t)middotsinprime(t) = cos(t)middotsin(t)minusint

minus sin(t)middotsin(t)

= cos(t) middot sin(t) +int

sin2(t) = cos(t) middot sin(t) +int

1minus cos2(t) =

cos(t) middot sin(t) + tminusint

cos2(t)dt

Oumlsszevonva

2

int

cos2(t) = cos(t) middot sin(t) + t

Ezt a legelejeacutere visszahelyettesiacutetveint

radic

1minus x2dx =

int

cos2(t)dt =cos(t) middot sin(t) + t

2=

1

2

radic

1minus x2 middot x+1

2arcsin(x)

Summaacutezvaint

radic

1minus x2 =1

2xradic

1minus x2 +1

2arcsin(x)

pl

int

radic

4minus x2dx = 2

int

radic

1minus (x

2)2 = 7

2

int

radic

1minus t22dt = x

radic

1minus (x

2)2 + 2 arcsin

radicx2

int

radic

3minus 2xminus x2 =

int

radic

4minus (x+ 1)2 =

2

int

radic

1minus (x+ 1

2)2t =

x+ 1

22dt = dx

int

radic

3minus 2xminus x2 =

int

radic

4minus (x+ 1)2 =

2

int

radic

1minus (x+ 1

2)2t =

x+ 1

28

5(x = sin(t)maacuter volt)6(dx

dt= cos(t))

7x = 2t

82dt = dx

4

int

radic

1minus t2dt =

2 a gt 0 exist valoacutes gyoumlk plint

radic

x2 minus 1 ch2 minus1 = sh2 =

int

sh(t) middot sh(t)dt =int

sh(t) middot chprime(t)

3 a gt 0∄ valoacutes gyoumlk plint

radic

x2 + 1 =

int

ch(t) middot ch(t)dt

2

Kalkulus II

Balog Daacuteniel

20100216

Inpropius integraacutel

Motivaacuteloacute peacuteldaacutek

11int

0

1xdx

A rendes Riemann-integraacutel mieacutert nem hasznaacutelhatoacuteAzeacutert mert a Riemann-integraacutel korlaacutetos fuumlggveacutenyekre vaneacutertelmezve az1

xpedig nem korlaacutetos a(0 1) intervallumban

A kiszaacutemiacutetaacutesi moacutedszer(Newton-Liebnitz) sem mukoumldik

1int

0

1

xdx = [lnx]

10 = ln 1 minus ln 0

Itt baj van ugyanisln 0 = minusinfin iacutegy az eredmeacuteny veacutegtelenJelenleg ez megoldhatatlan

2infinint

0

eminusxdx

Az a baj hogy az integraacutelt eddig csak korlaacutetos intervallumoneacutertelmeztuumlk

infinint

0

eminusxdx =[eminusx

]1

0= minuseminusinfin minus

(minuseminus0

)= 0 minus (minus1) = 1

Ez se stimmel

Ceacutel

Definiaacuteljunk integraacutelt nem korlaacutetos

intervallumra (1)fuumlggveacutenyre (2)

I

Korlaacutetos fuumlggveacuteny integraacutelja nem korlaacutetos intervallumonDef Legyenf [a +infin) rarr R olyan hogy Riemann-integraacutelhatoacute[a b]-nforallb gt a-raAz f fuumlggveacutenynekexist az [ainfin) intervallumon inpropius integraacutelja ha u

exist limbrarrinfin

bint

a

f(x)dx

Ennek jele+infinint

a

f(x)dx

Hasonloacuteanbint

minusinfin

f(x)dx = limararrminusinfin

bint

a

f(x)dx

Peacuteldaacutek

bullinfinint

0

eminusxdx = limbrarrinfin

bint

0


[minuseminusx]b

0 =

= limbrarrinfin

(minuseminusbminus (minuseminus0)


bullinfinint

1

1

xdx = lim

brarrinfin

bint

1

1

xdx = lim

brarrinfin[lnx]

b

1 =

= limbrarrinfin

[ln b minus ln 1] = infin

bullinfinint

1

1

x2dx = lim

brarrinfin

bint

1

1

x2dx = lim

brarrinfin

[

minus 1

x

]b

1

=

= limbrarrinfin

[

minus1

bminus(

minus1

1

)]

= 1

0

1

2

0 1 2 3 4

Eacuterdekes hogy a pirossal jeloumllt teruumllet (x2) veacuteges a keacutekkel jeloumllt (x)veacutegtelen

1

infinint

1

1xα

dx Milyen α eacuterteacutekre veacuteges

α 6= 1

infinint

1

1

xαdx = lim

brarrinfin

bint

1

1

xαdx = lim

brarrinfin

[

minus x1minusα

1 minus α

]b

1

=

= limbrarrinfin

(b1minusα

1 minus αminus 11minusα

1 minus α

)

Haα gt 1 0 minus 1αminus1

Haα lt 1 infin

Def Azinfinint

a

f(x)dx inpropinus integraacutelt konvergensnek nevezzuumlk ha

veacuteges Ha divergens akkor vagy veacutegtelen vagy nincs limespl

infinint

1

1x2 dx konvergens

infinint

0

eminusxdx konvergensinfinint

2

1x2 dx konvergens

infinint

1

1xdx divergens

infinint

0

exdx divergensinfinint

03

1xdx divergens

infinint

1

1radicxdx divergens

infinint

0

sinxdx divergensinfinint

2

1xdx divergens

Tanulsaacuteg Az alsoacute hataacuter a konvergens-divergens keacutereacutesben nemszaacutemiacutet

neheacutez keacuterdeacutesek

bullinfinint

0

eminusx2

dx

0

1

2

0 1 2 3 4

Laacutethatoacute hogyeminusx alatt van hax gt 1Tudjuk hogy

infinint

0

eminusxdx konvergens

dArrinfinint

1


dArrinfinint

1

eminusx2

dx konvergens

dArrinfinint

0

eminusx2

dx konvergens

bullinfinint

0

eminusx2

dx =

= lim0rarrinfin

infinint

0

eminusx2

dx = lim0rarrinfin

[]b0

Keacuteplettel nem adhatoacute meg a primitiacutev fuumlggveacuteny

Kideruumllinfinint

0

eminusx2

dx =radic

π

2

infinint

minusinfin

infinint

minusinfin

eminus(x2+y2)dx dy = π2

1

bullinfinint

0

sin xx

dx

bullinfinint

0

eminusx sinxdx

bullinfinint

0

sin(x2)dx

A gyoumlkoumlkradic

k middot π egyre surubben joumlnnek ezeacutert konvergens lesz

II

Intervallum korlaacutetos de a fuumlggveacuteny nem korlaacutetospl1int

0

1xdx

DefLegyenf [a b) rarr R olyan hogy azf [a c] intervallumon Riemann-integraacutelhatoacuteforallc isin (a b)-re

f inpropius integraacutelhatoacute[a b)-n halimcrarrb

cint

a

f(x)dx exist ekkor

bint

a

f(x)dx = limcrarrb

cint

a

f(x)dx

Ha aza-naacutel van baj akkor

bint

a

f(x)dx = limcrarra

bint

c

f(x)dx

bint

c

f(x)dx inpropius integraacutel konvergens haexist lim eacutes veacuteges

pl

1int

0

1

xdx = lim

crarr0+

1int

c

1

xdx = lim

crarr0+[lnx]

1c =

limcrarr0+

( ln 1︸︷︷︸

0

minus ln c) = +infin

1int

0

1radicx

dx = limcrarr0+

1int

c

xminus 12 dx = lim

crarr0+[]10 = lim

crarr0+

(

1 minusradic

012

)

= 2

1polaacuterkoordinaacuteta

2

Kalkulus II

Balog Daacuteniel

20100223

1 Fuumlggveacutenysorozatok eacutes fuumlggveacutenysorok

11 Fuumlggveacutenysorozat fogalma

pl fn(x) = xn n = 1 2 3

1

Ez egy fuumlggveacutenysorozat

x n = 1 n = 2 n = 3 n = 4 konvergencia

fn(

12

)

12

122

123

124 konvergens

fn(13 )

13

132

133

134 konvergens

fn(1) 1 1 1 1 divergens


fn(minus2) minus2 minus22 minus23 minus24 nem tart sehova

Hax-et fixaacuteljuk akkor egy szaacutemsorozatot kapunkHol lesz konvergens hol lesz divergensHax isin (minus1 1 ] akkorfn(x) konvergensexist lim eacutes veacuteges

Def Legyenekfn R rarr R fuumlggveacutenyekn = 1 2 Ezt fuumlggveacuteny-sorozatnak nevezzuumlk

Def LegyenfnR rarr R egy fuumlggveacutenysorozat Ennek konvergenciahal-mazaKH(fh) = x isin R fn(x)szaacutemsorozat konvergenspl

fn(x) = xn rArr KH(fn) = (minus1 1]

Def Az fn fuumlggveacutenysorozat az f fuumlggveacutenyhez tart pontonkeacutent halimnrarrinfin

= f(x)forallx isin KH(fn)

Keacuterdeacutes oumlroumlkli-e azf (limes-hataacuterfuumlggveacuteny) azfn tulajdonsaacutegaitFolytonossaacuteg derivaacutelt integraacutel

ffn

Vaacutelasz Aacuteltalaacuteban nem igaz de neacuteha igen MikorFolytonossaacuteg bukott integraacutel

1

1n

1int

0

fn(x)dx = 1minus1

2n

1int

0

f(x)dx = 1

Tehaacutet

limnrarrinfin

1int

0

fn(x)dx = 1minus1

2n=

1int

0

f(x)dx = 1

1

1

2n

1n

1int

0

fn(x)dx =2

nmiddot 1 middot

1

2= 1foralln

1int

0

f(x)dx = 0

Def azfn fuumlggveacutenysorozat egyenletesen tart azf fuumlggveacutenyhez egyHhalmazon haforallε gt 0existN isin N hogy|fn(x)minusf(x)| lt εforalln ge Nforallx isin H

x fix forallεexistN isin R hogy|fn(x) minus f(x)| lt εforalln ge N

H

fn

fn minus ε

fn + ε

AZ egyenletes konvergenciaacutenaacutelN nem fuumlggx-tol miacuteg a pontnaacutel magaacutetx-et vaacutelasztom eacutes hozzaacute egy kuumlszoumlbindexetplfn(x) = xn

fn(x) = xn fuumlggveacutenysorozat egyenletesen tart azf(x) = 0 fuumlggveacuteny-hez aH [0 12 ] halmazon

H

H=[01] maacuter nemH = [ 0 1 ) nem egyenletesen tartf -hez

deH = [ 0 q ) igen haq lt 1

Mindenfn kiloacuteg azε csobol

Teacutetel Legyenekfn-ek folytonos fuumlggveacutenyek eacutes aH-n egyenletesenkonvergaacutelnak azf hez akkor azf is folytonos aH-n

Teacutetel Legyenekfn-ek Riemann integraacutelhatoacutek egy[a b]-n eacutes egyen-letesen konvergaacutelnak az[a b] intervallumon azf -hez akkor azf isRiemann integraacutelhatoacute eacutes

limnrarrinfin

bint

a

f(x)dx =

bint

a

f(x)dx

Maacuteskeacutepp

limnrarrinfin

bint

a

f(x)dx =

bint

a

lim fn(x)dx

lim eacutesint

felcsereacutelehto ha egyenletes a konvergencia

2 Fuumlggveacutenysorok

Mi asor sorozatdArr dArr

sum

an anUgyanez a fuumlggveacutenysornaacutel is

Def Legyenfn egy fuumlggveacutenysorozat Az ezekbol keacutepezett fuumlggveacuteny-sorsum

fn azSn = f1 + f2 + fn

fuumlggveacutenysorozat

Def asum

fn fuumlggveacutenysor konvergenciahalmaza

KH(sum

fn) = x isin R

sum

fn(x)konvergens szaacutemsorsum

fn sor oumlsszegfuumlggveacutenye

f(x) =

infinsum

n=1

fn(x)

pl fn(x) = xn

s1(x) = x1 s2(x) = x1 + x2

sum

fn(x) milyenx-re konvergens

KH(sum

fn) = (minus1 1)sum

fn fuumlggveacutenysor egyenletesen konvergens aH halmazon haSn =f1 + f2 + fn fuumlggveacutenysorozat egyenletesen konvergens aH halmazonFontos speciaacutelis fuumlggveacutenysorhatvaacutenysorLegyenCn egy szaacutemsorozat eacutes legyena isin R egy szaacutem

sum

cn middot (x minus a)n

2

fuumlggveacutenysorta koumlzepu hatvaacutenysornak nevezik

sum

xn Cn = 1 a = 0

plsum

xn

nKH =

x =1

2sum 1

nmiddot1

2n

ez konvergens mertlt 12n Tehaacutet(minus1 1) benne van aKH-ban-

x = 2sum 2n

n

ez erosen divergensx = 1sum 1

ndivergens

x = minus1sum (minus1)n

nez konvergens

3

Kalkulus II

Balog Daacuteniel

20100302

1 Hatvaacutenysorok eacutes Taylor sorok

Def Hatvaacutenysor ilyen fuumlggveacutenysorsum

cn middot (xminus a)n

cn adott szaacutemsorozat(egyuumltthatoacutek)a isin R adott hatvaacutenysor koumlzepepl

sumxn cn = 1 a = 0

sumn middot xn cn = n a = 0

sumxn

n cn = 1n a = 0

AlapkeacuterdeacutesKonvergenciahalmaz (KH) milyenx-re konvergens

pl

sumxn

x = 12 rArr konvergens

x = 2 rarr divergens

Aacuteltalaacuteban milyenx-re konvergensHasznaacuteljuk a gyoumlkkriteacuteriumot

sumansor

konvergens ha lim n

radican lt 1

divergens ha lim n

radican gt 1

sum

xnpeacuteldaacutejaacutenan = xn

n

radic

|an| = |x|Ha negatiacutevok is vannak akkor megneacutezzuumlk az abszolulteacuterteacutekessort

lim n

radic

|an| = |x|

konvergens ha |x| lt 1divergens ha |x| ge 1

pl

sumn middot xn milyenx-re konvergens

an = n middot xn

n

radic

|an| = n

radicn|x|

limnrarrinfin

n

radic

|an| = 1 middot n|x|

Neheacutez hogylimnrarrinfin

n

radicn = 1

Gyoumlkkriteacuterium azt mondja meg hogy

ha

|x| lt 1 akkor

sumn middot xnkonvergens

|x| ge 1 akkorsum

n middot xndivergens

Aacuteltalaacutenos esetsum

cn middot (xminus a)n milyenx-re konvergens

an =sum


n

radic

|an| = n

radiccn middot |xminus a|

Kell ennek a sorozatnak a limese

limnrarrinfin

n

radic

|an| = limnrarrinfin

n

radiccn middot |xminus a| lt 1

hogy konvergens legyen

|xminus a| lt 1

lim n

radiccn

Def

sumcn middot (xminus a)n hatvaacutenysor konvergenciasugara

R =1

lim n

radiccn

(ha alim = 0rArr R = infin)

|xminus a| lt R lArrrArr x isin (aminusR a+R)︸︷︷︸

enneka a koumlzepe

Hogyha azx a-toacutel legfeljebbR taacutevolsaacutegra van akkor konvergens

Ezt fogalmazza meg a teacutetel

(Cauchy-Hadamard) Teacutetel

Asum

cn(xminus a)n hatvaacutenysor konvergenciahalmazaacutera fennaacutel

(aminusR a+R) sub KH sub [aminusR a+R]

ahol

R =1

lim n

radiccn

Azeacutert sugaacuter mert ez komplex szaacutemmokkal is szeacutepen mukoumldikBizonyiacutetaacutes fennt

Megj Az intervallum veacutegpontjai benne lehetnekKH-ban de ezcn-tolfuumlgg (a teacutetel arroacutel nem szoacutel ha x pont a hataacuteron van)

1

pl

sum

xn cn = 1 a = 0

R =1

lim n

radic1= 1

(minus1 1) sub KH sub [minus1 1]

veacutegpontok(minus1)n (nem tart sehova)1n divergens (infin-hez tart)

KH = (minus1 1)

sum xn

n2cn =

1

n2a = 0

R =1

lim n

radic1n2

= 1

segiacutetseacuteg

limn

radic

1

n2= lim

n

radic1

n

radicn2

= limn

radic1

( n

radicn)2

=1

12= 1


veacutegpontok

x = 1 rArrsum 1

n2konvergens

x = minus1 rArrsum (minus1)n

n2konvergens

KH = [minus1 1]

sum xn

ncn =

1

na = 0

R =1

lim n

radic1n

= 1


veacutegpontok

x = 1 rArrsum 1

ndivergens

x = 1 rArrsum (minus1)n

nkonvergens

KH = [ minus 1 1 )

Def

Legyensum

cn(xminus a)n egy hatvaacutenysor A hatvaacutenysor fuumlggveacutenye

f(x) =

infinsum

0

cn(xminus a)nx isin KH

pl

sumxnf(x) =

x isin (minus1 1)

Nsum

0

xn = 1 + x+ x2 + xN = 11 middot xN+1 minus 1

xminus 1

infinsum

n=0

xn = limNrarrinfin

(

Nsum

0

xn) = limNrarrinfin

1 middot xN+1 minus 1

xminus 12 =

1

1minus x

infinsum

n=0

xn =1

1minus x

Oumltletsum

nxn oumlsszegeacutere

infinsum

n=0

xn =1

1minus x

Derivaacuteljuk Keacutesobb megneacutezzuumlk hogy szabad-e

infinsum

n=0

nxnminus1 =minusminus 1

1minus x

2

middot x

infinsum

n=0

nxn =1

1minus x

n

pl

infinsum

n=0

n

2n=

1

2+

2

4+

3

8+

4

16+

Ez hagyomaacutenyos moacutedszerrel eseacutelytelen

12

(1minus 12 )

2=

1

2times 1

4= 2

Igaz-e hogy(sum

fn)prime =sum

fnprimeEz nem mindig igaz Kulcsszoacute egyenletes konvergenciaTeacutetel

Legyensum

cn(x minus a)n egy hatvaacutenysorLegyen az oumlsszegfuumlggveacuteny

f(x) =

infinsum

n=0

cn(xminus a)n x isin KH

Ekkor f akaacuterhaacutenyszor derivaacutelhatoacute eacutes tagonkeacutent lehet a sort derivaacutelniazaz

f prime(x) =infinsum

n=1

cnn(xminus a)nminus1(middot1)

f primeprime(x) =

infinsum

n=2

cnn(nminus 1)(xminus a)nminus23

1meacutertani sorozat2xN+1

0-hoz tart kicsi a sokadikon31-tol illetve 2-tol hiszen az 1 ill 2 tag 0 lesz

2

2 Az oumlsszegfuumlggveacuteny eacutes az egyuumltthatoacutekkapcsolata

Legyen

f(x) =

infinsum

n=0

cn(xminus a)n = c0 + c1(xminus a) + c2(xminus a)2 +

f(a) = c0

f prime(x) = c1 + 2c2(xminus a) + 3c3(xminus a)2 +

f prime(a) = c1

f primeprime(x) = 2c2 + 6c3(xminus a) + 4 middot 3(xminus a)2 +

f primeprime(a) = 2c2

f primeprimeprime(a) = 6c3

Raacutejoumlvuumlnk hogy1c0 = 0c06c3 = 3c3

Teacutetel

Legyen

f(x) =infinsum

n=0

cn(xminus a)n rArr f (n)(a) = cn middot n

21 Taylor sor

Eddig adott volt

cn eacutes(a)rArrsum

cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n

Eacuterdekes azf (n)(a) visszaadja acn-tMost legyen adott

fR rarr R

a isin R

akaacuterhaacutenyszor derivaacutelhatoacuteLegyen most

cn =f (n)a

nforalln

Neacutezzuumlkf(x) =

sum

cn(xminus a)n

hatvaacutenysortDef

Az f fuumlggveacutenya koumlzepu Taylor sora

sum f (n)a

n(xminus a)n

KeacuterdeacutesKH = valamint oumlsszegfuumlggveacuteny(de ezt nem illikf -el jeloumllni)

Joacute lenne ha az oumlsszegfuumlggveacuteny egyenlo lennef -elpl

f(x) =1

1minus x= (1minus x)minus1 legyena = 0

f (n) f (n)(a)f(x) = (1minus x)minus1 c0 = 1

f prime(x) = minus1(1minus x)minus2 minus 1 c1 = 1f primeprime(x) minus 2(1minus x)minus3 minus 1 c2 = 2f primeprimeprime(x)minus 6(1minus x)minus4 minus 1 c3 = 6

f (n)(a) = n

cn = 1

A kapott hatvaacutenysorsum

xn

IsmertKH = (minus1 1) oumlsszegfuumlggveacuteny 11minusx

Visszakaptuk azf -et de csak(minus1 1)-re Az f eacutertelmezve voltR1-reaz oumlsszegfuumlggveacuteny(minus1 1)-en eacutes ott =f

22 Fontos Taylor sorok

f (n) f (n)(a)f(x) = ex c0 = 1f prime(x) = ex c1 = 1f primeprime(x) = ex c2 = 1

2f primeprimeprime(x) = ex c3 = 1

6

Aacuteltalaacutenosan

cn =1

n

ex Taylor sorasum

xn

nKH oumlsszegfuumlggveacuteny

R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin

Magyarul akaacutermit iacuterok be konvergens lesz

Haacutenyadoskriteacuteriummal kijoumln aKHKH = R (0 koumlzepu jobbra-balra veacutegtelen intervallum)

sum xn

nforallx-re konvergens

Mi az oumlsszegfuumlggveacutenyTaylor polinom Lagrange maradeacutektaggalTeacutetelforallx isin R eseteacuten

infinsum

0

sum xn

n= ex

3

biz Lagrange maradeacutektagTeacutetel

f (n) f (n)(a)f(x) = sin(x) c0 = 0f prime(x) = cos(x) c1 = 1

f primeprime(x) = minus sin(x) c2 = 0f primeprimeprime(x) = minus cos(x) c3 = minus1

forallx isin R eseteacuteninfinsum

0

(minus1)n middot x2n+1

(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1

f prime(x) = minus sin(x) c1 = 0f primeprime(x) = minus cos(x) c2 = minus1f primeprimeprime(x) = sin(x) c3 = 0


0

(minus1)n middot x2n

(2n)

Taylor sor=infin fokuacute Taylor polinom

4

Kalkulus II

Balog Daacuteniel

20100309

Trigonometrikus eacutes Fourier sorok

A hatvaacutenysor eacutes Taylor sor ismeacutetleacutese

1 A hatvaacutenysor fogalmasum

Cn(xminus a)n

2 Konvergenciahalmaz(aminusR a+R) intervallum CH keacuteplet

3 Oumlsszegfuumlggveacuteny eacutes egyuumltthatoacutek kapcsolata

f(x) =

infinsum

n=0

Cn(xminus a)n Cn =fn(a)

n

4 f Taylor sorasum fn(a)

n(xminus a)n

5 peacuteldaacutekex sin(x) cos(x) forallx-re egyenlo

Trigonometrikus sor

1 Trigonometrikus sor definiacutecioacuteja

a0

2+sum

an middot cos(nx) + bn middot sin(nx)

2 A konvergenciahalmaza neheacutez uacutegyhogy itt nem taacutergyaljuk

3 Legyen

a0

2+sum

an middot cos(nx) + bn middot sin(nx) = f(x)

valamilyenx-ekre ahol ez konvergens

bull Hogyan szedjuumlk kiam-t eacutesbn-t

4 Elokeacuteszuumlletπint

minusπ

cos(nx) middot sin(kx)

eacutes hasonloacutek kiszaacutemiacutetaacutesa

bullπint

minusπ

sin(kx) = 0forall k-ra

πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull

cos(α+ β) = cos(α) cos(β) minus sin(α) sin(β)

+ cos(αminus β) = cos(α) cos(β) + sin(α) sin(β)

cos(α) cos(β) = 1

2(cos(αminus β) + cos(α+ β))


minus cos(αminus β) = cos(α) cos(β) + sin(α) sin(β)

minus sin(α) sin(β) = 1

2(cos(αminus β)minus cos(α+ β))

sin(α+ β) = sin(α) cos(β) + cos(α) sin(β)

+ sin(αminus β) = sin(α) cos(β) minus cos(α) sin(β)

sin(α) cos(β) = 1

2(sin(αminus β) + sin(α+ β))

bullπint

minusπ

cos( nx︸︷︷︸

α

) middot sin( kx︸︷︷︸

β

) =

1

2

πint

minusπ

(sin [(k + n)x] + cos [(k minus n)x]) dx =

1

2middot

[minus cos(k + n)x

n+ k+

minus cos(k minus n)x

k minus n

]π

minusπ

= 0

Mert acos fuumlggveacuteny periodikus2π-n keacutent

πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ

(cos [(k + n)x] + cos [(k minus n)x]) dx =

=

2π han = k = 0π han = k 6= 00 minden maacutes esetben

πint

minusπ

sin(nx) sin(kx) =

π han = k 6= 00 midnen maacutes esetben

1

bull

f(x) =a0

2+

infinsum

n=1

an cos(nx) + bn sin(nx)

middot sin(kx)

πint

minusπ

πint

minusπ

f(x) middot sin(kx) =a0

2

πint

minusπ

sin(kx)dx︸︷︷︸

=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=

A keacuterdeacutes az egyenletes konvergencia

= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ

f(x) cos(kx)dx = πak

ak =1

πmiddot

πint

minusπ

f(x) middot cos(kx)dx

k isin N

TEacuteTEL

Most nem iacuterjuk le de meg lehet fogalmazniMost azf adott

Def Legyenf 2π szerint periodikus fuumlggveacuteny (vagy adott(minusππ)-

n eacutes legyenak = 1

πmiddot

πint

minusπ

f(x) middot cos(kx)dx eacutes = bk middot π =

1

π

πint

minusπ

f(x) sin(kx)dx

Az f fuumlggveacuteny(minusππ) intervallumhoz tartozoacute Fourier-sora

a0

2+

infinsum

n=1


Def Az f fuumlggveacuteny Fouriet-sora sorba fejtheto (minusππ)-n ha a Fo-uriet sora ott konvergens eacutes eloaacutelliacutetj f -et azaz

f(x) =a0

2+

infinsum

n=1


forallx isin (minusππ)

MegjItt sok matematikai teacutetel van

5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1

kπ[cos(k middot 0)minus cos(k middot π)]minus

1

kπ[cos(k middot π)minus cos(k middot 0)] =

1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4

kπ ha k paacuteratlan

ak = 0 forallkMert f paacuteratlan fuumlggveacuteny

2

Kalkulus II

Balog Daacuteniel

20100316

Toumlbbvaacuteltozoacutes fuumlggveacutenyekRn

DefR

n = RtimesRtimes middot middot middot timesR︸︷︷︸

n db

Rn teacuter elemei szaacutem n-esek(n dimenzioacutes vektorok)pl x = (x1 x2 xn) xi isin RforalliR

n-ben muveletekx y isin Rn

x+ y = (x1 + y1 x2 + y2 xn + yn)

x isin Rn c isin Rc middot cx = (cx1 cx2 cxn)

x isin Rn hosszaMegjegyzeacutes a hosszra sokfeacutele definiacutecioacute van

|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1

Ezek metrikaacutet definiaacutelnak

P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r

Toumlbbvaacuteltozoacutes fuumlggveacutenyLegyenf Rn rarr R

1 toumlbbvaacuteltozoacutes fuumlggveacutenypl

f(x1 x2)=x1 + x2 n = 2 k = 1f(x1 x2)=(x1 middot x2 e

x1 middot cos(x2)) n = 2 k = 2f(x1 x2 x3)=(x1 + x2

2x1 + x3) n = 3 k = 2

Haf Rn rarr Rk rArr f1 fn fk koordinaacutetafuumlggveacutenyei

fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))

plf1(x1 x2 x3) = x1x

22 f2(x1 x2 x3) = x1 + x3

Toumlbbvaacuteltozoacutes fuumlggveacuteny aacutebraacutezolaacutesa

bull Rrarr R eset(x f(x)) isin R2

pontokat oumlsszegyujtuumlk

x

f(x)

bull R2 rarr R eset(x f(x)) isin R3

a pontok egy feluumlletet alkotnakviacutezszintesR2

fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)

bull Egyszerusiacutetett aacutebraacutezolaacutesSzintvonalak

1

1radic2

2

Az elozo peacuteldaacutenaacutel is hasonloacute de maacutesok a szaacutemok

bull R3 rarr R eset(x f(x)) isin R4 Neheacutez aacutebraacutezolni szintvonalas csakR3-hez kellpl f(x y z) = x2 + y2 + z2 szintfeluumlleteif(x y z) = x2 + y2 + z2 = c

b

c

bull R4 rarr R itt semmi sem segiacutet

bull Rrarr R2

(x f(x)) isin R3 pontokat gyujtuumlnkpl f(x) = (cos(x) sin(x)) x isin R

f1

f2

f1(x)

f2(x)

x

Jobbra a rendes balra az egyszerusiacutetett aacutebraacutezolaacutes

bull Rrarr R3

(x f(x)) ezt maacuter nem rajzoljuk de az egyszerusiacutetett rajzR3-banvan eacutes ez egy goumlrbe a teacuterbenf(x) = (x cos(x) sin(x)) ennek az egyszerusiacutetett rajza az elobbicsavarvonal

bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C

eset ez nagyjaacuteboacutel remeacutenytelen

bull R3 rarr R3

pl gravitaacutecioacutes eroteacuterMatematikus ezt vektormezonek hiacutevja

(NagysaacutegaγMmr2

iraacutenya M feleacute)

f(x) = minusx middot M middotm|x|3 γ

f1(x1 x2 x3) =minusMmγ

(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R

Hataacutereacuterteacutek eacutes folytonossaacuteg toumlbbvaacuteltozoacutes fuumlggveacutenyekre

Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent

Az f fuumlggveacuteny hataacutereacuterteacuteke aza isin Rn pontbanA isin Rk haforallε gt 0 existδ gt 0 hogy|xminus a| lt δ eacutesx 6= a rArr |f(x)minusA| lt ε

A

δ

x

f(x)

Rk

Rn

Def azf fuumlggveacuteny folytonos aza isin Rn pontban haforallε gt 0 existδ gt 0 hogy|xminus a| lt δ eacutes|f(x)minus f(a)| lt ε

Toumlbbvaacuteltozoacutes fuumlggveacutenyek derivaacutelaacutesa

Parciaacutelis derivaacutelt

Def LegyenRn rarr R Az f fuumlggveacuteny elso vaacuteltozoacute szerintiparciaacutelis derivaacuteltja azx isin Rn pontban

part1f(x) = limhrarr0

f(x1 + h x2 xn))minus f(x1 x2 xn))

h

Egy vaacuteltozoacute eseteacuteben volt

f prime(x) = limhrarr0

f(x+ h)minus f(x)

h

Az i-edik vaacuteltozoacute szerinti parciaacutelis derivaacutelt

partif(x) = limhrarr0

f(x1 xi + h xn))minus f(x1 xn))

h

Maacutes jeloumlleacutessel

partif(x) =partf

partxi

Peacuteldaacutek

bull f(x1 x2) = x1 middot x2

part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2

bull f(x1 x2) = x1 middot ex2

part1f(x1 x2) = ex2

part2f(x1 x2) = x1 middot ex2


2 + 2x1 + x2

part1f(x1 x2) = 2x1 middot x32 + 2

part2f(x1 x2) = x21 middot 3x2

2 + 1

bull f(x1 x2 x3) =1radic

x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2

part1f(x1 x2 x3) = minus1

2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2

part2f(x1 x2 x3) = minusx2 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2

(part1f part2f part3f) = minus x

|x|3

Ez a gradiens (grad)

Furcsa peacutelda

f(x1 x2) =

0 hax1 middot x2 = 01 minden maacutes esetben

Folytonos-e a fuumlggveacuteny(0 0)-banEz nem folytonos

part1f(0 0) = 0 part2f(0 0) = 0part1f(1 0) = 0 part2f(0 1)part1f(0 1) = ∄ part2f(1 0) = ∄ lim

hrarr0

1minus0h

Ez a limes divergens tehaacutet nincs limese

3

Kalkulus II

Balog Daacuteniel

20100323

Toumlbbvaacuteltozoacutes fuumlggveacutenyek derivaacutelaacutesaVoltparciaacutelis derivaacutelt

f Rn rarr R


f(x1 x2 xi + h xn)minus f(x1 xn)

h

Derivaacutelt vagy Jacobi maacutetrix

Egy vaacuteltozoacutes fuumlggveacutenyf Rrarr R


f(x+ h)minus f(x)

h

Mostf Rh rarr R

az a baj a keacuteplettel hogyh isin f Rn-el osztani kelleneAacutetalakiacutetva

0 = limhrarr0

f(x+ h)minus f(x)minus f prime(x) middot h

h

Ha ez a hataacutereacuterteacutek0 akkor az abszolulteacuterteacuteke is0

0 = limhrarr0

|f(x+ h)minus f(x)minus f prime(x) middot h|

|h|

Az abszolulteacuterteacutekes truumlkk csak akkor igaz ha 0 a hataacutereacuterteacutek hiszenplusmn0 = 0 (deplusmn2 6= 2)f prime(x) middot h isin R kell kuumlloumlnben nem joumln ki (szaacutem - vektor)

f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)

DefLegyenf Rn rarr R Ezx isin Rn helyen differenciaacutelhatoacute haexistA isin R1timesn

sormaacutetrix hogy

limhrarr0

|f(x+ h)minus f(x)minusA middot h|

|h|= 0

Ekkorf prime(x) = A derivaacuteltJacobi maacutetrixHonnan tudjuk az A maacutetrixotProacutebaacuteljuk meg neacutehaacuteny speciaacutelish vektornaacutelh = (1 0 0)t t isin R(t 0 0) t rarr 0

x+ h = (x1t+ x2 + x3 + middot middot middot+ xn)|h| = |t|

A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0

|(x+ h)minus f(x)minusA1t|

|t|=

limtrarr0

∣∣∣∣∣∣∣∣

f(x+ h)minus f prime(x)

t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣

TehaacutetA1 = part1f(x)HasonloacuteanA2 = part2f(x)

AacutelliacutetaacutesHaf Rh rarr R differenciaacutelhatoacutex-benrArr

f prime(x) = (part1f(x) partnf(x)) = grad f(x)

Ez a kiszaacutemiacutetaacutes moacutedjaBizonyiacutetaacutes a fentipl

f(x1 x2) = x1 middot ex2f R2 rarr R

f prime(x1 x2) = (ex2 x1ex2)

Lineaacuteris koumlzeliacuteteacutes Emleacutekezteto n = 1

f Rrarr R

x0

f(x0)l(x)minus f(x0)

l(x)minus l(x0)

xminus x0= f prime(x0)

Maacuteskeacuteppl(x) = f(x0) + f prime(x0) middot (xminus x0))

1

Ugyan ezt kellene toumlbbvaacuteltozoacutesran = 2

f Rn rarr R

1f(x0 = l(x0)

1

Szeretneacutenk egyl(x y) lineaacuteris koumlzeliacuteteacutest

b

(x)(y)

l(x y) = f(x0 y0) + f prime(x0 y0) middot (xminus x0 y minus y0)

l(x y)f(x0 y0) + (part1(x0 y0) part1(x0 y0)) middot

(xminus x0

y minus y0

)

l(x y) = f(x0 y0) + part1f(x0 y0) middot (xminus x0) + part2f(x0 y0) middot (y minus y0)

Ezt iacuterjuk aacutet siacutekegyenletbe

z = f(x0 y0) + part1f(x0 y0) middot (xminus x0) + part2f(x0 y0) middot (y minus y0)

(tehaacutetl(x y) helyettz)

Legyenz0 = f(x0 y0)

0 = (part1f(x0 y0) part2f(x0 y0)minus1︸︷︷︸

~n

) middot (xminus x0 y minus y0 minus z minus z0︸︷︷︸

rminusr0

plf(x y)x2 + y2

(x)

(y)

(x0 y0) = (1 2)-houmlz eacuterinto siacutekot akarunky0 = f(x0 y0) = 12 + 22 = 5A siacutek normaacutelvektora

~n = part1f(1 2) part2f(1 2)minus1)

part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)

[(2 4minus1) (xminus 1 y minus 2 z minus 5)] = 0

2(xminus 1) + 4(y minus 2)minus (z minus 5) = 0

2x+ 4y minus z = 5

Derivaacutelt maacutetrixf Rn rarr Rk

(eddig k=1 volt)Legyenf Rn rarr R

k Ez x isin Rh helyen differenciaacutelhatoacute haexistA isin

Rktimesn maacutetrix hogy

limhrarr0


|h|= 0

Ekkorf prime(x) = A Itt most vektorok vannak skalaacuterok helyettItt hogyan joumln ki azA maacutetrixAacutelliacutetaacutesHaf Rn rarr R

k differenciaacutelhatoacutex-ben akkor

f prime(x) =

part1f1(x) part2f1(x) partnf1(x)part1f2(x) part2f2(x) partnf2(x)

part1fk(x) part2fk(x) partnfk(x)

Ez az igazi Jacobi maacutetrix(ez elozo is az volt de az nem volt annyira szeacutep maacutetrix) (itt most nincsbizonyiacutetaacutes)

pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2

f1(x1 x2) = x1 middot x2

f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1

Maacutes jeloumlleacutesDf(x) = A

Lineaacuteris koumlzeliacuteteacutes

f Rn rarr Rkfuumlggveacutenyre

f(x+ h) = f(x) + f prime(x) middot h+ α(h)2

lim|α(h)|

|h|= 0

Olyan kicsi azα(h) hogy meacuteg egy kicsih-val leosztva is0-aacutet adMaacutes jeloumlleacutessel

x = x0

h = xminus x0

f(x) = f(x0) + f prime(x0) middot (xminus x0) +α(xminus x0)

α lehuacutezaacutesa a lineaacuteris koumlzeliacuteteacutes

l(x) = f(x0) + f prime(x0) middot (xminus x0)

2ez kicsi

2

(eddig is ez volt csak k=1-el)

l Rn rarr Rk

Alkalmazaacutes spec esetre

n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt

ennek a lineaacuteris koumlzeliacuteteacutese egy egyenes a teacuterben

l(t)f(t0) + f prime(t0) middot (tminus t0)

Konkreacutetan

f prime(t)

minus sin tcos t1

t0 = π-ben eacuterinto

x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)

x = minus1y = π minus t

z =π + tminusπ

t isin R

Maacutesodik derivaacuteltEgy vaacuteltozoacute eseteacutenf primeprime nem maacutes mint(f prime)prime

Han-vaacuteltozoacute van

f Rrarr RrArr f prime(x) isin R1timesn

Ez maacutes tipusuacute mint azfJoacute hiacuterf prime = R1timesn tekintheto uacutegyRn

Ekkor

f primeprime(x) isin Rntimesnf primeprime Rn rarr Rntimesn

pl

f(x y) = x3 + y4

R2 rarr R

f prime(x y) = (3x2 4y3)f prime R2 rarr R2

f primeprime(x y) =

(6x 00 12y2

)

3

Szeacutelsoeacuterteacutek toumlbbvaacuteltozoacuteban

Balog Daacuteniel

20100330

Ugyanezt szeretneacutenkf Rn rarr R eseteacuten

(Rn rarr Rk-ba nem akarjuk ugyanezt

Nem mertRk-ban nincslt jel)

Def

Legyenf Rn rarr R a isin Rn lokaacutelis minimumhaexistδ gt 0 |xminus a| lt δ rArr f(x) ge f(a)Pl

a

(n=2 a rajzon) ez egy szigoruacute lokaacutelis minimum

a

ez egy szigoruacute lokaacutelis maximum

gt szigoruacute lokaacutelis minimumle lokaacutelis maxikmumlt szigoruacute lokaacutelis maximumszeacutelsoeacuterteacutek= min cup max

Teacutetel

(Szuumlkeacuteges felteacutetel) Haa isin Rn lokaacutelis szeacutelso eacuterteacutekf Rn rarr R

differenciaacutelhatoacute fuumlggveacutenynek akkor apartif(a) = 0 foralli = 1 n azazf prime(a) = 0

[Emleacutek part1f(a) part2f(a) partnf(a)](8 35) - Aacutellandoacute rettegeacutesben eacutelek hogy egy vektor nem egyenlo-e5szaacutemmal

Nem bizonyiacutetom de rajzolok

a

Elmeteszem egy siacutekkalx1 vaacuteltx2 = a2x3 = a3

xn = an

a1

egyvaacuteltozoacutes fuumlggveacuteny ennek van lokaacutelis minimumaez a goumlrbe f(x1 a2 a3 an) ennek derivaacuteltja 0 azazpart1f(a1 a2 a3 an) = 0

1

Pl

f(x1 x2) = x2

1+ x2

2

Keacuterdeacutes sz eacute (szeacutelsoeacuterteacutek)part1f(x1 x2) = 2x1 part2f(x1 x2) = 2x2

ezeket0-val egyenloveacute teszem2x1 = 0 2x2 = 0x1 = 0 x2 = 0

Szeacutelsoeacuterteacutek lehet(0 0) pontban

Pl2

f(x1 x2) = 1minus x2

1 minus x2

2

1

Keacuterdeacutes sz eacutepart1f(x1 x2) = minus2x1 part2f(x1 x2) = minus2x2

minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0

Sz eacute lehet(0 0) pontban

(8 47) - Voltak meacuternoumlki tanulmaacutenyaim0 percig voltam meacuternoumlktehaacutet

Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet

Keacuterdeacutes sz eacutepart1f(x1 x2) = 2x1 part2f(x1 x2) = minus2x2

2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0

Szeacute lehet(0 0) pontban persze nincs sz eacute

Eleacutegseacuteges felteacutetelf primeprime segiacutetseacutegeacutevel

Eloszoumlr a peacuteldaacutekkal

1 lokaacutelis minimum

f(x1 x2) = x21 + x2

2 f R2 rarr R

f prime(x1 x2) = (2x1 2x2) f prime R2 rarr R2times2

f primeprime(x1 x2) =

(

2 00 2

)

f primeprime R2 rarr R2times2

2 lokaacutelis maximum


1 minus x2

2

f prime(x1 x2) = (minus2x1minus2x2)

f primeprime = (x1 x2) =

(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2

f prime(x1 x2) = (2x1minus2x2)(

2 00 minus2

)

(9 00) - Az izgalmas reacuteszneacutel joumln a reklaacutem

2

Teacutetel

(Eleacutegseacuteges felteacuteteln = 2 eseteacuten)Legyenf R2 rarr R

2times2 differenciaacutelhatoacute Legyenf prime(a) = 0

1 Ha detf primeprime(a) gt 0 eacutespart11f(a) gt 0 rArr a-ban lokaacutelis minimumpart11f(a) = part1part1f(a)

2 Ha detf primeprime(a) lt 0 eacutespart11f(a) lt 0 rArr a-ban lokaacutelis maximum

3 Ha detf primeprime(a) lt 0 rArr nincs sz eacute (nyereg)Nem bizonyiacutetom

Megjegyzeacutes

Ez megfogalmazhatoacuteRn rarr R fuumlggveacutenyre isf primeprime(a)pozitiacutev definitrarr lok min negatiacutev definitrarr lok max(lt Ax x gtgt 0 ez a pozitiacutev definit) - NEM KELL TUDNI

Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=

f prime(x1 x2) = (3x21 minus 3x2 3x

22 minus 3x1)

3x21 minus 3x2 = 0 eacutes 3x2

2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1

Sz eacute lehet(0 0) vagy(1 1)

Most joumln az eleacutegseacuteges felteacutetel


(

6x1 minus3minus3 6x2

)

Beiacuterjuk az egyik jeloumlltet eacutes megneacutezzuumlk mit kapunk

f primeprime(0 0) =

(

0 minus3minus3 0

)

rarr detf primeprime(0 0) = 0minus(minus3)middot(minus3) = minus9 lt 0

(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)

rarr det f primeprime(1 1) = 6 middot 6 minus (minus3) middot (minus3) =

27 gt 0part11f(1 1) = 6 gt 0 rArr (1 1) lok minHF f(x1 x2) = x4

1+ x2

2+ 4x1x2 sz eacute=

Primitiacutev fuumlggveacuteny (potenciaacutel) toumlbbvaacuteltozoacutesfuumlggveacutenyre

Emleacutek

1-vaacuteltozoacutebanf R rarr R

n adott keresemF Rrarr R melyreF prime = f F primitiacutevfuumlggveacutenye f -nek

Def

Legyenf Rn rarr R F Rn rarr R

primitiacutev fuumlggveacutenye f-nek haF prime = f

F prime = (part1F part2F partnF )F prime = Rn rarr R

n

Megjegyzeacutes

bajf R2 rarr R3 nincsF

PL

f(x y) = (2x+ y x+ 2y) f R2 rarr R2

F =partFpartx

F (x y) = 2x+ ypartFparty

F (x y) = x+ 2yKi az akit x szerint derivaacutelva2x+ y joumln ki

F (x y) = x2 + xy + C

ez aC x szerint konstans tehaacutet akaacutermilyeny fuumlggveacuteny lehet(plsin y utoacutelagC(y))

F (x y) = x2 + xy + C(y)

3

partF

partyF (x y) = x+ Cprime(y) = x+ 2y

x+ Cprime(y) = x+ 2yminus x

Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K

Ez aK valoacutedi konstans

Megoldaacutes

F (x y) = x2 + xy + y2 +K

Amikor nincsen tehaacutet Antipeacutelda )

f(x y) = (minusy x)

part1F (x y) = minusy part2F (x y) = x

F (x y) = minusyx+ C(y)

minusx+ Cprime(y) = part2F (x y) = x

Baj nem esik ki az x Nincs F

Ennek az eroteacuternek nincs potenciaacutelja

Hogyan laacutetszik f -en hogy van primitiacutev fuumlggveacutenye

Levezeteacutes(Bizonyiacutetaacutes)

F prime = f lArrrArr part1F = f1 part2F = f2 partnF = fn(keacutet vektor egyenlo ha koordinaacutetaacuteik egyenloek)oumltlet

part1F = f1

derivaacutelomx2 szerintpart2part1F = part2f1

part2F = f2


(9 55) - Ez egy matematikus volt Eloszoumlr fiatal volt aztaacuten nemtudom

Teacutetel(Young)

szuumlkseacuteges felteacutetel primitiacutev fuumlggveacuteny leacutetezeacuteseacutereLegyen egy fuumlggveacuteny melynek maacutesodik derivaacuteltja leacutetezik eacutesfolytonosEkkor

part2f

partxparty=

part2f

partypartx

Nem szaacutemiacutet a sorrend

Ebbol koumlvetkezikVan egy fuumlggveacutenyem aminek szeretneacutem kiszaacutemiacutetani a primitiacutev fuumlggveacute-nyeacutet Haexist primitiacutev fuumlggveacutenyrArr partipartj = partjparti forallij = 1 n

Young-teacuteteles feliacuteraacutesban

partfi

partdj=

partfj

partdi

Pl

f(x y) = (minusy x)part1f2(x y) = 1 part2f1(x y)1 6= minus1kuumlloumlnboumlzoekrArr nincsF

Megjegyzeacutes

R3 rarr R

3 esetben ez a rotaacutecioacutementesseacuteget jelenti

f prime =

part1f1 part2f1 part3f1part1f2 part2f2 part3f2part1f3 part2f3 part3f3

Haexist primitiacutev fuumlggveacuteny akkor ez a maacutetrix szimmetrikusrot = 0 rArr maacutetrix szimmetrikus

4

Kalkulus II

Balog Daacuteniel

20100413

Iacutevhossz eacutes vonalintegraacutel

Def Egyr [a b] rarr Rn fuumlggveacutenyt nevezuumlnk goumlrbeacutenek azRn-ben

bull plr(t) = (t 2t)t isin [0 1]r [0 1]R2

1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz

Ugyanaz a vonal toumlbbfeacutelekeacuteppen parameacuteterezheto

bull plr(t) = (cos t sin t)t isin [0 2π]

1

minus1

1minus1

r(t) = (cos t sin t)t isin [0 4π]Ugyanez a rajz de minden pontot 2x rajzolunk meg

bull plr(t) = (RtminusR sin t) R minusR cos t)

b

Ez a paacutelya iacuteve a paacutelya az uacuten ciklois

Iacutevhossz

Legyenr [a b] rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)

Hosszkoumlzeliacuteto oumlsszeg

σ(r t1 t2 tn) =

nsum

i=1

|r(ti)minus r(timinus1)|

Def az r goumlrbeacutenek van hossza (rektifikaacutelhatoacute) hasupσ(r t1 t2 tn) n isin Ntitetszoleges felosztaacutes veacuteges

A goumlrbe hossza

l(r) = supσ(r t1 t2 tn) n isin Ntitetszoleges felosztaacutes

bull pl (az egyszeru)r(t) = (t 2t)t isin [0 1]σ = l(r) forall felosztaacutesral(r) =

radic5

bull pl (a furcsa)r(t) = (t sin 1

t)t isin (0 1]

itt a sup = infin

TeacutetelHa r differenciaacutelhatoacute eacutesrprime folytonos akkor veacuteges hossza van eacutes

l(r) =b

infa|rprime(t)|dt

Ezt nem bizonyiacutetjuk be


1

minus1

1minus1

1

rprime(t) = (minus sin t cos t)|rprime(t)| =

radic

(minus sin t)2 + (cos t)2 = 1

l(r) =2πint

0

1dt = 2π

bull Mazohista peacuteldar(t) = (cos t2 sin t2)t isin [0

radic2π] A koumlr ugyanaz de ugyanaz e a

koumlriacutev

bull plr(t) = (t t2)t isin [minus1 1]

1

minus1

1minus1

bull plr(t) = (t R cos t R sin t)t isin [0 2π]

Vonalintegraacutel

(munkaveacutegzeacutes egy goumlrbe menteacuten)Adott r [a b] rarr R

n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)

Legyentlowasti isin [timinus1 ti]A vonalintegraacutel koumlzeliacuteto oumlsszege

σ(r f t1 tn tlowast

1 tlowast

n) ==

nsum

i=1

〈f(r(tlowasti )) r(ti)minus r(timinus1)〉︸︷︷︸

~F middot~smiddotcosα

Finomiacutetjuk a felosztaacutest eacutes megneacutezzuumlk hogy mihez tart

Def Az f fuumlggveacutenynek azr goumlrbe menteacuten van vonalintegraacutelja haexistI isin R hogyforallε gt 0existδ gt 0 ha |ti minus timinus1| lt δ akkor baacutermilyentlowasti isin [ti timinus1] vaacutelasztaacutes eseteacuten|I minus σ()| lt ε

Ez azI a vonalintegraacutel jeleint

r

f = I

Teacutetel (kiszaacutemiacutetaacutes moacutedja)Har differenciaacutelhatoacute azrprime folytonos eacutesf (eroteacuter) is folytonos

Ekkorexistint

r

=bint

a

〈f(r(t)) rprime(t)〉 dtNem bizonyiacutetjuk ugyanis neheacutez


f(x y) = x y

Centraacutelis eroteacuter

int

r

f =

2πint

0

lt (cos t sin t) (minus sin t cos t)) gt dt =

= f

2πint

0

minus cos t sin t+ sin t cos tdt = 0


f(x y) =(

minusyx2+y2

xx2+y2

)

Rotaacutecioacutes eroteacuter

int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π

Vonalintegraacutel eacutes potenciaacutel kapcsolata

Adott egy f Rn rarr R

n eroteacuter a potencaacutelF Rn rarr R hogy

F prime = f(minus gradF = f)Legyenr [a b] rarr R

n goumlrbeCeacutel Vonalintegraacutelt a potenciaacuteboacutel szaacutemolhatunk-e

Leacutenyeg

Van primitiacutev fuumlggveacutenym

int

r

f csak azr veacutegpontjaitoacutel fuumlgg

mzaacutert goumlrbeacuten(r(a) = r(b)) 0 a vonalintegraacutel

TeacutetelA vonalintegraacutel kiszaacutemiacutetaacutesa a primitiacutev fuumlggveacutennyelHa r differenciaacutelhatoacuterprime eacutesf folytonos valamintexistF Rn rarr R hogyF prime = f rArr

int

r

= F (r(b)) minus F (r(a))

Bizonyiacutetaacutesint

r

definiacutecioacute szerint=bint

a

〈f(r(t)) rprime(t)〉 =bint

a

〈F prime(r(t)) rprime(t)〉 =bint

a

F (r(t)) = F (r(b)) minus F (r(a))

Azeacutert van ez mert azr [a b]-bol Rn-be keacutepez viszontF ezt vissza-hozzaR-be iacutegyR rarr R a fuumlggveacuteny eacutes erre keacutenyelmesen hasznaacutelhatoacute aNewton-Leibnitz teacutetel(f g)prime = (f prime g) middot 1gprime ezt kiterjesztjuumlkRn esetre

1maacutetrixszorzaacutes lesz

2

Kalkulus II

Balog Daacuteniel

20100420

Toumlbbvaacuteltozoacutes fuumlggveacuteny Riemann-integraacutelja

Emleacutekezteto egyvaacuteltozoacutes

f Rrarr R

a ξ b

a = x0 lt x1 lt x2 lt xn = b

(A teacuteglalap magassaacutega legyenξ)Legyenξi isin [ximinus1 xi]A koumlzeliacutetooumlsszeg

σ =

nsum

i=1

f(ξi)(xi minus ximinus1)

Def HaexistI isin R hogyforallε gt 0existδ gt 0 hogyforallxi minus ximinus1 lt δforalli rArr forallξi vaacutelasztaacutes mellett|I minus σ| lt ε

Ekkorf Riemann integraacutelhatoacute[a b]-n eacutesbint

a

= I

A ceacutel

f R2 rarr R2 fuumlggveacutenyre a Riemann integraacutel definiaacutelaacutesa

Legyena lt b c lt d

b

c

d

a

d

ca b

A felosztaacutesoka = x0 lt x1 lt x2 lt xn = b

c = y0 lt y1 lt y2 lt ym = d

Ezzel beraacutecsozzuk a teacuteglalapot Az egyvaacuteltozoacutes fuumlggveacutenyekneacutel teruuml-letet szaacutemiacutetottunk most teacuterfogatot szeretneacutenk kihozni

Legyenξi isin [xi minus 1minus xi] ηj isin [yjminus1 minus yj ]i = 1 n j = 1 mLegyen

σ =

msum

j=1

nsum

i=1

f(ξi ηj)(xi minus ximinus1)(yj minus yjminus1)

A keacutet szumma szabadon foumllcsereacutelheto csak az oumlsszegzeacutes moacuted-szere vaacuteltozik meg Peacuteldaacuteul sorok helyett oszloponkeacutent oumlsszegzuumlnkde a veacutegoumlsszeg ugyan annyi

Def azf fuumlggveacuteny az[a b]times [c d] teacuteglalapon Riemann integraacutelhatoacuteha existI isin R hogy forallε gt 0existδ gt 0 hogy forallxi minus ximinus1 lt δ eacutesforallyi minus yiminus1 lt δforalliforallj rArr forallξi ηj vaacutelasztaacutes mellett|I minus σ| lt εEkkor f Riemann integraacutelhatoacute[a b] times [c d]-n eacutes

int

T

= I itt

T = [a b]times [c d] teacuteglalapEz azonban nem mukoumldik ha nem teacuteglalap a grafikon alatti teruumlletpl feacutelgoumlmbHa mostD sub R

2 egy halmaz (pl koumlr) akkorint

D

hogyan

eacutertelmezheto

TD

LegyenD sub R2 korlaacutetos tartomaacutenyLegyenT sub R2 olyan teacuteglalap ami magaacuteba fogalalja aD-tLegyenf T rarr R olyan hogy

f(x) =

f(x) hax isin D0 hax isin D

Def Az f fuumlggveacuteny Riemann integraacutelhatoacuteD-n haf Riemann integraacutel-hatoacuteT -n eacutes

int

D

f =int

T

f

(igazolhatoacute hogy megfelelo f vaacutelasztaacutesaacuteval toumlk mindegy milyen ateacuteglalap)Neheacutez keacuterdeacutes milyen lehet aDSzeacutepD pl az amit egy differenciaacutelhatoacute goumlrbe hataacuterol (pl koumlr)

Az integraacutel kiszaacutemiacutetaacutesa

Fubini-teacutetelLegyenf [a b]times [c d] rarr R folytonosEkkor

int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy

Ezeacutert gyakori jeloumlleacutesint

[ab]times[cd]

f helyettdint

c

bint

a

f Fizikusok kiiacuterjaacutek a toumlbb-

szoumlroumls integraacutelt hiszen nem dolgoznakn dimenzioacuteban A teacutetel bizonyiacute-taacutesaacutenak alapondolata

Legyeng(x)dint

c

f(x y)dy

g(x) egy egyvaacuteltozoacutes riemann integraacutel amitσ koumlzeliacutetooumlsszeggel defini-aacuteltunkEgy kis raacutecs meacuterete nagysaacutegrendileg az egyvaacuteltozoacutes fuumlggveacuteny σ-jadint

c

g(x)dx az oszlopok teruumlleteinek oumlsszege

PeacuteldaacutekLegyenf(x y) = x middot y T = [0 1]times [2 3]int

T

f =

Fubini teacutetel szerint1int

0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4

LegyenT = [1 2]times [1 2] f(x y) = 2xy(x2+y2)2

int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4

Integraacutelkiszaacutemiacutetaacutes normaacuteltartomaacutenyon

Most a krumpli helyett keacutet fuumlggveacuteny aacutelltal koumlzbezaacutert teruumllet lesz aDDe mi is az a normaacuteltartomaacutenyLegyen [a b] sub R egy zaacutert intervallum legyenα β [a b] rarr R

fuumlggveacutenyekα(x) le β(x) forallx isin [a b]Def Azα eacutesβ fuumlggveacutenyek aacutelltal meghataacuterozott normaacuteltartomaacuteny

d

ca

β(x)

α(x)

b

N

N(x y) isin R2 x isin [a b] y isin [α(x) β(x)]Szeretneacutenk

int

N

f -et megadni

Veszek egyc eacutesd-t uacutegy hogyc le α(x) le β(x) le (d)forallxLegyenf N rarr R folytonos

Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx

Tehaacutet a normaacuteltartomaacutenyon az integraacutel uacutegy adhatoacute meg hogybint

a

(

dint

c

f(x y)dy

)

dx

Peacuteldaacutekf(x y) = x middot yAz y = 1minus x2 eacutes azy = x2 minus 1 parabolaacutek koumlzoumltt fekvo tartomaacutenyon

1

minus1

1minus1

Mosta = minus1 b = 1 α(x) = x2 minus 1 β(x) = 1minus x2

N =

int

N

=

1int

minus1

x2

int

x2minus1

1minus x2(x middot y)dy

dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427

Integraacuteltranszformaacutecioacute

(helyettesiacuteteacuteseacutees integraacutelaacutes toumlbbvaacuteltozoacutes fuumlggveacutenyre)

Ceacutel Integraacutel kiszaacutemiacutetaacutesa nem teacuteglalapon

mertint

T

f =b1int

a1

b2int

a2

f(x y)dxdy de ez koumlroumln maacuter nem mukoumldik

EmleacutekeztetoHelyettesiacuteteacuteses integraacutel egyvaacuteltozoacutes fuumlggveacutenyref Rrarr R

g[α β] rarr [a b]Hag(α) = a

g(β) = bbint

a

f(x)dx =

βint

α

f(g(t)) middot gprime(t)dt

gyakorlatilagx = g(t) dx = gprime(t)dt Fontos hogy ag egy bijekcioacutelegyen(ne legyen hullaacutemzoacute)Ha szigoruacute monoton noumlvekvo helyett szigoruacute monoton csoumlkkeno akkora hataacuterok helyet csereacutelnekUacutegy is meg lehet oldani hogy egy minuszjeletkapErre kellene valami oumlsszevont dologEacuteszreveheto ha a g no grsquo(t) pozitiacutev ha g csoumlkken grsquo(t) negatiacutev de azegeacutesz kap egy minuszjelet (ha a hataacuterok maradnak) Tehaacutet az oumlsszevontkeacuteplet

bint

a

f(x)dx =

βint

α

f(g(t)) middot |gprime(t)|dt

Proacutebaacuteljuk meg ugyaneztf R2 rarr R fuumlggveacutenyreint

T

f -et akarjuk kiszaacutemiacutetani helyettesiacuteteacutessel ittT = [a1 b1]times [a2 b2]

Legyen ag fuumlggveacutenyuumlnkg [α1 β1]times [α2 β2] rarr T bijekcioacuteEloszoumlr specg fuumlggveacutenyreg(t1 t2) = (g1(t1) g2(t2))Ez azeacutert speciaacutelis mertg1 g2 csak egyfeacutelet-tol fuumlgg

int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx

A zaacuteroacutejelben leacutevo valamix fuumlggveacutenye (hiszeny szerint kiintegraacuteltukattoacutel maacuter nem fuumlgg)Mosty helyeacutere behelyettesiacutetjuumlk ag2(t2)-t

int

T

f =

b1int

a1

β2int

α2

f(x g2(t2))|gprime2(t2)|dt2

dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)

Ez oumlsszesiacutetve int

R

(f g) middot | det gprime|

Ahol R = [α1 β1]times [α2 β2]Ez a levezeteacutes keacutet iraacutenyban is aacuteltalaacutenosiacutethatoacute

bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))

bull aT lehet maacutes tartomaacuteny nem csak teacuteglalap

Ezeket nem vezetjuumlk le

Teacutetel

LegyenD sub R2 f D rarr R folytonosLegyenR sub R teacuteglalapg R rarr D bijekcioacuteEkkor igaz hogy

int

D

f =int

R


Ez az integraacuteltranszformaacutecioacute keacutepleteMegjEz a keacutepletn dimenzioacuteban is iacutegy van

Peacutelda

0 r0

2π

D = (x y) isin R2 x2 + y2 le 4 (koumlrlap)f(x y) = x2 + y2

Kell egyR teacuteglalap eacutes egyg R rarr D bijPolaacuterkoordinaacuteta-gyanuacutes

r isin [0 2]ϕ = [0 2π]1

1itt [0 2π) eacutesr = (0 2] kellene

1

x = r middot cosϕy = r middot sinϕ

g(r ϕ) =

(cos(ϕ) minusr sinϕsinϕ r cosϕ

)

det gprime(r ϕ) = r middot 1int

D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π

Maacutesodik peacutelda

minus30

60

A kuumllso sugaacuter 2 a belso 1f( xyradic

(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ

g(r ϕ) = (r middot cosϕ r middot sinϕ)int

D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic

r2 cos2 ϕ+ r2 sin2 ϕrdϕdr =

=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12

Az eddigi keacutet peacutelda a polaacutertranszformaacutecioacutet hasznaacuteltaMost joumln a 3D peacutelda Elso a hengerkoordinaacutetaacutekD = (x y z) isin R3 x2 + y2 le 1z isin [0 2]f(x y z) = zradic

x2+y2

Kell egyg R rarr D helyettesiacuteto fuumlggveacutenyg(x y z) = (r cosϕ r sinϕ z)

R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)

cosϕ minusr sinϕ 0sinϕ r cosϕ 00 0 1

det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic

r2 cos2 ϕ+ r2 sin2 ϕrdzdϕdr =

int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π

goumlmb teacuterfogata

Itt haacuterom dimenzioacutes polaacuterkoordinaacutetaacutekat kell hasznaacutelniD = (x y z) isin R3 x2 + y2 + z2 le R2f(x y z) = 1Az R sugaruacute houmlmb teacuterfogataVg =

int

D

1

Kell egyg fuumlggveacutenyz = r middot cosϑx = r middot sinϑ middot cosϕy = r middot sinϑ middot sinϕ

g(r ϕ ϑ) = (r middot sinϑ middot cosϕ r cot sinϑ middot sinϕ r middot cosϑ)

gprime(r ϕ ϑ) =

cosϕ sinϑ minusr middot sinϕ sinϑ r middot cosϕ cosϑsinϕ sinϑ r middot cosϕ sinϑ r sinϕ cosϑ

cosϑ 0 minusr middot sinϑ

det gprime = minus(r2 middot sinϑ)| det gprime| = (r2 middot sinϑ)

Vg =

int

D

1 =

Rint

0

πint

0

2πint

0

1 middot r2 middot sinϑdϑdϕdr =

=

Rint

0

πint

0

r2 middot [sinϑ]2π0 dϕdr = 2

Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek

Emleacutekezteto goumlrbeacutekA goumlrbe egyr [a b] rarr R

nt isin [a b] lekeacutepezeacutesu alakzat

a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)

Az egyenes vonalak oumlsszege egy koumlzeliacuteto oumlsszeg

nsum

i=1

|r(timinusrtiminus1|

Ha eleacuteg suru a felosztaacutes akkor a koumlzeliacutetooumlsszeg tart a hosszaacutehozTeacutetel

l(r)(goumlrbe hossza)=

bint

a

|r(t)|dt

(a felbontaacutes tart az integraacutelaacutes feacutele felbontaacuteshoz)Mi a feluumlletDef Legyen Ω sub R

2 tartomaacuteny (teacuteglalap szokott lenni) EgyΦ Ω rarr R

3 fuumlggveacutenyt feluumlletnek nevezuumlkpl

Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ

Pontokat keacutepezuumlnk le(dehogy mi a geacutep ) Ha eleacuteg suruen vannak apontok akkor feluumlletnek laacutetjukMatematikai eacutertelemben az oumlsszes pontot (xi minus ximinus 1 lt ε-szerumegoldaacutes) kell venniMegjegyzeacutes A goumlrbe eacutes a feluumllet doumlbbdimenzioacutes aacuteltalaacutenosiacutetaacutesa asokasaacuteg(mainfold)plGoumlmbfeluumllet

ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =

R middot sinu middot cos vR middot sinu middot sin v

R middot cosu

A lilaacuteval jeloumllt szakasz hossza azRa saacutergaacuteval jeloumllt szoumlg (a z tangely-tol valoacute elteacutereacutes) a ϑ eacutes a zoumllddel je-loumllt (az x tegelytol valoacute elteacutereacutes) aϕHa maacuter uacutegyis megvan a feluumllet egy teacuteg-lalapban eacutes az hogy ebbol hogy leszgoumlmb akkor eacuterdemes a felsziacutent megmon-dani

Felsziacuten megadaacutesa(nagyjaacuteboacutel)

bb

b

Φ(u v)Φ(∆u+ u v)

Φ(u∆v + v)LegyenΩ-nak felosztaacutesa(raacutecsozaacutes)

Egy kis feluumlletdarab koumlzeliacuteto nagysaacutegaacutetkell definiaacutelni Az eacuterintosiacutekon egy para-lelogramma teruumlleteacutevel lehet koumlzeliacuteteni

partuΦ(u v) middot∆u partvΦ(u v) middot∆v isin R3

A paralelogramma teruumllete a ketto vektoriaacutelis szorzataacutenak abszolulteacuterteacuteke

|partuΦ(u v)times partvΦ(u v)|∆u∆v

σ a felsziacuten koumlzeliacuteto oumlsszege

sumsum


10 = 2π

1

(bunkoacute autoacuteskeacutent nem indexeluumlnk )Teacutetel Legyen egyΦ Ω rarr R

3 feluumllet differenciaacutelhatoacute Ennek felsziacutene

S =

int

Ω

|partuΦ(u v)times partvΦ(u v)|dudv

plGoumlmbfelsziacuten kiszaacutemiacutetaacutes

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =

R middot cosu middot cos vR middot cosu middot sin vminusR middot sinu

partv =

minusR middot sinu middot sin vR middot sinu middot cos v

0

partuΦtimes partvΦ =

∣∣∣∣∣∣

i j k

R cosu cos v R cosu sin v minusR sinuminusR sinu sin v R sinu cos v 0

∣∣∣∣∣∣

R2 middot sin2 u middot cos vR2 middot sin2 u middot sin v

R2 middot sinu cos2 v +R2 sinu cosu sin2 v︸︷︷︸

R2 sinu cosu

Ennek a hossza

R2 sinu(sin2 u cos2 v + sin2 u sin2 v︸︷︷︸

sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π

Felsziacutenintegraacutel (skalaacuter eacuterteacuteku fuumlggveacuteny integraacutelja feluumlleten)

LegyenΩ sub R2Φ ω rarr R3 differenciaacutelhatoacute feluumllet

LegyenU R3 rarr R folytonos fuumlggveacutenyDefiniaacutelni akarjuk az

int

Φ

U

Kelleneσ eacutes Definiciacuteoacute de egyik sincs veacutegigvezetveTeacutetel

int

Φ

U =

int

Ω

U(Φ(u v)) middot |partuΦ(u v)times partvΦ(u v)|dudv

plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

U(x y z) = x+ y + zint

Φ

U =

Kiszaacutemiacutetaacutes a teacutetellel

partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2

|partuΦtimes partvΦ| =radic1 + 1 + 4 =

radic6

int

Phi

U =

1int

0

1int

0

(u+ v + uminus v + u)︸︷︷︸

U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2

Mi van a vektorokkalFeluumlleti integraacutel (vektoreacuterteacuteku fuumlggveacuteny integraacutelja feluumlleten)LegyenΩ sub R2Φ ω rarr R

3 differenciaacutelhatoacute feluumlletLegyenF R3 rarr R

3 folytonos fuumlggveacutenyDefiniaacutelandoacute

int

Φ

Kelleneσ Definiacutecioacute de ezek ismeacutet kimaradnakTeacutetel

int

Φ

F =

int

Ω

〈F (Φ(u v) partuΦ(u v)times partvΦ(u v)〉 dudv

Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ

F =partuΦtimes partvΦ =

11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((

(uminus v + u+ v minus 2u)︸︷︷︸

0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf

A grad f azf parciaacutelis derivaacuteltjaiboacutel keacutepezett vektorHaf R3 rarr R

3

lt nabla f gt= partxf1 + partyf2 + partzf3 = div f

f = (f1 f2 f3) fi R3 rarr R

f prime

jakobi =


nablatimes f = rot f

- ∣∣∣∣∣∣

i k k

partx party partzf1 f2 f3

∣∣∣∣∣∣

=

partyf3minuspartzf2partzf1minuspartxf3partxf2minuspartyf1

= rot f

Haacutezinakrot(gradf) = 0 forallf

∆div(grad f) = part21f + part2

2f + part23f

Eacuterdemes megneacutezni az Integraacutelaacutetalakiacutetoacute teacuteteleketGauss-teacutetel eacutes Stokes teacutetel

3

Kalkulus II

Balog Daacuteniel

20100209

1 Integraacutelaacutes

11 Racionaacutelis toumlrtfuumlggveacutenyek

polinompolinom Ezeket mindig lehet integraacutelni az ilyen primitiacutev fuumlggveacutenye

meghataacuterozhatoacute

Ellenpeacuteldaint

eminusx2

Ez szimpatikusan neacutez ki meacutegsem lehet kiszaacute-molni

1int

1ax+b

plint

1

x= ln(x) + c

int

1

x+ 3= ln |x+ 3|+ c

int

1

2xminus 8=

1

2middot ln(2xminus 8) + c

2int

cx+d

ax+bpl

int

x+ 2

x+ 1=

int

x+ 1 + 1

x+ 1=

int

x+ 1

x+ 1+

1

x+ 1dx = x+ ln(x + 1) + c

int

3x+ 1

2x+ 3= 1

int

(2x+ 3) middot 32 minus 7

2

2x+ 3=

int

3

2minus

72

2x+ 3dx =

3

2xminus 7

2middot 12ln(2x+ 3) + c

3int

1(xminusa)(xminusb) pl

int

3

x2 + xminus 22 =

int

3

(xminus (minus2)) middot (xminus 1)=

int minus1

x+ 2+

1

xminus 1dx3 = minus ln(x+2)+ln(xminus1)+c = ln

xminus 1

x+ 2+c

4int

cx+d

(xminusa)(xminusb) plint

3 + x

2 + 3xminus 2x2= 4

int

3 + x

(2minus x)(2x+ 1)=

int

A

2minus x+

B

2x+ 1dx

1a 3

2azeacutert kell hogy a 3x stimmeljen a7

2hogy az 1 stimmeljen

2 minusbplusmn

radicb2minus4ac

2arArr x1 = minus2 x2 = 1

3(koumlzoumls nevezore hozaacutes)42 + 3xminus 2x2 = 0 x1 = 2 x2 = 1

2

Truumlkk a kibontottat beszorzomminus2-vel minus2(x minus 2)(x + 12 ) iacutegy ki

is joumln A szeacutetbontaacutes koacutedneveparciaacutelis toumlrtekre bontaacutes

A(2x+ 1) +B(2minus x) = 3 + x(2AminusB)x+A+ 2B = 3 + x

2AminusB = 1 A+ 2B = 3

A = 1 B = 1

=

int

1

2minus x+

1

2x+ 1= minus ln(2minus x) +

1

2ln(2x+ 1) + c

5int

1(xminusa)2 pl

int

1

x2= minus 1

x+ c

6int

cx+d

(xminusa)2 pl

int

2x+ 3

(xminus 1)2=

int

(xminus 1) middot 2 + 5

(xminus 1)2=

int

2

xminus 1+

5

(x minus 1)2dx = 2 ln(xminus 1)minus 5

xminus 1+ c

7int

1(xminusc)2+d

plint

1

x2 + 1= arctg(x) + c

int

1

x2 + 4=

int

1

4middot 1

(x2 )2 + 1

=1

4middot 2 middot arctg(x

2)

int

1

x2 + 3=

1

3middotint

1

( xradic3)2 + 1

=1

3middotradic3 middot arctg( xradic

3)

int

1

x2 minus 2x+ 2=

int

1

(xminus 1)2 + 1= arctg(xminus 1)

A megoldaacutes meneteA valoacutes gyoumlkoumlket kell ellenorizni Ha van 2 darabvaloacutes gyoumlk laacutesd 4 peacutelda ha 1 db van ld 5 ha nincs ld 7

8int

ax+b

(xminusc)2+d2 pl

int

2x

1 + x2Felismerheto hogy ez

f prime

ftipusuacute

int

2x+ 3

1 + x2=

int

2x

1 + x2+

3

1 + x2= ln(1 + x2) + 3 arctg(x) + c

int

x+ 3

1 + x2=

int

2x middot 12 + 3

1 + x2=

1

2ln(1 + x2) + 3 arctg(x)

1


ax2 + bx+ c


1 a lt 0 plint

radic

1minus x2 = 5

intradic


int

cos2(t)dt




cos2(t) =

int





1minus cos2(t) =


cos2(t)dt

Oumlsszevonva

2

int



radic

1minus x2dx =

int


2=

1

2

radic


2arcsin(x)

Summaacutezvaint

radic

1minus x2 =1

2xradic

1minus x2 +1

2arcsin(x)

pl

int

radic

4minus x2dx = 2

int

radic

1minus (x

2)2 = 7

2

int

radic

1minus t22dt = x

radic

1minus (x

2)2 + 2 arcsin

radicx2

int

radic

3minus 2xminus x2 =

int

radic

4minus (x+ 1)2 =

2

int

radic

1minus (x+ 1

2)2t =

x+ 1

22dt = dx

int

radic

3minus 2xminus x2 =

int

radic

4minus (x+ 1)2 =

2

int

radic

1minus (x+ 1

2)2t =

x+ 1

28


dt= cos(t))

7x = 2t

82dt = dx

4

int

radic

1minus t2dt =


radic


int




radic

x2 + 1 =

int


2

Kalkulus II

Balog Daacuteniel

20100216



11int

0

1xdx




1int

0

1

xdx = [lnx]



2infinint

0

eminusxdx


infinint

0

eminusxdx =[eminusx

]1


(minuseminus0


Ez se stimmel

Ceacutel



I


exist limbrarrinfin

bint

a

f(x)dx

Ennek jele+infinint

a

f(x)dx

Hasonloacuteanbint

minusinfin


bint

a

f(x)dx

Peacuteldaacutek

bullinfinint

0


bint

0


[minuseminusx]b

0 =

= limbrarrinfin



bullinfinint

1

1

xdx = lim

brarrinfin

bint

1

1

xdx = lim

brarrinfin[lnx]

b

1 =

= limbrarrinfin


bullinfinint

1

1

x2dx = lim

brarrinfin

bint

1

1

x2dx = lim

brarrinfin

[

minus 1

x

]b

1

=

= limbrarrinfin

[

minus1

bminus(

minus1

1

)]

= 1

0

1

2

0 1 2 3 4


1

infinint

1

1xα


α 6= 1

infinint

1

1

xαdx = lim

brarrinfin

bint

1

1

xαdx = lim

brarrinfin

[

minus x1minusα

1 minus α

]b

1

=

= limbrarrinfin

(b1minusα


1 minus α

)


Haα lt 1 infin

Def Azinfinint

a



infinint

1

1x2 dx konvergens

infinint

0


2

1x2 dx konvergens

infinint

1

1xdx divergens

infinint

0


03

1xdx divergens

infinint

1

1radicxdx divergens

infinint

0


2

1xdx divergens



bullinfinint

0

eminusx2

dx

0

1

2

0 1 2 3 4


infinint

0


dArrinfinint

1


dArrinfinint

1

eminusx2

dx konvergens

dArrinfinint

0

eminusx2

dx konvergens

bullinfinint

0

eminusx2

dx =

= lim0rarrinfin

infinint

0

eminusx2

dx = lim0rarrinfin

[]b0


Kideruumllinfinint

0

eminusx2

dx =radic

π

2

infinint

minusinfin

infinint

minusinfin


1

bullinfinint

0

sin xx

dx

bullinfinint

0

eminusx sinxdx

bullinfinint

0

sin(x2)dx

A gyoumlkoumlkradic


II


0

1xdx



cint

a

f(x)dx exist ekkor

bint

a

f(x)dx = limcrarrb

cint

a

f(x)dx


bint

a

f(x)dx = limcrarra

bint

c

f(x)dx

bint

c


pl

1int

0

1

xdx = lim

crarr0+

1int

c

1

xdx = lim

crarr0+[lnx]

1c =

limcrarr0+

( ln 1︸︷︷︸

0


1int

0

1radicx

dx = limcrarr0+

1int

c

xminus 12 dx = lim

crarr0+[]10 = lim

crarr0+

(

1 minusradic

012

)

= 2


2

Kalkulus II

Balog Daacuteniel

20100223



pl fn(x) = xn n = 1 2 3

1



fn(

12

)

12

122

123

124 konvergens

fn(13 )

13

132

133

134 konvergens











ffn


1

1n

1int

0

fn(x)dx = 1minus1

2n

1int

0

f(x)dx = 1

Tehaacutet

limnrarrinfin

1int

0

fn(x)dx = 1minus1

2n=

1int

0

f(x)dx = 1

1

1

2n

1n

1int

0

fn(x)dx =2

nmiddot 1 middot

1

2= 1foralln

1int

0

f(x)dx = 0



H

fn

fn minus ε

fn + ε



H






limnrarrinfin

bint

a

f(x)dx =

bint

a

f(x)dx

Maacuteskeacutepp

limnrarrinfin

bint

a

f(x)dx =

bint

a

lim fn(x)dx

lim eacutesint




sum





Def asum


KH(sum

fn) = x isin R

sum



f(x) =

infinsum

n=1

fn(x)

pl fn(x) = xn

s1(x) = x1 s2(x) = x1 + x2

sum


KH(sum

fn) = (minus1 1)sum


sum


2


sum

xn Cn = 1 a = 0

plsum

xn

nKH =

x =1

2sum 1

nmiddot1

2n


x = 2sum 2n

n


ndivergens


nez konvergens

3

Kalkulus II

Balog Daacuteniel

20100302





sumxn cn = 1 a = 0


sumxn

n cn = 1n a = 0


pl

sumxn




sumansor

konvergens ha lim n

radican lt 1

divergens ha lim n

radican gt 1

sum


n

radic


lim n

radic

|an| = |x|


pl


an = n middot xn

n

radic

|an| = n

radicn|x|

limnrarrinfin

n

radic



n

radicn = 1


ha

|x| lt 1 akkor


|x| ge 1 akkorsum




an =sum


n

radic

|an| = n



limnrarrinfin

n

radic


n



|xminus a| lt 1

lim n

radiccn

Def


R =1

lim n

radiccn



enneka a koumlzepe




Asum



ahol

R =1

lim n

radiccn



1

pl

sum

xn cn = 1 a = 0

R =1

lim n

radic1= 1



KH = (minus1 1)

sum xn

n2cn =

1

n2a = 0

R =1

lim n

radic1n2

= 1

segiacutetseacuteg

limn

radic

1

n2= lim

n

radic1

n

radicn2

= limn

radic1

( n

radicn)2

=1

12= 1


veacutegpontok

x = 1 rArrsum 1

n2konvergens


n2konvergens

KH = [minus1 1]

sum xn

ncn =

1

na = 0

R =1

lim n

radic1n

= 1


veacutegpontok

x = 1 rArrsum 1

ndivergens


nkonvergens

KH = [ minus 1 1 )

Def

Legyensum


f(x) =

infinsum

0


pl

sumxnf(x) =

x isin (minus1 1)

Nsum

0


xminus 1

infinsum

n=0

xn = limNrarrinfin

(

Nsum

0

xn) = limNrarrinfin


xminus 12 =

1

1minus x

infinsum

n=0

xn =1

1minus x

Oumltletsum


infinsum

n=0

xn =1

1minus x


infinsum

n=0


1minus x

2

middot x

infinsum

n=0

nxn =1

1minus x

n

pl

infinsum

n=0

n

2n=

1

2+

2

4+

3

8+

4

16+


12

(1minus 12 )

2=

1

2times 1

4= 2

Igaz-e hogy(sum

fn)prime =sum


Legyensum


f(x) =

infinsum

n=0




n=1


f primeprime(x) =

infinsum

n=2




2


Legyen

f(x) =

infinsum

n=0


f(a) = c0


f prime(a) = c1





Teacutetel

Legyen

f(x) =infinsum

n=0


21 Taylor sor

Eddig adott volt


cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n


fR rarr R

a isin R


cn =f (n)a

nforalln


sum

cn(xminus a)n

hatvaacutenysortDef


sum f (n)a

n(xminus a)n



f(x) =1




f (n)(a) = n

cn = 1


xn






6


cn =1

n

ex Taylor sorasum

xn


R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin



sum xn



infinsum

0

sum xn

n= ex

3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3


ax2 + bx+ c


1 a lt 0 plint

radic

1minus x2 = 5

intradic


int

cos2(t)dt




cos2(t) =

int





1minus cos2(t) =


cos2(t)dt

Oumlsszevonva

2

int



radic

1minus x2dx =

int


2=

1

2

radic


2arcsin(x)

Summaacutezvaint

radic

1minus x2 =1

2xradic

1minus x2 +1

2arcsin(x)

pl

int

radic

4minus x2dx = 2

int

radic

1minus (x

2)2 = 7

2

int

radic

1minus t22dt = x

radic

1minus (x

2)2 + 2 arcsin

radicx2

int

radic

3minus 2xminus x2 =

int

radic

4minus (x+ 1)2 =

2

int

radic

1minus (x+ 1

2)2t =

x+ 1

22dt = dx

int

radic

3minus 2xminus x2 =

int

radic

4minus (x+ 1)2 =

2

int

radic

1minus (x+ 1

2)2t =

x+ 1

28


dt= cos(t))

7x = 2t

82dt = dx

4

int

radic

1minus t2dt =


radic


int




radic

x2 + 1 =

int


2

Kalkulus II

Balog Daacuteniel

20100216



11int

0

1xdx




1int

0

1

xdx = [lnx]



2infinint

0

eminusxdx


infinint

0

eminusxdx =[eminusx

]1


(minuseminus0


Ez se stimmel

Ceacutel



I


exist limbrarrinfin

bint

a

f(x)dx

Ennek jele+infinint

a

f(x)dx

Hasonloacuteanbint

minusinfin


bint

a

f(x)dx

Peacuteldaacutek

bullinfinint

0


bint

0


[minuseminusx]b

0 =

= limbrarrinfin



bullinfinint

1

1

xdx = lim

brarrinfin

bint

1

1

xdx = lim

brarrinfin[lnx]

b

1 =

= limbrarrinfin


bullinfinint

1

1

x2dx = lim

brarrinfin

bint

1

1

x2dx = lim

brarrinfin

[

minus 1

x

]b

1

=

= limbrarrinfin

[

minus1

bminus(

minus1

1

)]

= 1

0

1

2

0 1 2 3 4


1

infinint

1

1xα


α 6= 1

infinint

1

1

xαdx = lim

brarrinfin

bint

1

1

xαdx = lim

brarrinfin

[

minus x1minusα

1 minus α

]b

1

=

= limbrarrinfin

(b1minusα


1 minus α

)


Haα lt 1 infin

Def Azinfinint

a



infinint

1

1x2 dx konvergens

infinint

0


2

1x2 dx konvergens

infinint

1

1xdx divergens

infinint

0


03

1xdx divergens

infinint

1

1radicxdx divergens

infinint

0


2

1xdx divergens



bullinfinint

0

eminusx2

dx

0

1

2

0 1 2 3 4


infinint

0


dArrinfinint

1


dArrinfinint

1

eminusx2

dx konvergens

dArrinfinint

0

eminusx2

dx konvergens

bullinfinint

0

eminusx2

dx =

= lim0rarrinfin

infinint

0

eminusx2

dx = lim0rarrinfin

[]b0


Kideruumllinfinint

0

eminusx2

dx =radic

π

2

infinint

minusinfin

infinint

minusinfin


1

bullinfinint

0

sin xx

dx

bullinfinint

0

eminusx sinxdx

bullinfinint

0

sin(x2)dx

A gyoumlkoumlkradic


II


0

1xdx



cint

a

f(x)dx exist ekkor

bint

a

f(x)dx = limcrarrb

cint

a

f(x)dx


bint

a

f(x)dx = limcrarra

bint

c

f(x)dx

bint

c


pl

1int

0

1

xdx = lim

crarr0+

1int

c

1

xdx = lim

crarr0+[lnx]

1c =

limcrarr0+

( ln 1︸︷︷︸

0


1int

0

1radicx

dx = limcrarr0+

1int

c

xminus 12 dx = lim

crarr0+[]10 = lim

crarr0+

(

1 minusradic

012

)

= 2


2

Kalkulus II

Balog Daacuteniel

20100223



pl fn(x) = xn n = 1 2 3

1



fn(

12

)

12

122

123

124 konvergens

fn(13 )

13

132

133

134 konvergens











ffn


1

1n

1int

0

fn(x)dx = 1minus1

2n

1int

0

f(x)dx = 1

Tehaacutet

limnrarrinfin

1int

0

fn(x)dx = 1minus1

2n=

1int

0

f(x)dx = 1

1

1

2n

1n

1int

0

fn(x)dx =2

nmiddot 1 middot

1

2= 1foralln

1int

0

f(x)dx = 0



H

fn

fn minus ε

fn + ε



H






limnrarrinfin

bint

a

f(x)dx =

bint

a

f(x)dx

Maacuteskeacutepp

limnrarrinfin

bint

a

f(x)dx =

bint

a

lim fn(x)dx

lim eacutesint




sum





Def asum


KH(sum

fn) = x isin R

sum



f(x) =

infinsum

n=1

fn(x)

pl fn(x) = xn

s1(x) = x1 s2(x) = x1 + x2

sum


KH(sum

fn) = (minus1 1)sum


sum


2


sum

xn Cn = 1 a = 0

plsum

xn

nKH =

x =1

2sum 1

nmiddot1

2n


x = 2sum 2n

n


ndivergens


nez konvergens

3

Kalkulus II

Balog Daacuteniel

20100302





sumxn cn = 1 a = 0


sumxn

n cn = 1n a = 0


pl

sumxn




sumansor

konvergens ha lim n

radican lt 1

divergens ha lim n

radican gt 1

sum


n

radic


lim n

radic

|an| = |x|


pl


an = n middot xn

n

radic

|an| = n

radicn|x|

limnrarrinfin

n

radic



n

radicn = 1


ha

|x| lt 1 akkor


|x| ge 1 akkorsum




an =sum


n

radic

|an| = n



limnrarrinfin

n

radic


n



|xminus a| lt 1

lim n

radiccn

Def


R =1

lim n

radiccn



enneka a koumlzepe




Asum



ahol

R =1

lim n

radiccn



1

pl

sum

xn cn = 1 a = 0

R =1

lim n

radic1= 1



KH = (minus1 1)

sum xn

n2cn =

1

n2a = 0

R =1

lim n

radic1n2

= 1

segiacutetseacuteg

limn

radic

1

n2= lim

n

radic1

n

radicn2

= limn

radic1

( n

radicn)2

=1

12= 1


veacutegpontok

x = 1 rArrsum 1

n2konvergens


n2konvergens

KH = [minus1 1]

sum xn

ncn =

1

na = 0

R =1

lim n

radic1n

= 1


veacutegpontok

x = 1 rArrsum 1

ndivergens


nkonvergens

KH = [ minus 1 1 )

Def

Legyensum


f(x) =

infinsum

0


pl

sumxnf(x) =

x isin (minus1 1)

Nsum

0


xminus 1

infinsum

n=0

xn = limNrarrinfin

(

Nsum

0

xn) = limNrarrinfin


xminus 12 =

1

1minus x

infinsum

n=0

xn =1

1minus x

Oumltletsum


infinsum

n=0

xn =1

1minus x


infinsum

n=0


1minus x

2

middot x

infinsum

n=0

nxn =1

1minus x

n

pl

infinsum

n=0

n

2n=

1

2+

2

4+

3

8+

4

16+


12

(1minus 12 )

2=

1

2times 1

4= 2

Igaz-e hogy(sum

fn)prime =sum


Legyensum


f(x) =

infinsum

n=0




n=1


f primeprime(x) =

infinsum

n=2




2


Legyen

f(x) =

infinsum

n=0


f(a) = c0


f prime(a) = c1





Teacutetel

Legyen

f(x) =infinsum

n=0


21 Taylor sor

Eddig adott volt


cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n


fR rarr R

a isin R


cn =f (n)a

nforalln


sum

cn(xminus a)n

hatvaacutenysortDef


sum f (n)a

n(xminus a)n



f(x) =1




f (n)(a) = n

cn = 1


xn






6


cn =1

n

ex Taylor sorasum

xn


R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin



sum xn



infinsum

0

sum xn

n= ex

3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Kalkulus II

Balog Daacuteniel

20100216



11int

0

1xdx




1int

0

1

xdx = [lnx]



2infinint

0

eminusxdx


infinint

0

eminusxdx =[eminusx

]1


(minuseminus0


Ez se stimmel

Ceacutel



I


exist limbrarrinfin

bint

a

f(x)dx

Ennek jele+infinint

a

f(x)dx

Hasonloacuteanbint

minusinfin


bint

a

f(x)dx

Peacuteldaacutek

bullinfinint

0


bint

0


[minuseminusx]b

0 =

= limbrarrinfin



bullinfinint

1

1

xdx = lim

brarrinfin

bint

1

1

xdx = lim

brarrinfin[lnx]

b

1 =

= limbrarrinfin


bullinfinint

1

1

x2dx = lim

brarrinfin

bint

1

1

x2dx = lim

brarrinfin

[

minus 1

x

]b

1

=

= limbrarrinfin

[

minus1

bminus(

minus1

1

)]

= 1

0

1

2

0 1 2 3 4


1

infinint

1

1xα


α 6= 1

infinint

1

1

xαdx = lim

brarrinfin

bint

1

1

xαdx = lim

brarrinfin

[

minus x1minusα

1 minus α

]b

1

=

= limbrarrinfin

(b1minusα


1 minus α

)


Haα lt 1 infin

Def Azinfinint

a



infinint

1

1x2 dx konvergens

infinint

0


2

1x2 dx konvergens

infinint

1

1xdx divergens

infinint

0


03

1xdx divergens

infinint

1

1radicxdx divergens

infinint

0


2

1xdx divergens



bullinfinint

0

eminusx2

dx

0

1

2

0 1 2 3 4


infinint

0


dArrinfinint

1


dArrinfinint

1

eminusx2

dx konvergens

dArrinfinint

0

eminusx2

dx konvergens

bullinfinint

0

eminusx2

dx =

= lim0rarrinfin

infinint

0

eminusx2

dx = lim0rarrinfin

[]b0


Kideruumllinfinint

0

eminusx2

dx =radic

π

2

infinint

minusinfin

infinint

minusinfin


1

bullinfinint

0

sin xx

dx

bullinfinint

0

eminusx sinxdx

bullinfinint

0

sin(x2)dx

A gyoumlkoumlkradic


II


0

1xdx



cint

a

f(x)dx exist ekkor

bint

a

f(x)dx = limcrarrb

cint

a

f(x)dx


bint

a

f(x)dx = limcrarra

bint

c

f(x)dx

bint

c


pl

1int

0

1

xdx = lim

crarr0+

1int

c

1

xdx = lim

crarr0+[lnx]

1c =

limcrarr0+

( ln 1︸︷︷︸

0


1int

0

1radicx

dx = limcrarr0+

1int

c

xminus 12 dx = lim

crarr0+[]10 = lim

crarr0+

(

1 minusradic

012

)

= 2


2

Kalkulus II

Balog Daacuteniel

20100223



pl fn(x) = xn n = 1 2 3

1



fn(

12

)

12

122

123

124 konvergens

fn(13 )

13

132

133

134 konvergens











ffn


1

1n

1int

0

fn(x)dx = 1minus1

2n

1int

0

f(x)dx = 1

Tehaacutet

limnrarrinfin

1int

0

fn(x)dx = 1minus1

2n=

1int

0

f(x)dx = 1

1

1

2n

1n

1int

0

fn(x)dx =2

nmiddot 1 middot

1

2= 1foralln

1int

0

f(x)dx = 0



H

fn

fn minus ε

fn + ε



H






limnrarrinfin

bint

a

f(x)dx =

bint

a

f(x)dx

Maacuteskeacutepp

limnrarrinfin

bint

a

f(x)dx =

bint

a

lim fn(x)dx

lim eacutesint




sum





Def asum


KH(sum

fn) = x isin R

sum



f(x) =

infinsum

n=1

fn(x)

pl fn(x) = xn

s1(x) = x1 s2(x) = x1 + x2

sum


KH(sum

fn) = (minus1 1)sum


sum


2


sum

xn Cn = 1 a = 0

plsum

xn

nKH =

x =1

2sum 1

nmiddot1

2n


x = 2sum 2n

n


ndivergens


nez konvergens

3

Kalkulus II

Balog Daacuteniel

20100302





sumxn cn = 1 a = 0


sumxn

n cn = 1n a = 0


pl

sumxn




sumansor

konvergens ha lim n

radican lt 1

divergens ha lim n

radican gt 1

sum


n

radic


lim n

radic

|an| = |x|


pl


an = n middot xn

n

radic

|an| = n

radicn|x|

limnrarrinfin

n

radic



n

radicn = 1


ha

|x| lt 1 akkor


|x| ge 1 akkorsum




an =sum


n

radic

|an| = n



limnrarrinfin

n

radic


n



|xminus a| lt 1

lim n

radiccn

Def


R =1

lim n

radiccn



enneka a koumlzepe




Asum



ahol

R =1

lim n

radiccn



1

pl

sum

xn cn = 1 a = 0

R =1

lim n

radic1= 1



KH = (minus1 1)

sum xn

n2cn =

1

n2a = 0

R =1

lim n

radic1n2

= 1

segiacutetseacuteg

limn

radic

1

n2= lim

n

radic1

n

radicn2

= limn

radic1

( n

radicn)2

=1

12= 1


veacutegpontok

x = 1 rArrsum 1

n2konvergens


n2konvergens

KH = [minus1 1]

sum xn

ncn =

1

na = 0

R =1

lim n

radic1n

= 1


veacutegpontok

x = 1 rArrsum 1

ndivergens


nkonvergens

KH = [ minus 1 1 )

Def

Legyensum


f(x) =

infinsum

0


pl

sumxnf(x) =

x isin (minus1 1)

Nsum

0


xminus 1

infinsum

n=0

xn = limNrarrinfin

(

Nsum

0

xn) = limNrarrinfin


xminus 12 =

1

1minus x

infinsum

n=0

xn =1

1minus x

Oumltletsum


infinsum

n=0

xn =1

1minus x


infinsum

n=0


1minus x

2

middot x

infinsum

n=0

nxn =1

1minus x

n

pl

infinsum

n=0

n

2n=

1

2+

2

4+

3

8+

4

16+


12

(1minus 12 )

2=

1

2times 1

4= 2

Igaz-e hogy(sum

fn)prime =sum


Legyensum


f(x) =

infinsum

n=0




n=1


f primeprime(x) =

infinsum

n=2




2


Legyen

f(x) =

infinsum

n=0


f(a) = c0


f prime(a) = c1





Teacutetel

Legyen

f(x) =infinsum

n=0


21 Taylor sor

Eddig adott volt


cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n


fR rarr R

a isin R


cn =f (n)a

nforalln


sum

cn(xminus a)n

hatvaacutenysortDef


sum f (n)a

n(xminus a)n



f(x) =1




f (n)(a) = n

cn = 1


xn






6


cn =1

n

ex Taylor sorasum

xn


R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin



sum xn



infinsum

0

sum xn

n= ex

3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

infinint

1

1xα


α 6= 1

infinint

1

1

xαdx = lim

brarrinfin

bint

1

1

xαdx = lim

brarrinfin

[

minus x1minusα

1 minus α

]b

1

=

= limbrarrinfin

(b1minusα


1 minus α

)


Haα lt 1 infin

Def Azinfinint

a



infinint

1

1x2 dx konvergens

infinint

0


2

1x2 dx konvergens

infinint

1

1xdx divergens

infinint

0


03

1xdx divergens

infinint

1

1radicxdx divergens

infinint

0


2

1xdx divergens



bullinfinint

0

eminusx2

dx

0

1

2

0 1 2 3 4


infinint

0


dArrinfinint

1


dArrinfinint

1

eminusx2

dx konvergens

dArrinfinint

0

eminusx2

dx konvergens

bullinfinint

0

eminusx2

dx =

= lim0rarrinfin

infinint

0

eminusx2

dx = lim0rarrinfin

[]b0


Kideruumllinfinint

0

eminusx2

dx =radic

π

2

infinint

minusinfin

infinint

minusinfin


1

bullinfinint

0

sin xx

dx

bullinfinint

0

eminusx sinxdx

bullinfinint

0

sin(x2)dx

A gyoumlkoumlkradic


II


0

1xdx



cint

a

f(x)dx exist ekkor

bint

a

f(x)dx = limcrarrb

cint

a

f(x)dx


bint

a

f(x)dx = limcrarra

bint

c

f(x)dx

bint

c


pl

1int

0

1

xdx = lim

crarr0+

1int

c

1

xdx = lim

crarr0+[lnx]

1c =

limcrarr0+

( ln 1︸︷︷︸

0


1int

0

1radicx

dx = limcrarr0+

1int

c

xminus 12 dx = lim

crarr0+[]10 = lim

crarr0+

(

1 minusradic

012

)

= 2


2

Kalkulus II

Balog Daacuteniel

20100223



pl fn(x) = xn n = 1 2 3

1



fn(

12

)

12

122

123

124 konvergens

fn(13 )

13

132

133

134 konvergens











ffn


1

1n

1int

0

fn(x)dx = 1minus1

2n

1int

0

f(x)dx = 1

Tehaacutet

limnrarrinfin

1int

0

fn(x)dx = 1minus1

2n=

1int

0

f(x)dx = 1

1

1

2n

1n

1int

0

fn(x)dx =2

nmiddot 1 middot

1

2= 1foralln

1int

0

f(x)dx = 0



H

fn

fn minus ε

fn + ε



H






limnrarrinfin

bint

a

f(x)dx =

bint

a

f(x)dx

Maacuteskeacutepp

limnrarrinfin

bint

a

f(x)dx =

bint

a

lim fn(x)dx

lim eacutesint




sum





Def asum


KH(sum

fn) = x isin R

sum



f(x) =

infinsum

n=1

fn(x)

pl fn(x) = xn

s1(x) = x1 s2(x) = x1 + x2

sum


KH(sum

fn) = (minus1 1)sum


sum


2


sum

xn Cn = 1 a = 0

plsum

xn

nKH =

x =1

2sum 1

nmiddot1

2n


x = 2sum 2n

n


ndivergens


nez konvergens

3

Kalkulus II

Balog Daacuteniel

20100302





sumxn cn = 1 a = 0


sumxn

n cn = 1n a = 0


pl

sumxn




sumansor

konvergens ha lim n

radican lt 1

divergens ha lim n

radican gt 1

sum


n

radic


lim n

radic

|an| = |x|


pl


an = n middot xn

n

radic

|an| = n

radicn|x|

limnrarrinfin

n

radic



n

radicn = 1


ha

|x| lt 1 akkor


|x| ge 1 akkorsum




an =sum


n

radic

|an| = n



limnrarrinfin

n

radic


n



|xminus a| lt 1

lim n

radiccn

Def


R =1

lim n

radiccn



enneka a koumlzepe




Asum



ahol

R =1

lim n

radiccn



1

pl

sum

xn cn = 1 a = 0

R =1

lim n

radic1= 1



KH = (minus1 1)

sum xn

n2cn =

1

n2a = 0

R =1

lim n

radic1n2

= 1

segiacutetseacuteg

limn

radic

1

n2= lim

n

radic1

n

radicn2

= limn

radic1

( n

radicn)2

=1

12= 1


veacutegpontok

x = 1 rArrsum 1

n2konvergens


n2konvergens

KH = [minus1 1]

sum xn

ncn =

1

na = 0

R =1

lim n

radic1n

= 1


veacutegpontok

x = 1 rArrsum 1

ndivergens


nkonvergens

KH = [ minus 1 1 )

Def

Legyensum


f(x) =

infinsum

0


pl

sumxnf(x) =

x isin (minus1 1)

Nsum

0


xminus 1

infinsum

n=0

xn = limNrarrinfin

(

Nsum

0

xn) = limNrarrinfin


xminus 12 =

1

1minus x

infinsum

n=0

xn =1

1minus x

Oumltletsum


infinsum

n=0

xn =1

1minus x


infinsum

n=0


1minus x

2

middot x

infinsum

n=0

nxn =1

1minus x

n

pl

infinsum

n=0

n

2n=

1

2+

2

4+

3

8+

4

16+


12

(1minus 12 )

2=

1

2times 1

4= 2

Igaz-e hogy(sum

fn)prime =sum


Legyensum


f(x) =

infinsum

n=0




n=1


f primeprime(x) =

infinsum

n=2




2


Legyen

f(x) =

infinsum

n=0


f(a) = c0


f prime(a) = c1





Teacutetel

Legyen

f(x) =infinsum

n=0


21 Taylor sor

Eddig adott volt


cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n


fR rarr R

a isin R


cn =f (n)a

nforalln


sum

cn(xminus a)n

hatvaacutenysortDef


sum f (n)a

n(xminus a)n



f(x) =1




f (n)(a) = n

cn = 1


xn






6


cn =1

n

ex Taylor sorasum

xn


R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin



sum xn



infinsum

0

sum xn

n= ex

3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Kalkulus II

Balog Daacuteniel

20100223



pl fn(x) = xn n = 1 2 3

1



fn(

12

)

12

122

123

124 konvergens

fn(13 )

13

132

133

134 konvergens











ffn


1

1n

1int

0

fn(x)dx = 1minus1

2n

1int

0

f(x)dx = 1

Tehaacutet

limnrarrinfin

1int

0

fn(x)dx = 1minus1

2n=

1int

0

f(x)dx = 1

1

1

2n

1n

1int

0

fn(x)dx =2

nmiddot 1 middot

1

2= 1foralln

1int

0

f(x)dx = 0



H

fn

fn minus ε

fn + ε



H






limnrarrinfin

bint

a

f(x)dx =

bint

a

f(x)dx

Maacuteskeacutepp

limnrarrinfin

bint

a

f(x)dx =

bint

a

lim fn(x)dx

lim eacutesint




sum





Def asum


KH(sum

fn) = x isin R

sum



f(x) =

infinsum

n=1

fn(x)

pl fn(x) = xn

s1(x) = x1 s2(x) = x1 + x2

sum


KH(sum

fn) = (minus1 1)sum


sum


2


sum

xn Cn = 1 a = 0

plsum

xn

nKH =

x =1

2sum 1

nmiddot1

2n


x = 2sum 2n

n


ndivergens


nez konvergens

3

Kalkulus II

Balog Daacuteniel

20100302





sumxn cn = 1 a = 0


sumxn

n cn = 1n a = 0


pl

sumxn




sumansor

konvergens ha lim n

radican lt 1

divergens ha lim n

radican gt 1

sum


n

radic


lim n

radic

|an| = |x|


pl


an = n middot xn

n

radic

|an| = n

radicn|x|

limnrarrinfin

n

radic



n

radicn = 1


ha

|x| lt 1 akkor


|x| ge 1 akkorsum




an =sum


n

radic

|an| = n



limnrarrinfin

n

radic


n



|xminus a| lt 1

lim n

radiccn

Def


R =1

lim n

radiccn



enneka a koumlzepe




Asum



ahol

R =1

lim n

radiccn



1

pl

sum

xn cn = 1 a = 0

R =1

lim n

radic1= 1



KH = (minus1 1)

sum xn

n2cn =

1

n2a = 0

R =1

lim n

radic1n2

= 1

segiacutetseacuteg

limn

radic

1

n2= lim

n

radic1

n

radicn2

= limn

radic1

( n

radicn)2

=1

12= 1


veacutegpontok

x = 1 rArrsum 1

n2konvergens


n2konvergens

KH = [minus1 1]

sum xn

ncn =

1

na = 0

R =1

lim n

radic1n

= 1


veacutegpontok

x = 1 rArrsum 1

ndivergens


nkonvergens

KH = [ minus 1 1 )

Def

Legyensum


f(x) =

infinsum

0


pl

sumxnf(x) =

x isin (minus1 1)

Nsum

0


xminus 1

infinsum

n=0

xn = limNrarrinfin

(

Nsum

0

xn) = limNrarrinfin


xminus 12 =

1

1minus x

infinsum

n=0

xn =1

1minus x

Oumltletsum


infinsum

n=0

xn =1

1minus x


infinsum

n=0


1minus x

2

middot x

infinsum

n=0

nxn =1

1minus x

n

pl

infinsum

n=0

n

2n=

1

2+

2

4+

3

8+

4

16+


12

(1minus 12 )

2=

1

2times 1

4= 2

Igaz-e hogy(sum

fn)prime =sum


Legyensum


f(x) =

infinsum

n=0




n=1


f primeprime(x) =

infinsum

n=2




2


Legyen

f(x) =

infinsum

n=0


f(a) = c0


f prime(a) = c1





Teacutetel

Legyen

f(x) =infinsum

n=0


21 Taylor sor

Eddig adott volt


cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n


fR rarr R

a isin R


cn =f (n)a

nforalln


sum

cn(xminus a)n

hatvaacutenysortDef


sum f (n)a

n(xminus a)n



f(x) =1




f (n)(a) = n

cn = 1


xn






6


cn =1

n

ex Taylor sorasum

xn


R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin



sum xn



infinsum

0

sum xn

n= ex

3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

1

2n

1n

1int

0

fn(x)dx =2

nmiddot 1 middot

1

2= 1foralln

1int

0

f(x)dx = 0



H

fn

fn minus ε

fn + ε



H






limnrarrinfin

bint

a

f(x)dx =

bint

a

f(x)dx

Maacuteskeacutepp

limnrarrinfin

bint

a

f(x)dx =

bint

a

lim fn(x)dx

lim eacutesint




sum





Def asum


KH(sum

fn) = x isin R

sum



f(x) =

infinsum

n=1

fn(x)

pl fn(x) = xn

s1(x) = x1 s2(x) = x1 + x2

sum


KH(sum

fn) = (minus1 1)sum


sum


2


sum

xn Cn = 1 a = 0

plsum

xn

nKH =

x =1

2sum 1

nmiddot1

2n


x = 2sum 2n

n


ndivergens


nez konvergens

3

Kalkulus II

Balog Daacuteniel

20100302





sumxn cn = 1 a = 0


sumxn

n cn = 1n a = 0


pl

sumxn




sumansor

konvergens ha lim n

radican lt 1

divergens ha lim n

radican gt 1

sum


n

radic


lim n

radic

|an| = |x|


pl


an = n middot xn

n

radic

|an| = n

radicn|x|

limnrarrinfin

n

radic



n

radicn = 1


ha

|x| lt 1 akkor


|x| ge 1 akkorsum




an =sum


n

radic

|an| = n



limnrarrinfin

n

radic


n



|xminus a| lt 1

lim n

radiccn

Def


R =1

lim n

radiccn



enneka a koumlzepe




Asum



ahol

R =1

lim n

radiccn



1

pl

sum

xn cn = 1 a = 0

R =1

lim n

radic1= 1



KH = (minus1 1)

sum xn

n2cn =

1

n2a = 0

R =1

lim n

radic1n2

= 1

segiacutetseacuteg

limn

radic

1

n2= lim

n

radic1

n

radicn2

= limn

radic1

( n

radicn)2

=1

12= 1


veacutegpontok

x = 1 rArrsum 1

n2konvergens


n2konvergens

KH = [minus1 1]

sum xn

ncn =

1

na = 0

R =1

lim n

radic1n

= 1


veacutegpontok

x = 1 rArrsum 1

ndivergens


nkonvergens

KH = [ minus 1 1 )

Def

Legyensum


f(x) =

infinsum

0


pl

sumxnf(x) =

x isin (minus1 1)

Nsum

0


xminus 1

infinsum

n=0

xn = limNrarrinfin

(

Nsum

0

xn) = limNrarrinfin


xminus 12 =

1

1minus x

infinsum

n=0

xn =1

1minus x

Oumltletsum


infinsum

n=0

xn =1

1minus x


infinsum

n=0


1minus x

2

middot x

infinsum

n=0

nxn =1

1minus x

n

pl

infinsum

n=0

n

2n=

1

2+

2

4+

3

8+

4

16+


12

(1minus 12 )

2=

1

2times 1

4= 2

Igaz-e hogy(sum

fn)prime =sum


Legyensum


f(x) =

infinsum

n=0




n=1


f primeprime(x) =

infinsum

n=2




2


Legyen

f(x) =

infinsum

n=0


f(a) = c0


f prime(a) = c1





Teacutetel

Legyen

f(x) =infinsum

n=0


21 Taylor sor

Eddig adott volt


cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n


fR rarr R

a isin R


cn =f (n)a

nforalln


sum

cn(xminus a)n

hatvaacutenysortDef


sum f (n)a

n(xminus a)n



f(x) =1




f (n)(a) = n

cn = 1


xn






6


cn =1

n

ex Taylor sorasum

xn


R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin



sum xn



infinsum

0

sum xn

n= ex

3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3


sum

xn Cn = 1 a = 0

plsum

xn

nKH =

x =1

2sum 1

nmiddot1

2n


x = 2sum 2n

n


ndivergens


nez konvergens

3

Kalkulus II

Balog Daacuteniel

20100302





sumxn cn = 1 a = 0


sumxn

n cn = 1n a = 0


pl

sumxn




sumansor

konvergens ha lim n

radican lt 1

divergens ha lim n

radican gt 1

sum


n

radic


lim n

radic

|an| = |x|


pl


an = n middot xn

n

radic

|an| = n

radicn|x|

limnrarrinfin

n

radic



n

radicn = 1


ha

|x| lt 1 akkor


|x| ge 1 akkorsum




an =sum


n

radic

|an| = n



limnrarrinfin

n

radic


n



|xminus a| lt 1

lim n

radiccn

Def


R =1

lim n

radiccn



enneka a koumlzepe




Asum



ahol

R =1

lim n

radiccn



1

pl

sum

xn cn = 1 a = 0

R =1

lim n

radic1= 1



KH = (minus1 1)

sum xn

n2cn =

1

n2a = 0

R =1

lim n

radic1n2

= 1

segiacutetseacuteg

limn

radic

1

n2= lim

n

radic1

n

radicn2

= limn

radic1

( n

radicn)2

=1

12= 1


veacutegpontok

x = 1 rArrsum 1

n2konvergens


n2konvergens

KH = [minus1 1]

sum xn

ncn =

1

na = 0

R =1

lim n

radic1n

= 1


veacutegpontok

x = 1 rArrsum 1

ndivergens


nkonvergens

KH = [ minus 1 1 )

Def

Legyensum


f(x) =

infinsum

0


pl

sumxnf(x) =

x isin (minus1 1)

Nsum

0


xminus 1

infinsum

n=0

xn = limNrarrinfin

(

Nsum

0

xn) = limNrarrinfin


xminus 12 =

1

1minus x

infinsum

n=0

xn =1

1minus x

Oumltletsum


infinsum

n=0

xn =1

1minus x


infinsum

n=0


1minus x

2

middot x

infinsum

n=0

nxn =1

1minus x

n

pl

infinsum

n=0

n

2n=

1

2+

2

4+

3

8+

4

16+


12

(1minus 12 )

2=

1

2times 1

4= 2

Igaz-e hogy(sum

fn)prime =sum


Legyensum


f(x) =

infinsum

n=0




n=1


f primeprime(x) =

infinsum

n=2




2


Legyen

f(x) =

infinsum

n=0


f(a) = c0


f prime(a) = c1





Teacutetel

Legyen

f(x) =infinsum

n=0


21 Taylor sor

Eddig adott volt


cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n


fR rarr R

a isin R


cn =f (n)a

nforalln


sum

cn(xminus a)n

hatvaacutenysortDef


sum f (n)a

n(xminus a)n



f(x) =1




f (n)(a) = n

cn = 1


xn






6


cn =1

n

ex Taylor sorasum

xn


R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin



sum xn



infinsum

0

sum xn

n= ex

3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Kalkulus II

Balog Daacuteniel

20100302





sumxn cn = 1 a = 0


sumxn

n cn = 1n a = 0


pl

sumxn




sumansor

konvergens ha lim n

radican lt 1

divergens ha lim n

radican gt 1

sum


n

radic


lim n

radic

|an| = |x|


pl


an = n middot xn

n

radic

|an| = n

radicn|x|

limnrarrinfin

n

radic



n

radicn = 1


ha

|x| lt 1 akkor


|x| ge 1 akkorsum




an =sum


n

radic

|an| = n



limnrarrinfin

n

radic


n



|xminus a| lt 1

lim n

radiccn

Def


R =1

lim n

radiccn



enneka a koumlzepe




Asum



ahol

R =1

lim n

radiccn



1

pl

sum

xn cn = 1 a = 0

R =1

lim n

radic1= 1



KH = (minus1 1)

sum xn

n2cn =

1

n2a = 0

R =1

lim n

radic1n2

= 1

segiacutetseacuteg

limn

radic

1

n2= lim

n

radic1

n

radicn2

= limn

radic1

( n

radicn)2

=1

12= 1


veacutegpontok

x = 1 rArrsum 1

n2konvergens


n2konvergens

KH = [minus1 1]

sum xn

ncn =

1

na = 0

R =1

lim n

radic1n

= 1


veacutegpontok

x = 1 rArrsum 1

ndivergens


nkonvergens

KH = [ minus 1 1 )

Def

Legyensum


f(x) =

infinsum

0


pl

sumxnf(x) =

x isin (minus1 1)

Nsum

0


xminus 1

infinsum

n=0

xn = limNrarrinfin

(

Nsum

0

xn) = limNrarrinfin


xminus 12 =

1

1minus x

infinsum

n=0

xn =1

1minus x

Oumltletsum


infinsum

n=0

xn =1

1minus x


infinsum

n=0


1minus x

2

middot x

infinsum

n=0

nxn =1

1minus x

n

pl

infinsum

n=0

n

2n=

1

2+

2

4+

3

8+

4

16+


12

(1minus 12 )

2=

1

2times 1

4= 2

Igaz-e hogy(sum

fn)prime =sum


Legyensum


f(x) =

infinsum

n=0




n=1


f primeprime(x) =

infinsum

n=2




2


Legyen

f(x) =

infinsum

n=0


f(a) = c0


f prime(a) = c1





Teacutetel

Legyen

f(x) =infinsum

n=0


21 Taylor sor

Eddig adott volt


cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n


fR rarr R

a isin R


cn =f (n)a

nforalln


sum

cn(xminus a)n

hatvaacutenysortDef


sum f (n)a

n(xminus a)n



f(x) =1




f (n)(a) = n

cn = 1


xn






6


cn =1

n

ex Taylor sorasum

xn


R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin



sum xn



infinsum

0

sum xn

n= ex

3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

pl

sum

xn cn = 1 a = 0

R =1

lim n

radic1= 1



KH = (minus1 1)

sum xn

n2cn =

1

n2a = 0

R =1

lim n

radic1n2

= 1

segiacutetseacuteg

limn

radic

1

n2= lim

n

radic1

n

radicn2

= limn

radic1

( n

radicn)2

=1

12= 1


veacutegpontok

x = 1 rArrsum 1

n2konvergens


n2konvergens

KH = [minus1 1]

sum xn

ncn =

1

na = 0

R =1

lim n

radic1n

= 1


veacutegpontok

x = 1 rArrsum 1

ndivergens


nkonvergens

KH = [ minus 1 1 )

Def

Legyensum


f(x) =

infinsum

0


pl

sumxnf(x) =

x isin (minus1 1)

Nsum

0


xminus 1

infinsum

n=0

xn = limNrarrinfin

(

Nsum

0

xn) = limNrarrinfin


xminus 12 =

1

1minus x

infinsum

n=0

xn =1

1minus x

Oumltletsum


infinsum

n=0

xn =1

1minus x


infinsum

n=0


1minus x

2

middot x

infinsum

n=0

nxn =1

1minus x

n

pl

infinsum

n=0

n

2n=

1

2+

2

4+

3

8+

4

16+


12

(1minus 12 )

2=

1

2times 1

4= 2

Igaz-e hogy(sum

fn)prime =sum


Legyensum


f(x) =

infinsum

n=0




n=1


f primeprime(x) =

infinsum

n=2




2


Legyen

f(x) =

infinsum

n=0


f(a) = c0


f prime(a) = c1





Teacutetel

Legyen

f(x) =infinsum

n=0


21 Taylor sor

Eddig adott volt


cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n


fR rarr R

a isin R


cn =f (n)a

nforalln


sum

cn(xminus a)n

hatvaacutenysortDef


sum f (n)a

n(xminus a)n



f(x) =1




f (n)(a) = n

cn = 1


xn






6


cn =1

n

ex Taylor sorasum

xn


R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin



sum xn



infinsum

0

sum xn

n= ex

3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3


Legyen

f(x) =

infinsum

n=0


f(a) = c0


f prime(a) = c1





Teacutetel

Legyen

f(x) =infinsum

n=0


21 Taylor sor

Eddig adott volt


cn(xminus a)n

Ebbol legyen

f(x) =

infinsum

n=0

cn(xminus a)n


fR rarr R

a isin R


cn =f (n)a

nforalln


sum

cn(xminus a)n

hatvaacutenysortDef


sum f (n)a

n(xminus a)n



f(x) =1




f (n)(a) = n

cn = 1


xn






6


cn =1

n

ex Taylor sorasum

xn


R =1

lim n

radic1n

= infin

Segiacutetseacutegn

radicn = infin



sum xn



infinsum

0

sum xn

n= ex

3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3





0


(2n+ 1)

f (n) f (n)(a)f(x) = cos(x) c0 = 1



0


(2n)


4

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Kalkulus II

Balog Daacuteniel

20100309




Cn(xminus a)n



f(x) =

infinsum

n=0


n


n(xminus a)n


Trigonometrikus sor


a0

2+sum



3 Legyen

a0

2+sum





minusπ



bullπint

minusπ


πint

minusπ

cos(nx) =

2π hak = 00 hak 6= 0

bull



cos(α) cos(β) = 1








sin(α) cos(β) = 1


bullπint

minusπ

cos( nx︸︷︷︸

α


β

) =

1

2

πint

minusπ


1

2middot

[minus cos(k + n)x

n+ k+


k minus n

]π

minusπ

= 0


πint

minusπ

cos( nx︸︷︷︸

α

) cos( kx︸︷︷︸

β

) =

1

2

πint

minusπ


=


πint

minusπ

sin(nx) sin(kx) =


1

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

bull

f(x) =a0

2+

infinsum

n=1


middot sin(kx)

πint

minusπ

πint

minusπ


2

πint

minusπ


=0

+

infinsum

n=1

an middot

πint

minusπ

cos(nx) sin(kx)dx+

bn middot

πint

minusπ

sin(nx) sin(kx)dx

=


= bk middot π =1

π

πint

minusπ

f(x) sin(kx)dx

πint

minusπ


ak =1

πmiddot

πint

minusπ


k isin N

TEacuteTEL




πmiddot

πint

minusπ


1

π

πint

minusπ

f(x) sin(kx)dx


a0

2+

infinsum

n=1



f(x) =a0

2+

infinsum

n=1




5 Peacutelda

-1

1 πminusπ

bk =1

π

πint

minusπ

f(x) sin(kx)dx =

minus1

π

0int

minusπ

f(x) sin(kx)dx +1

π

πint

0

f(x) sin(kx)dx =

minus1

π

[minus cos(kx)

k

]0

π

+1

π

[minus cos(kx)

k

]π

0

=

1


1


1

kπ

(1minus (1)k

)minus

1

kπ

((1)k minus 1

)=

2π

kminus

2π

k(1)k =

0 ha k paacuteros4



2

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Kalkulus II

Balog Daacuteniel

20100316


DefR


n db



x+ y = (x1 + y1 x2 + y2 xn + yn)



|x| =radic

(x21 x

22 x

2n) P = 2

||x|| =infinsum

i=1

|xi| P = 1

||x||p = p

radicinfinsum

i=1

|xi|p P ge 1


P = 2b

s =y isin R2 ||y||2 = r

P = 1b

s =y isin R2 ||y||1 = r

P = infinb

|x|+ |y| = r





2x1 + x3) n = 3 k = 2


fi Rn rarr R

1 f(x) = (f1(x) f2(x) f3(x) fn(x))


22 f2(x1 x2 x3) = x1 + x3




x

f(x)



fuumlgguumllegesR

y

x

z

y0x0

Pl x2 + y2

(x)

(y)


1

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

1radic2

2



b

c


bull Rrarr R2


f1

f2

f1(x)

f2(x)

x


bull Rrarr R3


bull R2

︸︷︷︸

C

rarr R2

︸︷︷︸

C


bull R3 rarr R3






(x21 + x2

2 + x23)

3

2

x1

f1 R3 rarr R


Legyenf Rn rarr Rk

Def lim f(x) = A

Ez mit jelent


A

δ

x

f(x)

Rk

Rn







h



f(x+ h)minus f(x)

h




h


partif(x) =partf

partxi

Peacuteldaacutek


part1f(x1 x2) = x2︸︷︷︸

(cmiddotx)prime=c

part2f(x1 x2) = x1

bull f(x1 x2) = x1 + x2

part1f(x1 x2) = 1︸︷︷︸

(c+x)prime=1

part2f(x1 x2) = 1

2


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3


part1f(x1 x2) = ex2



2 + 2x1 + x2



2 + 1


x2

1+x2

2+x2

3

= (x21 + x2

2 + x23)

minus1

2


2middot (x2

1 + x22 + x2

3)minus

3

2 middot 2x1 =

= minusx1 middot1

(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


(x21 + x2

2 + x23)

3

2


|x|3


Furcsa peacutelda

f(x1 x2) =




hrarr0

1minus0h


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Kalkulus II

Balog Daacuteniel

20100323


f Rn rarr R



h




f(x+ h)minus f(x)

h

Mostf Rh rarr R


0 = limhrarr0


h


0 = limhrarr0


|h|


f prime(x) = (( ) ( ) ( )︸︷︷︸

nminusdarab

)


sormaacutetrix hogy

limhrarr0


|h|= 0



A middot h =(A1 An

)middot

t

0

= A1t

limtrarr0


|t|=

limtrarr0

∣∣∣∣∣∣∣∣


t︸︷︷︸

part1f(x)

minusA1

∣∣∣∣∣∣∣∣








f Rrarr R

x0


l(x)minus l(x0)



1


f Rn rarr R

1f(x0 = l(x0)

1


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3


b

(x)(y)



(xminus x0

y minus y0

)





Legyenz0 = f(x0 y0)


~n


rminusr0

plf(x y)x2 + y2

(x)

(y)



part1f(1 2) = 2x = 2

part2f(1 2) = 2x = 4

~n = (2 4minus1)



2x+ 4y minus z = 5





limhrarr0


|h|= 0



f prime(x) =




pl

f(x1 x2) =

(x1 middot x2

sin(x1 + 3x2)

)

f R2 rarr R2


f2(x1 x2) = sin(x1 + 3x2)

f prime(x) =

(x2 x1

cos(x1 + 3x2) 3 cos(x1 + 3x2)

)

f(x) =

cosxsinxx

f R1 rarr R3

f prime(x) =

minus sinxcosx1





lim|α(h)|

|h|= 0


x = x0

h = xminus x0




2ez kicsi

2


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3


l Rn rarr Rk


n = 1(t)k = 3(x y z)

(x) (y)

b

pl

cos tsin tt



Konkreacutetan

f prime(t)

minus sin tcos t1


x

y

z

=

minus10π)

+

0minus11

middot (tminus t0)


z =π + tminusπ

t isin R





Ekkor


pl

f(x y) = x3 + y4

R2 rarr R


f primeprime(x y) =

(6x 00 12y2

)

3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3


Balog Daacuteniel

20100330




Def


a


a



Teacutetel





a


xn = an

a1


1

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Pl

f(x1 x2) = x2

1+ x2

2




Pl2


1 minus x2

2

1


minus2x1 = 0 minus2x2 = 0x1 = 0 x2 = 0



Pl3

f(x1 x2) = x2

1minus x2

2

nyereg feluumllet


2x1 = 0 minus2x2 = 0x1 = 0 minusx2 = 0





f(x1 x2) = x21 + x2

2 f R2 rarr R



(

2 00 2

)




1 minus x2

2



(

minus2 00 minus2

)

part1part1f

part2part1f

3 nyereg

f(x1 x2) = x2

1 minus x2

2


2 00 minus2

)


2

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Teacutetel






Megjegyzeacutes


Pl

f(x1 x2) = x31 + x3

2 minus 3x1x2

Szeacute=


22 minus 3x1)


2 minus 3x1 = 0x2

1= x2 x2

2= x1

x4

1= x1 rArr x1(x

3

1minus 1) = 0

x1 = 1 x2 = 1

x1 = 0 x2 = 0

1

1




(


)


f primeprime(0 0) =

(

0 minus3minus3 0

)


(0 0) nem sz eacute

f primeprime(1 1) =

(

6 minus3minus3 6

)



1+ x2

2+ 4x1x2 sz eacute=


Emleacutek



Def




n

Megjegyzeacutes


PL


F =partFpartx



F (x y) = x2 + xy + C


F (x y) = x2 + xy + C(y)

3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

partF



Cprime(y) = 2y

C(y) =

int

2y dy = y2 +K

F (x y) = x2 + xy + C(y) = x2 + xy + y2 +K


Megoldaacutes

F (x y) = x2 + xy + y2 +K


f(x y) = (minusy x)









part1F = f1


part2F = f2



Teacutetel(Young)


part2f

partxparty=

part2f

partypartx




partfi

partdj=

partfj

partdi

Pl


Megjegyzeacutes

R3 rarr R


f prime =



4

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Kalkulus II

Balog Daacuteniel

20100413




1

2

1

r(t) = (2t 4t)t isin [0 1

2] ugyanaz a rajz



1

minus1

1minus1



b


Iacutevhossz


a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)


σ(r t1 t2 tn) =

nsum

i=1



A goumlrbe hossza



radic5


t)t isin (0 1]

itt a sup = infin


l(r) =b

infa|rprime(t)|dt



1

minus1

1minus1

1


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3


radic


l(r) =2πint

0

1dt = 2π



koumlriacutev


1

minus1

1minus1


Vonalintegraacutel


n f Rn rarr Rn

a t1 t2 t3 t4 t5 br(a) r(t1)

r(t2) r(b)



1 tlowast

n) ==

nsum

i=1






r

f = I


Ekkorexistint

r

=bint

a



f(x y) = x y


int

r

f =

2πint

0


= f

2πint

0



f(x y) =(

minusyx2+y2

xx2+y2

)


int

r

f =

2πint

0

lang((minus sin t

1cos t

1

)

(minus sin t cos t)

)rang

dt =

=

2πint

0

sin2 t+ cos2 tdt =

2πint

0

1dt = 2π






Leacutenyeg


int

r




int

r



r


a


a


a




2

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Kalkulus II

Balog Daacuteniel

20100420



f Rrarr R

a ξ b



σ =

nsum

i=1




a

= I

A ceacutel


Legyena lt b c lt d

b

c

d

a

d

ca b





σ =

msum

j=1

nsum

i=1




int

T

= I itt



D

hogyan

eacutertelmezheto

TD


f(x) =



int

D

f =int

T

f




int

[ab]times[cd]

f =

1

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

=

bint

a

dint

c

f(x y)dy

dx =

dint

c

bint

a

f(x y)dx

dy


[ab]times[cd]

f helyettdint

c

bint

a



Legyeng(x)dint

c

f(x y)dy


c



T

f =


0

3int

2

xydy

dx =

=

1int

0

[

x middoty2

2

]3

2

=

=

1int

0

(9

2xminus 2x)dx =

1int

0

(5

2x) =

5

2

[

x2

2

]1

0

=5

4


int

T

=

2int

1

2int

1

2xy

(x2 + y2)2dy

dx =

=

2int

1

[

minusx

x2 + y2

]2

1

=

2int

1

(

x

x2 + 1minus

x

x2 + 4

)

dx =

=

[

1

2ln(x2 + 1)minus

1

2ln(x2 + 4)

]2

1

=

=1

2

[

lnx2 + 1

x2 + 4

]2

1

=1

2ln

5825

=1

2ln

25

16= ln

5

4




d

ca

β(x)

α(x)

b

N


int

N

f -et megadni


Legyenf(x) =


int

N

f =

int

f[ab]times[cd] =

bint

a

dint

c

fdy

dx =

=

bint

a

dint

c

f(x y)dy

dx


a

(

dint

c

f(x y)dy

)

dx


1

minus1

1minus1


N =

int

N

=

1int

minus1

x2

int

x2minus1


dx =

=

1int

minus1

[

x middoty2

2

]1minusx2

x2minus1

dx =

x(1 minus x2)2

2minus x

(x2 minus 1)2

2=

1int

minus1

0

2

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Kalkulus II

Balog Daacuteniel

20100427




mertint

T

f =b1int

a1

b2int

a2




g(β) = bbint

a

f(x)dx =

βint

α



bint

a

f(x)dx =

βint

α



T



int

T

f =

b1int

a1

b2int

a2

f(x y)dy

dx


int

T

f =

b1int

a1

β2int

α2


dx

y helyettg1(t1)

int

T

f =

β1int

α1

β2int

α2

f(g1(t1) g2(t2))︸︷︷︸

f(g(t))

|gprime2(t2)|dt2

|gprime1(t1)|dt1

Mi is a gprime(t) =

(gprime1(t1) 0

0 gprime2(t2)

)


R



bull g(t1 t2) = (g1(t1 t2) g2(t1 t2))



Teacutetel


int

D

f =int

R



Peacutelda

0 r0

2π



r isin [0 2]ϕ = [0 2π]1


1


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3


g(r ϕ) =


)


D

f =

int

[02]times[02π)

(r2 cos2 ϕ+ r2 sin2 ϕ)︸︷︷︸

fg

middotrdrdϕ =

2int

0

2πint

0

r3dϕdr =

2int

0

2πr3dr = 2π[r4

4]20 = 2π middot 4 = 8π


minus30

60


(x2+y2))

int

D

f =

R = [1 2]︸︷︷︸

r

times [minusπ

6+

π

3]

︸︷︷︸

ϕ


D

f =

2int

1

+π

3int

minusπ

6

r2 cosϕ sinϕradic


=

2int

1

+π

3int

minusπ

6

= r21

2sin(2ϕ)dϕdr =

=

2int

1

r2

2

[

minuscos(2ϕ)

2

]+π

3

minusπ

6

dr =

=

2int

1

r2

4(1

2minus (minus1

2))dr =

1

4[r3

3]21 =

7

12


x2+y2


R = [0 1]︸︷︷︸

r

times [0 2π]︸︷︷︸

ϕ

times [0 2]︸︷︷︸

z

gprime(x y z)


det gprime = r

int

D

f =

1int

0

2πint

0

2int

0

zradic


int

D

f =

1int

0

2πint

0

2int

0

z

dzdϕdr = [

z2

2]20 =

1int

0

2πint

0

2dϕdr =

1int

0

4πdr = 4π



int

D

1



gprime(r ϕ ϑ) =




Vg =

int

D

1 =

Rint

0

πint

0

2πint

0


=

Rint

0

πint

0


Rint

0

πint

0

r2dϕdr =

2

Rint

0

2πr2dr = 4π[r3

r]R0 = 4π

R3

3

2

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

Kalkulus II

Balog Daacuteniel

20100504

Feluumlletek



a

b

r(t) r(t+ 1)

A goumlrbe hossza

ab

r(t) r(t+ 1)


nsum

i=1




bint

a

|r(t)|dt




Ω = [0 1]︸︷︷︸

u

times [0 1]︸︷︷︸

v

01

1

0 1

1Φ


ϑ isin [0 π]

ϕ isin [0 2π]1

u = ϑ

v = ϕ

Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu



bb

b








sumsum


10 = 2π

1



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3



S =

int

Ω



Ω = [0 π]times [0 2π]

Φ(u v) =


R middot cosu

partuΦ =


partv =


0


∣∣∣∣∣∣

i j k


∣∣∣∣∣∣



R2 sinu cosu

Ennek a hossza


sin2 u

+cos2 u

︸︷︷︸

1

)1

2 = R2 sinu

s =

πint

0

2πint

0

R2 sinudvdu = 2πR2

πint

0

sinudu = 4R2π




int

Φ

U


int

Φ

U =

int

Ω


plΩ = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u


Φ

U =


partuΦ ==

111

partvΦ =

1minus10


∣∣∣∣∣∣

i j k

1 1 11 minus1 0

∣∣∣∣∣∣

=

11minus2


radic6

int

Phi

U =

1int

0

1int

0


U(Φ(uv))

middotradic6dudv

= 3 middotradic

(6)

1int

0

1int

0

udu

dv =3radic6

2




int

Φ


int

Φ

F =

int

Ω


Ω = [0 1]times [0 1]

Φ(u v) =

u+ v

uminus v

u

F =

y

x

z

int

Φ


11minus2

int

Ω

F =

1int

0

1int

0

lang

u+ v

uminus v

u

11minus2

rang

dudv

1int

0

1int

0

((((((((((


0

dudv

2

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

nabla div grad rot

nabla

nabla =

partxpartypartz

Haf R3 rarr R

nablaf =

partxf

partyf

partzf

= gradf


3



f prime

jakobi =



- ∣∣∣∣∣∣

i k k


∣∣∣∣∣∣

=


= rot f



2f + part23f


3

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Readme - Fizwebfizweb.elte.hu/download/Fizika-BSc/Kalkulus-2/Kalkulus2...Kalkulus II Balog Dániel 2010.02.16 Inpropius integrál Motiváló példák: 1. R1 0 1 x dx A rendes Riemann-integrál