Bai Giang Toan 1 Hai

8/3/2019 Bai Giang Toan 1 Hai

1/98

Bi ging Ton 1 Ths L Th Minh Hi

Bi s 1Gii hn v tnh lin tc ca hm s

1.1. Hm s mt bin s1. nh ngha hm s

Cho 2 tp hp D v E l cc tp con ca R. Tng ng f D E: cho tng ng

mi phn t x D vi duy nht mt phn t y E c gi l hm s mt bin sthc.+ Tp D c gi l min xc ca f.+ Tp f(X) c gi l min gi tr ca f.+ x D c gi l bin sc lp ( hay i s ).+ f x x D ( ), c gi l bin s ph thuc ( hay hm s ).

2. th ca hm s: ( ){ }fG x f x x A= , ( ) | + Cch nhn bit th theo phng php kim tra ng thng ng: Mt ngcong trong mt phng xyl th ca mt hm ca xnu v chnu ng thngsong song vi Oy ct ng cong ti nhiu nht mt im.

th hm s Khng l th hm s

1.2 Gii hn hm s:1. V d 1: Xt hm s y f x x x = = +2( ) 2 . Ta lp bng cc gi tr ca hm s tinhng im x gn x =0 2.


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Nhn thy khi x tin gn n x =0 2 th cc gi tr cc hm s f x( ) tin gn n 4.Ta ni rng hm s c gii hn bng 4 khi x x =0 2 .2. nh ngha gii hn hm snh ngha 1: Ta ni hm s f x( ) c gii hn L (hu hn) khi x x 0 v vit

x xf x L

=

0

lim ( ) nu vi bt k dy { }nx m nx x 0 th nn

f x L

=lim ( ) .

nh ngha 2: theo ngn ng .

x xf x L x x f x L

= > > < 0 , chn = . Ta c: x lim ( ) 0 , N > 0 ln, sao cho x N f x L > lim ( ) 0 , N > 0 ln, sao cho x N f x L < 2

1 10 .

+ Ta c: > 0 , chn N

= 21

. Khi x N f x >


4/98


V d 5: Chng minh rngx

x

x=

sinlim 0 .

Ta c:x

x x

sin 10 . M

x x=

1lim 0 nn

x

x

x=

sinlim 0 , hay ta c pcm.

5. Mt s phng php kh dng v nh: 0 , , , 1 .0

+ Phn tch a thc thnh nhn t hoc nhn biu thc lin hp kh dng vnh.

+ S dng gii hn kp

+ S dng mt s gii hn c bn sau:

x

x

x=

0

sinlim 1,

x

x

aa

x

=0

1lim ln ,

x

x

x+

=0

ln( 1)lim 1,

x

a

x

ae

x

+ = lim 1 ,

( )

x

x

a a+

= <


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V d 8: Tmx

x x

x++

+lim

1

Gii: Dng

.

+x

x x

x++ =

+lim

1

+ KQ: 1.

V d 9: Tm ( )x

x x x+

+ 2lim

+ Dng

+ ( )x

x x x+

+ =2lim .

+ KQ: .

V d 10: Tmx x

x

x

x

+

+

+

2 22

2

1lim

1,

+ Dng 1

+

( )

( )x

x x

x x x xx x

x

x x

xe e

x x+

+ ++

+ +

+ = + = =

2

2 2 22

2

2 2

1 1 2 222 lim221

2 2

1 2lim lim 1

1 1.

V d 11: Tm gii hn saux

x x

x

0

1 cos .cos2lim

1 cos.

+ Dng00

.

+( )

x x

x x xx x

x x

+ = =

0 01 cos cos2 1 cos21 cos .cos2

lim lim1 cos 1 cos

=

+ KQ: 5.V d 12: Tm gii hn sau ( )x

xx

2

1

0lim cos .

+ Dng 1

+ Ta c: ( )x

x x= = 2cos 1 1 cos 1 2sin2


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+ ( )xx

x

=21

0lim cos

+ KQ:e

12

6. Gii hn mt phaa. nh ngha: Gii hn ca f(x) khi x a x a , ) nu tn ti

gi l gii hn tri ( hoc gii hn phi ). K hiu x a x a f x f a f x f a + +

= =lim ( ) ( ), lim ( ) ( ) .K hiu khc:

x a x a f x f a f x f a

+= = +

0 0lim ( ) ( 0), lim ( ) ( 0) .

b. nh l: Tn tix a

f x L

=lim ( ) khi v chkhix a

x a

x a x a

f x

f x

f x f x L

+

+

= =

lim ( )

lim ( )

lim ( ) lim ( )

V d 13: Xt s tn ti cax

x

x0lim .

Ta c:x x

x x

x x+ + = =

0 0

lim lim 1,x x

x x

x x

= =

0 0

lim lim 1. Vyx

x

x0lim khng tn ti.

V d 14: Nux x

f xx x

>= , chng ta c:

( )4 4

lim lim 4 4 4 0x x

f x x+ +

= = =

V ( ) 8 2 f x x= vi 4x < , chng ta c :

( ) ( )4 4

lim lim 8 2 8 2.4 0x x

f x x

= = =

Gii hn tri v gii hn phi bng nhau. V vy, gii hn tn ti v ( )4

lim 0x

f x

=.

th ca fc chra trong Hnh 3.


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HNH 3

7. V cng ln, v cng bnh ngha: Hm s f(x) c gi l v cng b, vit tt l VCB khi x x 0 nu

x xf x

=

0

lim ( ) 0 . Hm s f(x) c gi l v cng ln, vit tt l VCL khi x x 0 nu

x xf x

= +

0

lim ( ) .

Ch :+ x0 c th hu hn hoc v hn.

+x x x x

f xf x

= =0 0

1lim ( ) lim 0( )

. xf x x= +1

( ) (1 )

+ so snh tc dn n 0 ca cc VCB f(x), g(x) khi cng x x 0 th xt

x x

f x

g x 0

( )lim

( ). Ta c cc trng hp sau:

Nux x

f x

g x=

0

( )lim 0

( )ta ni rng f(x) bc cao hn g(x), k hiu

f x o g x x x = 0( ) ( ( )), .

Nux x

f xC

g x=

0

( )lim 0

( )

ta ni rng f(x) cng bc vi g(x).

Nux x

f x

g x=

0

( )lim 1

( )ta ni rng f(x) tng ng vi g(x), k hiu f x g x ( ) ( ) .

Mt s VCB cng bc khi x 0 : xx x x x e x + sin , ln(1 ) , 1 .

ln(1+x) x khi x 0 vx

x

x+

=0

ln(1 )lim 1

nh l: Nu f(x) f*(x), g(x) g*(x) khi x x 0 . Khi :x x x x

f x f x

g x g x =

0 0

*

*

( ) ( )lim lim

( ) ( ).

V d 15: Tnhx

x

e

x

+

2

0

1lim

ln(1 sin3 ).

Ta c: xe 2 1 2x khi x 0; ln(1+sin3x) sin3x 3x khi x 0

Do :x

x x

e x

x x

= =+

2

0 0

1 2 2lim lim

ln(1 sin3 ) 3 3.


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1.3. Tnh lin tc ca hm s1. nh nghanh ngha 1: Hm s f(x) lin tc ti im x0 nu

x xf x f x

=

00lim ( ) ( ) .

Hm s y = f(x) lin tc trn min D nu n lin tc ti mi im thuc min D.Ch : Tnh ngha 1, ta thy y = f(x) lin tc ti im x0 cn n 3 iu kin:

1. x0 thuc tp xc nh ca hm s.2. Tn ti

x xf x

0lim ( ) .

3.x x

f x f x

=0

0lim ( ) ( )

Nhn xt:+ Cc a thc, hm phn thc, hm hu t, hm lng gic, hm m, hm logarit lcc hm s lin tc trn min xc nh ca n.+ Hm s y = f(x) lin tc trn (a, b) th th ca n l mt ng cong trn trnkhong ny (tc l khng b gy, khng bt on).

nh ngha 2: Hm sf(x) c gi l lin tc phi ti x0 nu ( ) ( )x x

f x f x +

=0

0lim .

Hm sf(x) c gi l lin tc tri ti x0 nu ( ) ( )x x

f x f x

=0

0lim .Hm s y = f(x) lin tc ti x0 khi v chkhi n va lin tc tri, va lin tc phi ti

x0 .

V d 16: Xt tnh lin tc ca hm s ( )x x

xf x x

x

= =

2 22

21 2

+ Ta thy hm s lin tc ti mi im x 2 .

+ Xt ti x = 2.

( )( )( )

( )x x x x

x xx xf x x f

x x

+ = = = + = =

2

2 2 2 2

2 12lim lim lim lim 1 3, (2) 1

2 2

Nhng ( ) ( )x

f x f

2

lim 2 . Nn fkhng lin tc ti 2.

V d 17: Tm a hm s sau lin tc trn R

ax

xx

f x x

ae x x

>= +

2

sin20

( )

1 0

+ Hm s lin tc vi mi x 0 , hm s lin tc trn R th n phi lin tc tix= 0 .

+ Ti x= 0

( )x x

xf x

x+ + = =

0 0

sin2lim lim 2

,( ) ( )ax

x xf x ae x a f

= + = =2

0 0lim lim 1 1 (0)


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hm s lin tc ti x = 0 th f f f a a + = = = =(0 ) (0 ) (0) 1 2 3 .

V d 18: Hm s f(x) khng xc nh ti x = 0, hy xc nh f(0) hm s f(x) lintc ti x = 0 vi :

( )xf x x= +1

( ) 1 2

Gii: hm s lin tc ti x = 0 th x

x xf f x x e

= = + =

12

0 0(0) lim ( ) lim(1 2 ) .

2. im gin on ca hm snh ngha: Hm s f(x) c gi l gin on ti x = a nu ti x = a hm s

khng lin tc.Nu tn ti f a f a + ( ), ( ) v f a f a + ( ) ( ) th x = a c gi l im gin on loi

1.Nu f a f a + =( ) ( ) th x = a c gi l im gin on khc.im gin on khc (khng phi loi 1) gi l gin on loi 2.

V d 19: Tm v phn loi im gin on ca cc hm s sau:

a.x

f xx

=( ) b.x

x

f x

e=

1

1( )

1

Gii: a. Xt ti x = 0( )

xf x

+=

0lim

( )

xf x

=

0lim

nn x = 0 l gin on loi 1.b. Ti x = 1.

( ) ( )x x

f x f x +

= =1 1

lim lim

nn x = 1 l gin on loi 1.Ti x = 0.

( ) ( )x x

f x f x +

= =0 0

lim lim

nn x = 0 l gin on loi 2.V d 20: Kho st s lin tc ca hm s v tnh cht im gin on

xx

f x

x x

= >

cos 12( )

1 1

(S: x = - 1 l im gin on loi 1)

Bi tp v nh:Trang 87 ( Bi 1 19), trang 91 ( bi 18 62), 25 trang 251, Trang278 ( Bi 33 - 43).


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Bi s 2

o hm ca hm s mt bin

2.1 nh ngha vo hm

1. nh ngha o hmCho hm s y f x= ( ) , o hm f x'( ) ca hm s f x( ) l mt hm mi c gi tr tiim x c xc nh bi gii hn sau (khi gii hn tn ti):

x

f x x f x f x

x +

=0

( ) ( )'( ) lim .

+ Nu gii hn tn ti vi x = a, th hm sy = f(x) c gi l kh vi ti a.+ Hm kh vil hm s kh vi ti mi im trong tp xc nh ca n.

y = f(x)

P

x

x + x0 x0 x

y

Q

f(x + x) - f(x )0 0

Ch :+ f(x) l dc ca tip tuyn ca ng cong y = f(x) ti P.+ C nhiu cch k hiu khc nhau ca o hm hm s y f x= ( ) :

f x'( ) , y ,dy

dx,df x

dx

( ),

df xdx ( ) .

+ Nu y f x= ( ) thdy

dxcn c gi l sut bin i ca y theo x.

+ Nu ta mun vit gi tr s ca o hm ti mt im c th x = 3, ta vit :

x

dy

dx =

3hoc

x

dy

dx =3, hoc f(3) .

+ x x x = 0 nnx x x

f x f x f x x f x f x

x x x

+ = =

00

00

( ) ( )( ) ( )'( ) lim lim .

2. Quy tc tm o hm ti mt im theo nh ngha:Bc 1. Tm s gia f(x +x) - f(x) v tin hnh rt gn

Bc 2. Thit lp t s:f x x f x

x

+

0 0( ) ( )

Bc 3. Tnh gii hn ca t s trn khi x0. Nu gii hn tn ti th chnh l o hm ca hm s ti im cn tm :


11/98


x

f x x f x f x

x +

=0

( ) ( )'( ) lim .

V d 1. Tm f(x) nu f(x) =x

1

Bc 1:x x x x

f x x f x x x x x x x x x x

+ + = = =+ + + 1 1 ( )( ) ( )

( ) ( )

Bc 2.

f x x f x

x x x x

+ =

+ 0 0( ) ( ) 1

( )

Bc 3. Kt lunx

f xx x x x

= =

+ 201 1

'( ) lim( )

V d 2 : V th hm s y f x x x = =( ) v cho bit n khng kh vi ti im no

Gii : x xy f x x x x x

= = =

2

2, 0( )

, 0nn x xf x

x x >=


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+ Mt hm kh vi ti mtim th lin tc tiimv:

x x x x

y y dy y x x

x x dx

= = = = 0 0 0 0lim lim lim lim 0 0 .

+Mt hm c th lin tc ti mt im m khng kh vi ti im .+ Mt hm skhng lin tcti x0 th skhng kh vi ti im .

2. MT S CNG THC V QUY TC TNH O HM HC

1.d

cdx

= 0 2.d du

u udx dx

= 1

3. u ud du

a a adx dx

= ln 4.d du

udx u dx

=1

ln .

5.d du

u udx dx =sin cos . 6.d du

udx u dx = 21

tan .cos

7.d du

u udx dx

= cos sin . 8.d du dv

u vdx dx dx

+ = +( )

9.d du dv

uv v u dx dx dx

= +( ) 10.d u u v v u

dx v v

= 2

' '

11.

dy dy du

dx du dx = . (Quy tc dy chuyn hay o hm hm hp).

V d 5. Tnh y ca hm s

a. y x= + +1 1 . b. ( )y x = ln sin ln

Gii:

a. y ='

+ KQ:x

x x+ + +2 22 1 . 1 1.

b. y ='


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+ KQ:( )xx

cot ln

2.2. Hm n v o hm hm na. Hm n Hu ht cc hm ta gp c dng y f x= ( ) , trong y biu din trc

tip (hoc tng minh) theo x. Ngoi ra y thng nh ngha l hm ca x bngphng trnh F x y =( , ) 0 (1), khng gii c i vi y, nhng trong xv y clin quan vi nhau. Khi , ta ni phng trnh (1) xc nh ynh l mt (hocnhiu) hm n ca x.V d 6.+ P/trnh xy= 1 xc nh mt hm n ca xm ta c th vit mt cch tng

minh l yx

=1

.

+ P/trnh 2x2 - 2xy = 5 - y2 xc nh hai hm n:

y x x v y x x = + = 2 25 5 .

b. o hm ca hm n :

V d 7 (i) Xt p/trnh xy= 1. Ly o hm hai v theo x:

dyx y

dx+ = 0 hoc

dy y

dx x=

+ T phng trnh ta c yx

=1

nn:dy y

ydx x x x x x

= = = = 21 1 1 1

.

+ Nu o hm trc tip yx

=1

, cng cdy

dx x= 2

1.

(ii) T phng trnh x2 + y2 = 25 o hm 2 v ta c :dy dy x

xdx y dx dx y

+ = = 2 2 0 .

2.3. Hm ngc v o hm hm ngc

1. nh ngha: Cho hm s :f X Y

x y f x

=

:

( )

Nu tng ng ngc : Y X sao cho y x y f x =| ( ) cng l mt hm sth ta ni rng hm s y f x= ( ) c hm s ngc

f Y X

y x f y

=

1

1

:

( )


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C hm s ngc f x1( ) Khng c hm s ngc2.Cng thc hm s ngc :Xt hm s : f X Y x y f x =: , ( )

Xt phng trnh n x: f x y=( ) (*)Nu vi mi y Y phng trnh (*) c duy nht mt nghim x X th hm s

y f x= ( ) c hm s ngc:

f Y X

y x f y

=

1

1

:

( )

trong x f y= 1( ) chnh l cng thc nghim duy nht ca phng trnh (*).

3 iu kin tn ti hm s ngca. Khi nim : Hm s y f x= ( ) c gi l hm mt mt nu vi x x1 2 th tac f x f x 1 2( ) ( ) .

Khng l hm 1-1

b. iu kin : Nu y f x= ( ) l hm mt - mt c TX l Xv MGT l Y. Khi

tn ti hm ngc f1 vi TX l Y v MGT l X, hn na y f x x f y = = 1( ) ( ) .


15/98


Ch :: + Nu y f x= ( ) c hmngc f1 th

+ ( )f f x x x X = 1 ( ) , .

+ ( )f f y y y Y

= 1

( ) ,

c. th : Nu y f x= ( ) c hm s ngc y f x= 1( ) th th ca hai hm s s i xng nhau qua ngphn gic th nht y x= .

4. Hm ngc ca mt s hm s cp

a) Hm s:f

x y f x x

+ +

= =

:

( )

c hm s ngc l y x= 2 .

b)Hm s ngc ca hm y x= sin :

Nu xt hm s :[ ] [ ]

x y x

=

sin : / 2, / 2 1,1

sin

Khi tn ti hm s ngc :

[ ] [ ]

y x y

=

1

1

sin : 1,1 / 2, / 2

sin

K hiu khc : x x =1sin arcsin .Ch : a b b= =1sin arcsin chnh l so gc m

a b=sin .

V d 8: = = =1

1 1 1sin sin arcsin

6 2 6 2 2.

c) Hm ngc ca hm cosine : Tng t, nu xt

[ ] [ ]

x y x

=

cos : 0, 1,1

cos

Khi s tn ti hm ngc : y x= 1cos .

d. Hm ngc ca hm tang

Xt hm s :

Hnh 9.19


16/98


x y x

=

tan : , ( , )2 2

tan

khi tn ti hm s ngc +

1tan : ( , ) ,2 2

Ch : a b= 1tan chnh l so ca gc m a b=tan . th ca hm y = tg 1x l ng m nt hnh 9.19.

e. Hm ngc ca hm cotang :Khi xt

( )

x y x

=

cotan : 0, ( , )

cot

Tng t: Hm ( )y x x

= =1

arccot cot ( )

5. o hm hm ngc

a. nh l : Cho hm y f x= ( ) l hm lin tc, mt mt trn khong a b( , ) . Khi tn ti hm ngc x f y= 1( ) xc nh trong ln cn ca y0 vi y f x=0 0( ) . Gi s

y f x= ( ) c o hm ti x0 v f x 0( ) 0 , th hm ngc x f y= 1( ) s c o hm

ti y0 v ( )f y

f x

= '1

00

1( )

'.

V d 9: Hm s y f x x = =( ) c hm ngc x f y y = =1 2( )

Ta c : f x xx= >1

'( ) , 02 ; f y y x x f xx

= = = = > '1 1 1( ) 2 2 , 01 '( )

2.

b. o hm hm lng gic ngc

Cho u l hm kh vi ca x, ta c :

d duu

dx dx u

=

1

2

1(sin )

1

d duu

dx dx u

=

1

2

1(cos )

1

d duu

dx u dx

=+

12

1(tan )

1

d duu

dx u dx

= +

12

1(tan )

1

V d 10: Tnh dy/dx ca hm s y x x x = +1 2tan ln 1 .

Gii : y x x x = +1 21

tan ln(1 )2

nn

dy

dx= x= 1tan .


17/98


V d 11: Tnh dy/dx ca hm s y x x x = + 1 2sin 1 .

Gii :dy

xdx

= = 22 1 .

2.4 .VI PHN

a. nh ngha :Cho hm s y f x= ( ) , tch s f x x'( ). gi l vi phn ca f(x) tiim x, k hiu dy f x x = '( ). .

Khi y f x x = =( ) th f x ='( ) 1 nn dy dx x = = , do : dy df f x dx = = '( ) .

Nu hm s f(x) kh vi ti x th ( )y f x x f x f x x o x = + = + ( ) ( ) '( ). b. Cng thc vi phn. Quy tc tnh o hm dn n cc cng thc vi phn

tng ng.

dc

dx= 0 d(c) = 0

n nd duu nudx dx

= 1 d(xn) = nxn-1dx

d ducu c

dx dx =( ) d(cu) = cdu

d du dv u v

dx dx dx + = +( ) d(u + v) = du + dv

d dv du uv u v

dx dx dx

= +( ) d(uv) = vdu + udv

du dv v u

d u dx dx dx v v

=

2( ) d(

u vdu udv

v v

=

2)

n nd duu nudx dx

= 1 d(un) = nun-1du

V d 12. Gi s y = x4 + 3x2 + 7 tm dyGii :+ Cch 1 :Tm o hm

dy x xdx

= +34 6

v nhn vi dx, ta c dy x x dx = +3(4 6 )+ Cch 2: Chng ta cng c th dng cc cng thc vi phn trn

dy = d(x4 + 3x2 + 7) = dx4 +3 d(x2) + d(7)= 4x3dx + 3.2xdx+0= (4x3 + 6x) dx


18/98


V d 13. Tnh d (x

x +

2

2 1)

Gii: + Dng cng thc vi phn ca mt thng:

d(x x d x x d x

xx

+ +=

++

2 2 2 2 2

22

1. ( ) ( 1 ))

11

x dxx x dx

x x xxd dx

x xx

+ ++ = = + + +

32

2 32

2 2 3/22

2 121

1 ( 1)1.

V d 14: Gi thit rng yl mt hm kh vi i vi x v tha mn phng trnh:x2y3- 2xy + 5 = 0.

Hy s dng vi phn tmdy

dx.

Gii:

+ Kt qu :dy y xy

dx x y x

=

3

2 2

2 23 2

vi /k x y x 2 23 2 0 .

Ch : rng tip tuyn vi ng cong m st ng cong gn tip im.iu ny c ngha rng khi dx nh, th ng cong thc s gn vi tip tuynca n, v v th vi phn dy d dng c tnh ton, n cho xp xtt i vi s gia

y .

c. ng dng ca vi phn trong tnh gn ng

Xt hm s y f x= ( ) kh vi trong ln cn ca x a b0 ( , ) . Theo cng thc s giaca hm kh vi ta c

f x x f x f x x + + 0 0 0( ) ( ) '( ).

V d

15: Tnh x

p x

ln11.Gii: + Xt f x x x x = = =0( ) ln , 10, 1+ p dng f x x f x f x x + + 0 0 0( ) ( ) '( ). ta c

= + + 1

ln11 ln(10 1) ln10 .1 1,04310.ln10

.

2.5 o hm v vi phn cp cao1. nh ngha:


19/98


Cho hm s f(x) xc nh trong khong (a, b). Gi s hm s y = f(x) c o hmy = f(x) v f(x) c o hm th ta gi o hm ca f(x) l o hm cp hai ca hmf(x). K hiu y = f(x) = [f(x)].

Tng t ta c o hm cp n ca hm y = f(x): n ny x y x = ( ) ( 1)( ) ( )

V d 16: Cho hm s xy e=2

. Tnh y .Ta c:

y ='

( )

( ) xy y

y y e x x

= =

= = 2 3

' '

" ' 4 (3 2 ).

2. Cc quy tc ly o hm cp cao:a. Vi f, g l cc hm s c o hm cp n v R , , ta c:

n n nf x g x f x g x + = +( ) ( ) ( )( ( ) ( )) ( ) ( ) b. Quy tc Leibniz: Vi f, g l cc hm s c o hm cp n, ta c:

nn k n k k

n

k

fg C f g

=

= ( ) ( ) ( )0

( )

V d 17: Tm cng thc tnh o hm cp n ca:

a. yx

=1

1, b. ky x= , vi k R

Gii:

a. y xx

= =

11 ( 1)1

, nn y y= =' , " ,

,Nn

n n

ny n

x += =

( )

11( 1) !.

( 1).

b. Nu k N th y y= =' , " ,,

ny =( )

Nu k N th

k n

n

k k k n x k n

y k k n

k n

+ >= =


20/98


n n n n n n n

n n

n n n n n

n n

n x y n y y x n x y n y x y

y y

n x y x y n a x

y y

+ = =

+ = =

2 1 2 1 2 1 1 2 2

2 2 2 2

2 1 2

2 2 2 1

( 1) ( 1) . '. ( 1) ( 1) .''

( 1) ( ) ( 1)

3. Vi phn cp cao:

Vi phn cp 2 ca hm s f(x) ti mt im no (nu c) l vi phn ca vi phncp mt df. K hiu d f d df =2 ( ) .Quy np ta c: Vi phn cp n, k hiu l nd f l vi phn cp mt ca vi phn cp (n-1): n nd f d d f = 1( ) .Ch : Khi tm vi phn cp cao ca hm s, ta lun coi dx nh l hng s.

n n n n

d y d dy d y dx y dx y dx

d y d d y y dx

= = = =

= =

2 2 2

1 ( )

( ) ( ' ) "( ) "

....................

( )

V d 19: Cho hm s y x= ln . Tm d y5

Ta c: d y y dx dx dx x x

= = =5 (5) 5 5 55 54! 24

.

Bi tp v nh: Trang 90 ( Bi 6 - 13), trang 254 ( Bi 1 4), Trang 287 ( Bi 1 -9). Trang 104 ( bi 1- 5), Trang 108 ( bi 1 - 8), Trang 112( bi 1 - 12).


21/98


Bi ging s 3

CC NG DNG CA O HM

3.1. BI TON GI TR LN NHT, NH NHT

a) nh ngha: Hm sy=f(x) xc nh trn on [a,b], ta ni :

i. Hm st GTLN nu :f x M x

( ) , [a,b]

c [a,b]: f(c)=M

ii. Hm st GTNN nu :f x m x

( ) , [a,b]

c [a,b]: f(c)=m

b) Cch tm : + Tm cc im ti hn ca f(x) trong on [a,b ]: chng hn l c

+ Khi : { }m f x m =[a,b]ax ( ) ax f(a), f(b), f(c) hoc { }f x =

[a,b]min ( ) min f(a), f(b), f(c) .

Ch : 1) Nu hm s y f x= ( ) lin tc trnD, v trn n c duy nht mtcc tr

+ Nu cc tr l cc tiuth cng l GTNN ca hm s trn min .

+ Nu cc tr l ccith cng l GTLN ca hm s trn min .

2) Nu hm s y f x= ( ) ng bin trn [ ]a b, th [ ] [ ]a ba b f x f b f x f b = =,,max ( ) ( ), min ( ) ( ) .

Nu hm s y f x= ( ) nghch bin trn [ ]a b, th[ ] [ ]a ba b

f x f b f x f a = =,,

max ( ) ( ), min ( ) ( ) .

3) Hm s y f x= ( ) lin tc trn [ ]a b, th lun tn ti GTLN v GTNN trn min .

V d 1: Tm hai s dng m tng ca chng bng 16 v tch ca chng t gitr ln nht.

Gii: + Gi sxv yl hai s dng m tng ca chng bng 16

+ V vy: x + y = 16+ Tch ca chng : P = xy+ Ta c : y = 16 x , khi :

P = xy = x(16 - x) = 16x - x2, vi 0


22/98


+ Lp bng bin thin, suy ra GTLN ca P: max P=64, ti x = 8. Vy x = y =8.

V d 2: Mt mnh vn hnh ch nht 450m2c ro li.Nu mt cnh ca mnh vn c bo v bi bc tng camt kho thc, th kch thc chiu di ca tng ro ngn nhtl bao nhiu?

Gii:+ Gi x l chiu rng ca vn, y l chiu di ca mnhvn, L l chiu di tng cng ca hng ro, vi x y L >, , 0 .+ Chng ta cn tm GTNN ca :

L = vi rng buc

+ L c thc vit nh l hm ca mt bin x :

+ V vy mnh vn c hng ro ngn nht l 15 v 30.

V d 3: Tm kch thc ca hnh ch nht c din tch ln nht m n c th ni

tip trong na ng trn bn knh l a.Gii:+ Xt na trn ca ng trn : x2+ y2= a2.+ Chng ta phi tm GTLN ca: A = vi iu kin :+ a A v hm mt bin s x:

(9)

x

y450ft

2

Barn


23/98


+ V vy kch thc ca hnh ch nht ni tip ln nht l x a=2 2 va

y=2

2,

hnh ch nht ny c chiu di gp i chiu rng v kch thc lm cho A tGTLN v GTLN l : A a= 2 .

V d 4: Mt ci dy di Lc ct thnh hai on. Mt on b ni thnh dng hnhvung v on kia thnh hnh trn. Ci dy s b ct nh th no sao cho tng dintch bao gm bi 2 on dy:a) L ln nht b) L nh nhtGii: + Gi sxl cnh ca hnh vung v rl bn knh ca hnh trn ( x> 0) , khi tng din tch ca hai hnh c to thnh l:

A =

Hnh 4.18

+ VyL

A

= + +

2

min 14 4khi

Lx

=

+4;

LA

=

2

max 4khi x = 0.


24/98


V d 5: Mt ngi bn hng dnh bn 500kg khoai ty bc v vi gi 1,5USD/kg (gi gc l 70 cent /kg). Tuy nhin nu c h gi mt cent th s bn thmc 25 kg. Hi ngi bn hng nn bn vi gi no t li nhun ln nht?

Gii: + Gi xl s cent m ngi bn hng h gi,

+ Li nhun ca mi mt kg khoai ty gt v l (80 - x) cent+ S lng bn c l 500 + 25x.+ V vy ton b li nhun s l (bng cent)

P =

+

+ Gi bn thun li nht l 1,2 la/kg.

V d 6: Mt nh my sn xut cc hp ng x phng hnh tr nhn an thng i vi cc hp c th tch c ch r V0. Vi kch thc no th din tchton phn ca mt ci hp nh vy st GTNN v s lng kim loi cn n chonh my l bao nhiu?

Gii:+ Gi srl bn knh ca y v hl chiu cao ca hp hnh tr+ Khi th tch l: V r h= 20 . (1)

v din tch mt ton phn l: A r r h = +2

2 2 (2)+ a A v hm 1 bin s r, ta c :

A =


25/98


+ Kt qu : h r= 2 .

3.2 . NH L V GI TR TRUNG BNH

Nhn xt hnh hc : Gia hai im bt k Pv Q trn th ca hm s kh vi, tn ti t nht mtim m ti c ng tip tuyn song song vidy cung ni hai im P v Q, ni cch khc : Tnti t nht mt im c nm gia a v b (a < c < b)tho mn iu kin:

f b f a f c b a=

/( ) ( )

( ) .

a)nh l 1 (nh l Rolle). Nu hm s f(x ) lin tc trnon [a,b] v kh vitrong khong m (a,b) v nu f(a) = f(b) = 0 th khi tn ti t nht mt s c nmgia a v b tho mn f(c) = 0.

ngha hnh hc:nh l ny pht biu rng nu mt ng cong trn cttrc Oxti 2 im, th khi s c t nht mt im ca ng cong ny nm gia 2im trn m ti tip tuyn c phng nm ngang.

V d 7. Hm s : x xf x x x

=

0 1( ) 2 1 2

Hm s ny c gi tr bng 0 ti x = 0v x = 2, vlin tc trn khong ng 0x2. Hm s kh vi trongkhong m0 < x < 2, trim x = 1 v khi o hmca n khng tn ti. o hm f(x) r rng l khngbng 0 ti bt kim no trn khong . y l mttht bi trong kt lun ca nh l Rolle v thc t lhm s khng kh vi ti im x = 1.

V d 8. Hm s:

Hm s bng 0 ti x = 0v x = 1, v kh vi trong khong 0< x < 1. Hm s lin tc trn 0x < 1, khng lin tc ti x =1. o hm f(x) khng bng 0 ti bt k im no trn

0

y

x21

0 1x

x xf

x


26/98


khong ny, v trong trng hp ny kt lun ca nh l Rolle khng cn ng.

V d 9. Gi sa b

c+ + = 03 2

. Chng minh rng phng trnh

ax bx c + + =2 0 c nghim trong ( )0,1 .

Gii: Xt hm s a bf x x x cx = + +3 2( ) 3 2 :

+ Xc nh lin tc trn [ ]0,1 , kh vi trn ( )0,1

+ Ta c f f= =(0) (1) 0

+ Do tn ti x 0 (1,0) sao cho f x =0'( ) 0 , tc l ax bx c + + =20 0 0 .

Vy ta c cn chng minh.

b) nh l 2 (nh l gi tr trung bnh).Nu hm s f(x) lin tc trn [a,b] v khvi trn (a,b), khi tn ti t nht mt s c nm gia a v b tho mn:

f b f a f c

b a

=

/ ( ) ( )( )

V d 10. CRM nu b a<


27/98


Nu f(x) v g(x) u bng 0 ti x = a v kh vi th:x a x a

f x f x

g x g x

=

( ) ( )

lim lim( ) ( )

.

V d

11 : x xx x x

x x x

+

= =+ +

2

22 2

3 7 2 6 7 5

lim lim5 14 2 5 9 .

x x

c

= =

0 0

sinx osxlim lim 1

x 1.

x xx x

x x

e e = =

2

2 1 20 0

tan6 6sec 6lim lim 3

2.

V d 12: p dng qui tc LHospital hai ln ta c:

x x x

c c

= = =

20 0 0

1 osx sinx osx 1lim lim lim

x 2x 2 2

.

x x x

x xx x

x x

+ + + + = = =

1/2 3/2

20 0 0

11 (1 ) 1/ 2( 1) 1/ 2 1/ 4( 1) 1/ 2 12lim lim lim

2 2 8

.

Quy tc LHospital cho gii hn dng

v mt s dng khc:

Nu x af x

g x( )

lim ( ) c dng ; cc hm s f x g x ( ), ( ) kh vi, th:

x a x a

f x f x

g x g x

=

( ) ( )

lim lim( ) ( )

V d 13: Chng t rng :p

xx

x

e=lim 0 vi mi hng sp.

Gii: + Nu p 0 : gii hn xc nh v gi tr gii hn bng 0.+ Vi p>0gii hn c dng khng xc nh : /,+ p dng Qui tc LHospital:

p p

x xx x

x px

e e

=

1

lim lim

+ Qu trnh ny tip tc tng bc th chng ta s gim thiu s m ti 0 hoc timt s m, v ta nhn c iu cn chng minh.

Nhn xt : khi x + th xe nhanh hn bt ka thc no.


28/98


V d 14: Chng minh rng:px

x

x=

lnlim 0 vi mi hng p>0.

Gii: + Gii hn c dng /, theo qui tc LHospital ta c

p p px x x

x x

x px px = = =

1

ln 1/ 1lim lim lim 0

Nhn xt : khi x + th xln tng chm hn mi hm m dng ca xchod hm m nhn c no.

Ch : Ch p dng quy tc LHospital cho dng

0,

0, cc dng v nh khc

0 00. , , 0 , , 1 c tha v dng

0,

0bng cc bin i sau :

+ ( )f x

f x g x g x

( ) 0

( ). ( ) 0. ,1/ ( ) 0

.

+ ( ) [ ]g x f x g x g x f x f f e e = ( )/ 1/ln ( )( ) 0 0 ( ) ln ( )( ) 0 , , 1 v s dng ( )yyy e e= = l im lnlnlim lim

V d 15. Tm:x

x x+0lim ln

Gii: + Gii hn c dng 0.+ Bin i v dng / v p dng qui tc LHospital nh sau:

x x

xx x

x+ + = = =

0 0

lnlim ln lim 0

1/

V d 16. Tnh: x x x /2lim (sec - tan )+ Gii hn c dng dng -.+ Chng ta s chuyn qua dng 0/0 v p dng qui tc LHospital:

x x xc

= = = = /2 /2 /21 sinx 1 sinx

lim (secx-tanx) lim lim 0osx cosx cosx

V d 17. Tm xx

x+0

lim

Gii: + Gii hn c dng : 00+ a v dng 0. bng cch s dng php ton loga.+ t xy x= , ta c:

xy x x x = =ln ln ln

+ Theo V d 3 : xx x x

y x x x + + +

= = =0 0 0

lim ln lim ln lim ln 0

+ T :x y

x x xx y e e

+ + + = = = =ln 0

0 0 0lim lim lim 1


29/98


V d 18. Tm xx

x

1/lim

Gii: + Gii hn c dng 0.

+ t xy x= 1/ ta c: xx

y x khi x x

= = 1/ln

ln ln 0

+ Nh vy :x

xx

= =1/lim 1

V d 19. Hy chng t rng : x ax

ax e

+ =1/0

lim(1 ) vi mi hng sa.

Gii:+ Nu a=0th y l mt dng xc nh : khng nh ng bi v c hai v c gi

tr l 1.+ Nu nha0, gii hn c dng 1.

t : xy= + 1/(1 ax) ta c:

x x x

ay a

x + +

= = =0 0 0

ln(1 ax) / (1 ax)limln lim lim

1

iu ny ngha l:x y a

x x xax y e e

+ = = =1/ ln

0 0 0lim(1 ) lim lim .

Bi tp v nh:

Trang133 ( bi 1 - 30), trang 362 ( bi 1 - 25), trang 367 (bi 1 - 44).


30/98


Bi s 4

NGUYN HM V TCH PHN XC NH

I. NGUYN HM (TCH PHN KHNG XC NH)

1. nh ngha: Cho hm s f x( ) xc nh trong khong (a, b).

Hm sF x( ) gi l nguyn hm ca f x( ) nu tho mnd

F x f x dx

=( ) ( ) .

Ch :+ Mt nguyn hm ca f(x) th thng c gi l tch phn khng xcnhca

f(x) v php ly nguyn hm c gi l php ly tch phn.+ Nu F x( ) l nguyn hm ca f x( ) th F x C+( ) cng l nguyn hm ca f x( ) ,

vi mi hng s C.

K hiu : f x dx F x C = +

( ) ( ) .

2. Mt s tnh cht :

1. f x dx f x = '

( ) ( ) hay l d f x dx f x dx = ( ) ( ) .

2 : Gi s F x( ) kh vi, ta c : dF x F x C = + ( ) ( ) .

3 : Vi c l hng s th : cf x dx c f x dx = ( ) ( ) .

4 : [ ]f x g x dx f x dx g x dx + = + ( ) ( ) ( ) ( ) .

3. Phng php tm nguyn hm

Nhc li mt s cng thc :

ec ec

= = + = + =2 2 21 1

cos ; sec ; 1 tan sec , 1 cot cossin cos

d du d du u u u u u

dx dx dx dx

d du d duu ec u ecu ecu u dx dx dx dx

= =

= =

2

2

tan sec sec sec tan

cot cos cos cos cot

a. S dng trc tip bng cc nguyn hm c bn

1.u

u du c

+

= + +

1

, 11

2. lndu

u cu

= +


31/98


3. dueu = eu+ c 4.

ln

uu a

a du ca

= +

5. duucos = sin u + c 6. sin u du = - cos u + c

7. 21 tancosdu u c

u= + 8. 21 cotsin

du u cu

= +

9.2 2

1ln

2

du u ac

a u au a

= +

+ 10. 22 ua

du

= sin-1

a

u+ c

11. 22 uadu

+=

a

1tan-1

a

u+ c 12. tan u du = -ln|cos u| + c

13. cot u du =ln|sin u| + c 14. sec u du =ln|sec u + tan u| + c

15. csc u du = -ln|csc u + cot u| + c

V d 1.

x dx x dx x c

x xdx x x dx x x c

x

= = +

= = +

3 2 2/3 5/3

1/3 1/51/6 7/10 5/6 3/10

35

5 3[5 3 ] 6 10

V d 2 :

x xdx dx x dx x x x c x x x = = = + +

2 2

sin 1 cos 1 cos ln sec tan sincos cos cos .

b. Phng php th

Csca phng php: Nu f x dx F x C = + ( ) ( ) v u g x= ( ) th :

[ ]f g x g x dx F g x C = + ( ) '( ) ( ( ))

V d 3. Tm xxe2

dx.

Gii: t u= - x2, th du x dx xdx du = =- 2 . - , v

xxe2

dx = -12 e

udu = -12

eu= -12

xe2 + c


32/98


V d 4. Tm1tan

21

xe

dxx

+

Gii: t 12

1tan

1u x du dx

x

= =+

. Nn

11

tantan

2

1

xxe

dx e cx

= = ++

V d 5. Tm 2x49dxx

Gii: + t: u = 9 - 4x2do du = - 8x dxv

2x49dxx

= = -

4

1 249 x +c.

c. Phng php tch phn tng phn

Cng thc : ' 'udv uv vdu uv dx uv vu dx= = Mt s dng cbn :

Dng : I p x H x dx = ( ). ( ) trong { }nxH x ax bx e ( ) sin , cos , , p x( ) l a thc:

khi ta chn :u p x

v H x

= =

( )

' ( ).

Dng : I p x q x dx = ( ).ln ( ) ta tu q x

v p x

= =

ln ( )

' ( )

Dng : axI e K x dx = . ( ) trong { }K x ax bx ( ) sin , cos

khi ta chn :axu e

v K x

= = ' ( )hoc

ax

u K x

v e

= =

( )

'..

V d 6: Tnh cc tch phn sau:

a. I xdx = = 1sin 2

KQ: x sin x x C +1 212 1 48


33/98


b. ( )2

ln I x dx= =

KQ: ( )x ln x x ln x x C + +2 2 2

c.x2exdx =

KQ: x x xx e xe e C + +2 2 2

d. cosx J e xdx= =

KQ: ( )x xe cos x e sin x C + +12

V d 7. Tm cng thc gim bc cho Jn=sinnx dx.

Gii: Bng PPTP tng phn v quy nap ta s c :

sinnx dx = -

n

1sinn-1 x cos x +

n

n )1( sin

n-2x dx .

4. Nguyn hm ca mt s hm s c bn.

a. Tch phn cc hm s lng gic

Xt tch phn dng I f x x dx = (sin ,cos )

Phng php chung: tx

t= tan2

Nu f x x f x x = ( sin ,cos ) (sin ,cos ) t t x= cos .

Nu f x x f x x = (sin , cos ) (sin ,cos ) t t x= sin .

Xt cc tch phn dng m nI x x dx =

sin cos (*)

Nu n l, m chn:

+ Tch ra tha s xdx d x =cos (sin )

+ V s m ca cos xl chn, nn ta c th s dng tnh cht :cos2x = 1 - sin2x biu din phn cn li ca tch phn ban u di dng t hp ca sin x.

Nu m l, n chn:


34/98


+ Tch ra tha s xdx d x = sin (cos )

+ S dng tnh cht: sin2x = 1 - cos2xtng t nh trn. Nu n v m l s nguyn chn, khng m: S dng cng thc gc chia i

.

V d 8.( )x x dx x x xdx x x d x

x x d x x x C

= =

= = +

2 3 2 2 2 2

2 4 3 5

sin cos sin cos cos sin 1 sin (sin )

1 1(sin sin ) (sin ) sin sin .

3 5

V d 9.

( )x dx x xdx x d x x x C = = = + + 3 2 2 31

sin sin sin 1 cos (cos ) cos cos3

.

Ch :Nu mt trong cc s m trong (*) l s l dng v rt ln, th cn phi

s dng cng thc nh thc. V d, vi bt k s m dng l no ca cos x:cos2n +1 x = cos2nx cos x = (cos2x)ncos x = (1 - sin2x)ncos x,

Ta t u = sin x th du = cos x dx, v khi :n nxdx x xdx u du + = = 2 1 2 2cos (1 sin )cos (1 )

V d 10. Ta c:

xdx x dx dx xdx x xd x x x = + = + = + = +

2

1 1 1 1 1 1cos (1 cos2 ) cos2 cos2 2 sin2

2 2 2 2 2 4

V d 11. p dng lin tip hai ln cng thc gc chia i i vi csin cho ta

xdx x x x C = + + + 43 1 1

cos sin2 sin44 4 32

Xt cc tch phn dng tan secm n I x x dx= vi n l mt s nguyn dngchn v ml mt s nguyn dng l.

+ Da trn cc biu thc: d(tan x) = sec2x dx

d(sec x) = sec x tan x dx,tan2x + 1 = sec2x.

V d 12. Ta c


35/98


4 6 4 4 2 4 2 2

4 2 4 8 6 4

9 7 5

tan sec tan sec sec tan (1 tan ) (tan )

tan (1 2 tan tan ) (tan ) (tan 2 tan tan ) (tan )

1 2 1tan tan tan

9 7 5

= = +

= + + = + +

= + + +

x xdx x x xdx x x d x

x x x d x x x x d x

x x x C

V d 13.3 5 2 4 2 4

6 4 7 5

tan sec tan sec .sec tan (sec 1)sec (sec )

1 1(sec sec ) (sec ) sec sec

7 5

x xdx x x x xdx x xd x

x x d x x x C

= =

= = +

Tch phn dng : cot cscm n I x xdx=

Cng c cho cc trng hp ny l cc cng thc:

d(cot x) = -csc2x dx

d(csc x) = -csc x cot x dx

1 + cot2

x = csc2

x.

b. Tch phn cc hm phn thc hu t

tng c bn l phn tch hm phn thc hu t cho thnh tng ccphn thc n gin hn (gi l cc phn thcn gin).

Mt hm hu tc gi l chnh quynu s m ln nht ca t s nh hn sm ln nht ca mu s, ngc li n c gi l khng chnh quy.

+ Mt hm khng chnh quy bt k)(

)(

xQ

xPc th biu din thnh:

)(

)(

xQ

xP= a thc +

)(

)(

xQ

xR

bc ca R(x) nh hn bc ca Q(x).

V d 1. Tnh )3)(1(75

+

+

xx

xdx

Tch:5 7

( 1)( 3) 1 3

+= =

+ +

x A B

x x x x(3)

+ Chng ta c th tm c A v B bng cch ng nht cc h s ca xv nhnc:

5 3

3 7 2

A B A

A B B

+ = =

= =

+ )3)(1(75

+

+

xx

xdx = ( 1

3

x+

3

2

+x)dx= 3 ln (x - I) + 2 ln (x + 3) + c.


36/98


V d 2. Tm 1122

4

23

++

x

xxxdx

Gii:Ta c

1

1224

23

++

x

xxx= =

++

++

)1)(1)(1(

1222

23

xxx

xxx

1+x

A+

1x

B+

12 +

+

x

DCx

1x

1x2xx24

23

++=

1x

1

++

1x

1

+

1x

12 +

.

+ Vy nn: 1122

4

23

++

x

xxxdx = ln |x + 1| + ln|x 1| + tan

-1x + C.

c. Tch phn hm s v t

Mt s php thcbn :

+ Nu tch phn cha 22 xa , t x =a sin , 22 xa = a cos.

+ Nu tch phn cha 22 xa + , t x= a tan , 22 xa + = a sec.

+ Nu n c cha 22 ax , t x = a sec , 22 ax = tan.

V d 1. Tm2 2

= a x

I dxx

Gii: +Ta t : x = a sin, dx = a cos d,22

xa = a cos

+ Khi dxxxa 22

=

sina

cosaa cosd = a

sina

cosa 2d

= a

sin

sin12

d = a(csc - sin)d

=- a ln(csc + cot) + a cos (7)

+ Tr li bin gc x: Xt Hnh 10.1

csc=xa

, cot= xxa 22

, v cos= axa 22

,nn t (7) ta c

dxxxa 22

= 22 xa - a ln(x

xaa 22 +) + C

Hnh 10.1

a2 - x2

a x


37/98


V d 2. Tm2 2

=+

dx

I

a x

Gii:+ Ta t: x = a tan, dx = a sec2d, 22 xa + = a sec.

+ Khi : 22

xa

dx

+= a

sec

sec2

a

a d = secd = ln(sec + tan) +C .

+ T: x = a tan hay tan = x/a ta c: sec =a

xa 22 +v tan=

a

x.

+ Do : 22xa

dx

+= ln(

a

xxa 22 ++) + C = ln( 22 xa + + x) + C.

V d 3. Tm 223)2(

xx

dxx

+

+

Gii:+ Ta c: 3 + 2x - x2 = 4 - (x2 - 2x + 1)= 4 - (x - 1)2 = a2 - u2,

+ V x = u + 1, nn ta c dx = duv x + 2 = u + 3, v do

223)2(

xx

dxx

+

+= 22

)3(

ua

duu

+= 22

ua

udu

+ 3 22

ua

du

= 22 ua + 3sin-1a

u

= 223 xx + + 3sin-1 )

2

1(

x+ c

V d 4. Tm 522 + xx

xdx

Gii:+ Ta vit : x2 - 2x + 5 = (x2 - 2x + 1) + 4 = (x - 1)2 + 4 = u2 + a2

+ t: x = u + 1, dx = du, ta c 522 + xx

xdx= 22

)1(

au

duu

+

+= 22

au

udu

++

22 audu

+.

+ Ta c: + 22 audu

= ln(u + 22 au + )

+ + 5x2x

xdx

2= 22 au + + ln(u + 22 au + )= 522 + xx + ln(x - 1

+ 522 + xx ) + C.

II. TCH PHN XC NH


38/98


1. Din tch hnh thang cong

Bi ton: Tm din tch ca mt min nm di th ca hm s lin tc( ) y f x= vi a x b , nm gia hai ng thng thng ng x = av x = b, nm

pha trn trc honh.

Cch tnh :+ Cho n l s nguyn dng, chia on [ ]a b, thnh n phn bng nhau bi (n-1)

im chia: x0 = a < x1 < x2 < . < xn = b

+ Ta c n on con: kx , k k kb a

x x x k n n

= = =1 , 1,2,...,

+ Gi [ ]k k kx x x 1, , ta c tng xp xsnca tt c nhng hnh ch nht to thnhl:

nn ns f x x f x x f x x = + + + 1 21 2( ) ( ) ... ( ) hay sn=n

K k

k

f x x=

1

( ) (1)

Thay cho n bngk

x max 0 ta c din tch ca hnh thang cong l:

k

n

k kx

k

f x x

=

max 0

1

lim ( ) .

nh ngha : Gii hn (nu tn ti): ( )n

k km

k

f x x

=iax x 0 1lim

c k hiu bi :

b

a

dxxf )( v c gi l tch phn xc nh tan bca f(x)dx. Nh vy :

Hnh 6.11 : Tn di Hnh 6.12 : Tn trn


39/98


( )b n

k km

ka

f x dx f x x

=

= iax x 0 1

( ) lim .

Ch :+ Khi tani rng hm y f x= ( ) kh tchtrn [ ]a b, . Gi tr ca gii

hn khng ph thuc vo cch chia on [ ]a b, .

+ Mi hm lin tc u kh tch.

V d 1. Xt hm s y f x x = =( ) trn on [a,b]. Min nm di th ny ltam gic vung c chiu cao l bv cnh y l b. Tnh din tch ca ca min .

Gii : +Chia on [0, b] thnh (n-1) phn bng nhau bi cc im chia :

n n

b b n b x x x x x b

n n n

= = = = =0 1 2 1

2 ( 1)0, , ,..., ,

+ Cc cnh ca hnh ch nht l k b n = / , v chiu cao ca cc hnh ch nhtl:

n

bxf =)( 1 ;

n

bxf

2)( 2 = ; ..;

n

nbxf n =)(

+ Ta c : [ ] )1

1(22

)1(....21......

2.

22

2 n

bnn

n

bn

n

b

n

b

n

nb

n

b

n

b

n

b

n

bSn +=

+=+++=+++=

+ Nh vy: Din tch ca min =2

)1

1(2

blimSnlim

22

n

b

nn=+=

+ Theo nh ngha tch phn xc nh ta c: 2

2

0bxdxb = .

2. Cc tnh cht ca tch phn xc nh.

a). Din tch i s


40/98


Nu y f x= ( ) 0 vi x a b [ , ] , din tch min c gii hn bi:

y f x

a x b

y f x

= =

( )

( ) 0

:b

a

S f x dx = ( )

Nu y f x= ( ) 0 vi x a b [ , ] , din tch min c gii hn bi:

y f x

a x b

y f x

= =

( )

( ) 0

:

S = b

a

dxxf )(

Xt hm s y f x= ( ) vi x a b [ , ] , trong a b b b b < < <


41/98


+ f x ( ) 0 vi x 0 1: din tch min c gii hn biy f x

x

=

( ) 0

0 1l

A x x dx = =1

22

0

1( 1)

4

+ Din tch i s cn tm l: I A A= =1 2 0 .

b). Din tch hnh hc :b

a

S f x dx A A A A= = + + + 1 2 3 4( )

V d 2: Tnh din tch gii hn bi cc trc to v ng cong x y a+ = .

Gii: Ta c x y a 0 , nn ( )x y a y a x + = = 2

Nn din tch cn tm l:

( ) ( )aa a x a

S a x dx a a x x dx ax a x = = + = + =

2 223/ 2

0 0 0

22 2 .3 2 6

.

c) Mt s tnh cht ca tch phn xc nh:

a

a

f x dx = ( ) 0

Khi a b : ta cb a

a b

f x dx f x dx = ( ) ( )

Ta cb c b

a a c

f x dx f x dx f x dx = + ( ) ( ) ( )

Tnh tuyn tnh: [ ]b b b

a a a

f x g x dx f x dx f x dx + = + ( ) ( ) ( ) ( )

Nu f x x ( ) 0, [a,b] th ta c:b

a

f x dx ( ) 0

Nu f x g x x ( ) ( ), [a,b] th ta c:b b

a a

f x dx g x dx ( ) ( )

Nu m f x M x ( ) , [a,b] th ta cb

a

m b a f x dx M b a ( ) ( ) ( ) .

b b b

a a a

f x dx f t dt f u du = = ( ) ( ) ( ) , tc l tch phn xc nh khng ph thuc

vo tn bin di du tch phn.


42/98


Nu f(x) kh tch trn [a, b] f x ( ) cng kh tch trn [a, b] v :

b b

a a

f x dx f x dx ( ) ( )

Nu f(x) lin tc trn [a, b] th tn ti sao chob

a

f x dx f b a = ( ) ( )( ) .

Tch phn vi cn thayi:

Nux

a

F x f t dt = ( ) ( ) , x [a,b] thx

a

d dF x f t dt f x

dx dx = =( ) ( ) ( ) .

V d:x

a

ddt

dx t x =

+ + 2 21 1

1 1.

3 NH L C BN CA GII TCH

nh l : Nu f x( ) l hm lin tc trn mton [a, b], v F(x) l mt nguyn

hm bt kca f(x) thb

b

a

a

f x dx F x F b F a = = ( ) ( ) ( ) ( ) .

.

Ch : Nh vy, ta cng c cc phng php tm tch phn xc nh tngng vi cc phng php tm nguyn hm (tch phn khng xc inh) :

+ p dng trc tip bng cc nguyn hm c bn,

+ Phng php th (Cn ch thm ti qu trnh th cn)+ Phng php tch phn tng phn,

V d 1. Tnh +

1

03 5

4

7dx

x

x

Gii:+ t u x du x dx = + =5 47 5 , x u x u = = = =0 7, 1 8 .

Trc ht tm nguyn hm

( )x dx u du u x

= = =

+ 1 84

81/3 2/3 3

7530 7

1 1 3 3. 4 49

5 5 2 107

.

V d 2: Tnh : I x xdx

= /4

0

arctan


43/98


+ tdx

duu xx

dv dx v x

== + = =

2arctan 1

+ Khi :xdx d x

x xdx x x x x

x x

+

= =

+ +

/4 /4 /4 2/4 /4

2 20 00 0 0

1 (1 )arctan arctan arctan

1 2 1

x x x

= + = +

242

0

1 1arctan ln(1 ) ln 1

2 4 2 16.

Bi tp v nh:Trang 280 (bi 1 - 30), trang 282 (bi 13 - 24), Trang 304 ( bi 1 -30), trang 308 (bi 1 - 23), trang 312 (bi 1 - 28), trang 315( bi 1 - 16), trang 321(2 - 13), trang 326

(bi 1 - 25).

c trc cc mc: 12.4 chun b cho Bi s 5Tch phn suy rng


44/98


CU TRC KIM TRA GIA K MN TON I

Hnh thc thi: T lun - Thi gian: 50 pht

Cu 1 (3,5 im) Gii hn v tnh lin tc ca hm s

+ Tm gii hn dng v nh.+ Hm lin tc ti 1 im, lin tc tng pha, lin tc trn1 khong, hmgin on.

Cu 2 (3,5 im) o hm v vi phn hm mt bin

+ o hm v vi phn cp 1,2, 3, cp n (khng cn chng minh bng quynp), o hm hm n, hm hp.+ ng dng kho st cc tr trong cc bi ton thc t.

Cu 3 (3,0 im) Tch phn.

+ Tnh tch phn xc nh hm a thc, phn thc hu t, cn


45/98


BI GING S 5

TCH PHN SUY RNG

Chng ta bit mt tch phn ( )b

a

f x dx xc nh khi hm f(x) l lin tc trn min

xc nh a x b , trong ,a b l hu hn.

Vn:Nu t nht mt trong cc s ,a b bng v cng hoc hm ( )f x khng b

chn trn on [ ],a b th ta c th ni g v tch phn ?

a b

y = f(x)y = f(x)

a t ->

I. Trng hp cn ly tch phn l v hn

1. nh ngha: Cho hm s f(x) xc nh trong khong [ ],a+ , kh tch trn mi

on [ ], ta vi mi s hu hn t > a.

Ta gi gii hn lim ( )

t

ta

f x dx + (1) l tch phn suy rng ca hm f(x) trong on

[ ],a+ v k hiu ( )a

f x dx

+

(2). Vy ( ) lim ( )t

ta a

f x dx f x dx

+

+=

Nu gii hn (1) tn ti (hu hn) ta ni tch phn (2) hi t. Nu gii hn (1)

khng tn ti hoc bng v cng, ta ni tch phn (2) phn k.

Ch : Nu tn ti gii hn th c th vit

lim ( ) ( )t

F t F+ = + hay ( ) ( ) ( ) ( )a

a

f x dx F t F F a

+

+= = + .

V d 1: Xt s tn ti ca tch phn0

xI e dx

+= .


46/98


Gii:0

0 0

1lim lim lim 1 1

tt

x x x

tt t te dx e dx e

e

+

+ + +

= = = + = .

Gii hn tn ti v hu hn nn tch phn suy rng ny hi t.

Tng t ta cng c cc nh ngha:

( ) lim ( )a a

tt

f x dx f x dx

=

( ) ( ) ( )a

a

f x dx f x dx f x dx a

+ +

= + , hoc ( ) lim ( )A

AB B

f x dx f x dx

+

+

=

Tch phn suy rng v tri hi t khi v chkhi c hai tch phn v phi hi t.

2. Cch tnh: Xt ( ) lim ( )t

t

a a

I f x dx f x dx

+

+= =

+ Dng cng thc Newton Leibniz tnh ( ) ( ) ( )t

a

f x dx F t F a =

+ Tnh ( )lim ( ) lim ( ) ( ) lim ( ) ( )t

t t ta

f x dx F t F a F t F a + + +

= =

3. Mt s v d

V d 2: Xt s hi t ca cc tch phn sau:

a.

1

dx

x

+

b.0

cos xdx

a. [ ]1

1 1

lim lim ln lim lnt

t

t t t

dx dx x t

x x

= = = = . Tch phn phn k.

b.0 0

cos lim lim sint

t tx dx cosx dx t

= = . Tch phn phn k v gii hn khng tn

ti.

V d 3 Nu p l mt hng dng, chra rng cc tch phn suy rng1

p

dx

x

s hi

t nu p > 1 v phn k nu p 1.

Gii: Nu p= 1 th c th hin trong v d 3a.


47/98


+ Nup 1:

1 1

1 1 1

111

1lim lim lim1 1

1

tt p p

p pt t t

khi p dx dx x t p

x x p p khi p

> = = = = Tch phn suy rng Riemann loi 1.

Vi 1 : phn k

Vi 1 > : hi t v1

1a

dx aI

x

+

= = .

V d 4: Xt s hi t v phn k ca tch phn:2

1

dxI

x

+

=

+.

Gii: +0

2 2 201 1 1dx dx dx

x x x

= +

+ + +

+ =0

2 201 1

t

t tt

dx dx lim lim

x x ++

+ + =01 1

0

t

tt tlim tan x lim tan x

+

= ( ) ( )

+ = + =

1 1lim tan lim tan2 2t t

t t .

V d 5: Xt s hi t v phn k ca tch phn a>( 0)

a.axI e bxdx

+= =

0

sin

b.axJ e bxdx

+=

0

cos =

II. Trng hp hm s tch phn khng b chn1. nh ngha: Cho hm s f(x) khng b chn trn on hu hn [a, b]. Gi s f(x)

b chn v kh tch trong on [a, b ], vi 0 > b tu v khng gii ni khi


48/98


x b (x = b gi l im bt thng). Khi 0

( ) lim ( )b b

a a

f x dx f x dx

= (3)c gi l

tch phn suy rng ca f(x) trn [a, b].

Nu gii hn l hu hn ta ni tch phn suy rng hi t, nu gii hn l v cng

hoc khng tn ti ta ni tch phn suy rng phn k.

Tng t, ta nh ngha cho tch phn suy rng vi x = a l im bt thng:

0( ) lim ( )

b b

a a

f x dx f x dx

+

=

V nu f(x) khng b chn ti c, a < c < b, x = c l im bt thng:

( ) ( ) ( )b c b

a a c

f x dx f x dx f x dx = +

TPSR v tri hi t khi v chkhi c 2 tch phn bn v phi hi t.

2. Cch tnh: Xt:0

( ) lim ( )b b t

ta a

I f x dx f x dx

= =

+ Dng cng thc Newton Leibnitz tnh: ( ) ( ) ( )b t

a

f x dx F b t F a

= vi 0t>

+ Tnh ( )0 0 0

lim ( ) lim ( ) ( ) lim ( ) ( )b t

t t ta

f x dx F b t F a F b t F a

= =

Tch phn I hi t khi 0lim ( )t F b t hu hn .

3. Mt s v d: Tnh cc tch phn suy rng sau

V d 6:1

0 1

dx

x

Gii:1

1 1 10 0 0

lim lim 2 1 lim 2 1 2 21 1

tt

t t t

dx dxx t

x x = = = + =

V d 7:dx

x=

1

21 1


49/98


V d 8: Chng minh rng tch phn1

0p

dx

x hi t nu hng s p < 1 v phn k nu

1p .

Gii: + Nu p = 1: [ ]

1 11

0 0 00

lim lim ln lim lntt t tt

dx dx x tx x+ + + = = = = , TP phn k.

+ Nu 1p :11 1 1 1

0 0 00

111

1lim lim lim1 1

1

p p

p pt t t

tt

pdx dx x t p

x x p p p

+ + +

Ta c pcm.

Tng qut: Tch phn suy rng( )

b

a

dxI

b x =

Tch phn suy rng Riemann loi 2.

+ Vi 1 : phn k

+ Vi 1 < : hi t v1( )

( ) 1

b

a

dx b a I

b x

= =

.

Ch : Cng nhi vi tch phn xc nh thng thng, vi tch phn suy rng

cng c th thc hin php i bin v ly tch phn tng phn.

V d 9: Tnh( )( )

b

a

dxx a x b , x = a v x = b l im bt thng.

+ t 2 2cos sin sin2 ( )x a b dx b a d = + =

2 2 2( )( ) ( ) sin cos ( )cos sinx a x b b a b a = =

+ Vi 0; / 2x a x b = = = = .

+ Khi /2

/2

00

2 2( )( )

b

a

dxd

x a x b

= = = .

III. Tiu chun so snh

nh l 1: Nu 0 ( ) ( ),f x g x x trong ln cn ca im bt thng b (b c th

bng v cng). Khi :


50/98


+ Nu ( )b

a

f x dx phn k th ( )b

a

g x dx cng phn k.

+ Nu ( )b

a

g x dx hi t th ( )b

a

f x dx cng hi t.

V d 10: Xt tch phn2

0

xI e dx

+= .

+ Ta c th vit2 2 2

1

0 0 1

:x x xe dx e dx e dx A B + +

= + = +

+2

1

0

xI e dx = : hi t

+ V2

1: 0 x xx e e < v

( )1 11 1

lim limt

x x t

t te dx e dx e e e

+

+ += = = : suy ra B hi t

+ Vy tch phn cho hi t.

nh l 2 Cho [ )( ) 0, ( ) 0 , ,f x g x x a b tc l b l im bt thng (b c th

bng v cng) sao cho:( )

lim , 0( )x b

f xk k

g x= < < + . Khi TPSR: ( )

b

a

f x dx v

( )b

a

g x dx cng hi t hoc cng phn k.

V d 11: Xt tch phn2

1 1

dxI

x x

+

=+

.

+ Ta c2

2

1lim 1

x

x x

x++

=

+ M21

dx

x

+

l tch phn Riemann vi 2 1 = > nn hi t.

+ Theo nh l 2 suy ra tch phn cho hi t.

Bi tp v nh: Trang 372 (bi 1-24)c trc cc mc:17.1, 16.1, 16.2, 16.3, 17.2 chun b cho Bi s 6Phng trnh tham s ca ng cong. H ta cc. ng cong trong ta cc


51/98


Bi s 6

PHNG TRNH THAM S. TA CC

I. PHNG TRNH THAM S

1. Dng: x f t t t ty g t

= =1 2

( ) ,( )

2. Vng cong dng tham s

V d 1. V ng congx t

y t

= =

2

3v tm

phng trnh vung gc.

GiiCch 1 :+ Ta tnh x v y vi mt vi gi tr ca t.

t -2 -1 0 1 2x 4 1 0 1 4y - 8 - 1 0 1 8

+ Ni nhng im ta sc ng cong.Cch 2 : Kho st s bin i ca x v y khi t bin i.

dy t dt t

dx tdt = =

23 32 2

.

T - 0 + y - - 0 + + x + +

0

y0 +

-

Ta c : y x y x = =2 3 3/2hay l phng trnh vung gc ca ng cong.

3. Cch thit lp v chuyn i gia phng trnh ng cong tham s v vunggc

+ Chn mt i lng trung gian m x, y u d dng tnh qua n.

+ Mt ng cong c th c nhiu phng trnh tham s nhng chc duy nht mt

phng trnh vung gc.+ Mt s gi chuyn ng cong dng tham s sang vung gc :

Coi x hoc y nh l tham s : y f ( x )= , chuyn thnhy f ( x )

x x

= =.

Dng h toc cc :x r cos

y r sin

= =.

Hnh 17.1

t =t1

P = (x,y)

t =t2


52/98


S dng nghing ca tip tuyn ti im ( )x y, nh mt tham s :

dyt

dx= .

V d 2. Vit phng trnh tham s Parabol x py=2 4 .

S dng nghing ca tip tuyn ti im ( )x y, nh mt tham s :dy

tdx

= .

T :dy dy x

x pdx dx p

= =2 4 hay ,2

Phng trnh tham s trong trng hp ny l :

x pt

y pt

= =2

2.

Xt tham s :

y

m x= l nghing ca ng thngbn knh ni im ( )x y, vi gc O. Khi : y mx= v x py pmx = =2 4 4

Phng trnh tham s:x mp

y mp

= =2

4

4.

V d 3. Mt vin n c bn ra khi tm ti thi im t= 0 vi tc ban uv0 ft/s (hay m/s) v c nh hng bi mt gc , v gi s rng chc lc hpdn tc dng ln vin n. Xem xt quo ca chuyn ng.

Gii : Gi to ca vin n l P(x, y).

Ta xt cc thnh phn x v y ca gia tc : ( )x ya a a= ,

Do lc hp dn hng xung ta c:

yxx y

dvdva a g

dt dt = = = = 0,

vi g s= 232ft/ (hay 9,80m s2/ ) l gia t c trng trng.

+ Suy ra: xy

v cv gt c

= = +

1

2

+ Khi t= 0 , ta c x

y

v v

v v

= =

0

0

cos

sin

x y

dx dy v v v gt v

dt dt = = = = +0 0cos , sin

+ Ly nguyn hm cho ta : ( ) ( )x v t c y gt v t c = + = + +210 3 0 42cos , sin


53/98


+ Khi t= 0 th x y= = 0 suy ra c c= =3 4 0 . Vy

( ) ( )x v t y gt v t = = +210 02cos , sin y l phng trnh tham s ca quo

vin n.

+ Kh tham s ta c:

( ) ( )x x gy g v x x v v v

= + = +2

202 2 2 2

0 0 0

1 sin tan .2 cos cos 2 cos

V d 4. Xt ng trn bn knh a v tm ti gc.

+ Phng trnh tham s :x a

y a

= =

cos

sin

+ Hoc s dng cng thc + =2 2cos sin 1, ta c :

x yx y a

a a+ = + =

2 22 2 2

2 21 hay .

V d 5. Elip:x y

a b+ =

2 2

2 21 c thc tham s ho nh sau.

Tx y

a b

+ =

2 2

1, nn tn ti mt gc sao cho :

x a

y b

= =

cos

sin.

V d 6. Vng cong( )

( )

x t

y t

= =

2

2

cos / 2

sin / 2, v tm phng trnh vung gc.

Li gii. T ( ) ( )t t + =2 2cos / 2 sin / 2 1, im ( )P x y= ,

chuyn ng trn ng thng x y+ = 1 (hnh 17.7). Nhngkhng ch x m y cng khng th m, v vy ta ch xt mtphn ca ng ny nm trong gc phn t th nht.

II. TA CC

1. Khi nim:

Cho n trc phn ny chng ta mi ch bit h to cc vung gc,trong ta t trong mt phng hai trc vung gc. Tuy nhin thng xy ra trnghp l ng cong xut hin mi quan hc bit vi gc to, nh l ng ica mt hnh tinh quay xung quanh quo ca n c xc nh bi lc hp dnca mt tri. ng cong nh vy c m t tt nht nh l mt chuyn ng


54/98


im m v tr ca n c ch r bi hng n gc to v khong cch ngc to. chnh xc l nhng g m ta cc s lm.

Mt im c hon ton xc nh bi: khong cch v hng ca n ngc to

+ Hngc ch r bng mt gc (tnh bng radian), c o t chiudng trcOx. Gc ny c m t theo chiu ngc chiu kim ng h nu dng v theo chiu kim ng h nu m, nh trong lng gic.

+ Khong cch c tnh bi khong cchnh hng ro t gc dc tiim cui ca gc .

Hai srv , vit theo th t (r, ) : gi l to ccca im . Hng =0 (hng dng trc Ox) c gi l trc cc.Miim c nhiu cp to cc:

+ Chng hn, im P trong c cc to cc l:

( )3, / 4 , ( ) ( ) + 3, / 4 2 , 3, / 4 4 ,Tc l nu ta cc ca im Pl : ( )r , th ta cng c cc ta cc l:

( )r k +, 2 .

Thut ngkhong cchnh hng tc l r c th l s m khi m vihng cho trc, khi ta chuyn ngc qua gc mt khong r theo hng

ngc li.

Ta c hai im: ( )Q = 2, / 6 v ( )R = 2, / 6 , rng hai im ny ixng nhau qua gc ta .

Hnh 16.3

Gi tr r= 0 chnh l gc, khng cn kn gi tr ca . Chng hn, cc cp( )0,0 , ( )0, / 2 , ( )0, / 4 u l to cc ca gc to.

2. Mi lin h gia to vung gc v to cc.

Khi bit r , , ta c:x r

y r

= =

cos

sin

2

2

R = (-2,

Q = (2, /6)

/

6

/6)


55/98


Khi bit xv yta cng c:r x y

y

x

= + =

2 2 2

tan

Khi s dng cc cng thc ny cn phi cn thn xcnh chnh xc ducar v chn thch hp vi gc phn t m im ( )x y, nm trn.

V d 7. To vung gc ca mt im l ( )1, 3 . Tm to cc ca imny.

Gii. Ta c r , tan = + = = 1 3 2 3 .

V im ny nm trong gc phn t th 2 nn ta chn r ,

= =2

23

.

Vy cp to cc l ( )2,2 / 3 . Mt cp to cc khc: ( ) 2, / 3 .

th ca phng trnh cc : F r =( , ) 0 : l tp hp tt c cc im ( )P r = ,

sao cho to cc ny tho mn phng trnh.

+ M mt im ( )P r = , c nhiu cp to khc nhau nn Pnm trn thnu mt cp to bt ktrong cc to ca im tho mn phng trnh.

V d 8. Chng minh rng im ( )1, / 2 v im ( )0, / 2 u nm trn th

ca r = 2sin .

3. ng trong ta cc.

Xt hm s trong ta cc dng: ( )r f = + Nu hm ( )f l mt hm n gin, th ca n kh d v :

- Ta chn mt dy cc gi tr ca ,

- Mi gi tr xc nh mt hng t gc ta tnh ton gi tr tng ng ca r.

V d 9. Vng c phng trnh = , trong l mt hng s.Gii: th l mt ng thng qua gc to v to vi trc dng x mt gc .

(-1, 3)

32

1

2/3

/3


56/98


V d 10. Vng c phng trnh r a= , trong a l mt hng s dng.

Gii: th l mt ng trn tm ti gc v c bn knh bng a.

Hnh 16.5 Hnh 16.6

V d 11. Ch ra rng ng c phng trnh r = 2cos biu din mt ngtrn.

+ kim tra kt qu: Ta c r = 2cos suy ra

( )r r x y x x x y x y = + = + = + =22 2 2 2 2 22 cos , 2 , 2 0, 1 1.

y l mt ng trn tm ( )1,0 v bn knh bng 1.

Nhn thy vic nhn 2v ca pt vi r, lm cho gc to lunnm trn th. Tuy nhin, trong pt ny gc to nm trnpt nn vic nhn ny khng nh hng n th hm s.

Chuyn p/trnh cc v p/trnh trong h to vung gc.

=

r =

r

r


57/98


Ta s dng mi lin h:

r x y

y

r

x

r

y

x

= + = = =

2 2 2

sin

cos

tan

4. th ca mt s phng trnh cc dng r f ( )=

V d 12: Vng cong cardioid ( )r a = +1 cos , vi a > 0 (ng hnh tim).

Vit phng trnh ng cong ny trong h to vung gc.

Gii:

+ Hm cos tun hon vi chu k 2 nn ta chcn ly [ ], 0 2 .

+ Tnh to ca mt sim

0 /2 3 /2 2

r 2a 2a

a a

0

C th s dng tnh cht ca hm cos , i xng qua trc Ox, nn ta chcn v th trong [ ], 0 ri ly i xng qua trc Ox.

+ i sang h vung gc.

S dng mi lin h:y x y

r x yr r x

= + = = =2 2 2, sin , cos , tan .

Ta c:

( )x

r a r a r x x y ax ar r

= + = + + = 2 2 21 ,

+ Cui cng: ( ) ( )x y ax a x y + = +22 2 2 2 2 .

V d 13. Vng ng c sn (limacon hay

snail) ( )r a = +1 2cos vi a> 0 .

Gii:

+ th i xng qua trc Ox.

+ Xt khi r = 0 tc l = 2cos 1 = 2 / 3 .

Hnh 16.9

r = a

r = - a

r = 3a

= 43


58/98


0 /2 2/3 4 /3 3/22

r

3a 2a

a a

0 0- a

+ Khi tng, r gim, bng 0 khi = 2cos 1, tc l khi = 2 / 3 .

+ Khi tng lin tc n , r gim lin tc t 0 n gi tr m a , v ta vc na di ca vng trong c chtrong Hnh 16.9.

+ Khi tng lin tc qua gc phn t th ba v th t, r nhn li gi tr ca nvi v tr o ngc; vng trong hon thnh ti = 4 / 3 , v vng ngoi hon thnhti = 2 .

Ch :+ th ca ng cong r f (cos )= (r l mt hm ca cos ) i xng

qua trc Ox do c ( ) =cos cos . Tng t, nu r f (sin )= th ng cong i

xng qua trc Oy.

+ Phng trnh dng ( )r f =2 :

- Nu l mt gc sao cho ( )f < 0 th khng c mt im no trn ngcong.

- Nu gc l mt gc sao cho ( )f > 0 th c hai him tng ng trn

ng cong vi ( )r f = . Cc im ny c khong cch bng nhau ti tmnhng hng ngc nhau, v vy th ca ( )r f =2 lun lun i xng qua gcto.

V d 14. ng cong r a =2 22 cos2 c gi l mt ng lemniscate.

+ Hm s tun hon vi chu k .

+ Vi mi c hai gi tr ca r: r a = 2 cos2 . (1)

0 /4 /2 3/4

r a 2 0 || 0 a 2

+ Khi tng t 0 n /4 , 2 tng t 0 n /2 nn cos 2 gim t 1 n 0.


59/98


+ Hai gi tr r trong (1) ng thi v nn hai phn ca ng cong trong Hnh16.10.

+ Khi tip tc tng qua na th hai ca gc phn t th nht v na th nhtca gc phn t th hai, 2 bin i qua gc phn t th hai v th ba v cos2m, v vy khng c th cho tp im ny.

Qua na th hai ca gc phn t th hai, cos2 li dng, v hai gi tr ca rcho bi (1) ng thi hon thnh hai vng bt u bn tri hnh v.

Hnh 16.10

V d 15.ng cong r a = sin2 vi a> 0 c gi l mt bng hng bncnh.

+ Hm s tun hon vi chu k , v thi xng qua Oy.

0 /4 /2 3/4

r a - a

0 0 0

+ rng khi tng t 0 /4, 2 tng t 0 /2v r tng t 0 n a;

+ Khi tng t / 4 n / 2 , 2 tng t / 2 n v rgim t 0 n a.

+ T cho ta l th nht trong gc phn t th nht (Hnh 16.11).

+ Gi tr ca gia / 2 v ( 2 gia v 2 ) cho gi tr r m, ta vc ltrong gc phn t th t;

+ gia v 3 / 2 ( 2 gia 2 v 3 ) cho gi tr r dng, ta vc l tronggc phn t th ba;

+ gia 3 / 2 v 2 ( 2 gia 3 v 4 ) cho gi tr r m, ta vc l tronggc phn t th hai.

= /4

r = 2a

2ar =

= 3/4

Hnh 16.11

= /4

r = a


60/98


III. PHNG TRNH CC CA CC NG TRN, NG CNIC V CCNG XON C

V d16: Xt mt ng trn vi tm ti ( )a,0 v bn knh a

( )x a y a x y ax + = + =2 2 2 2 2hay 2 (2)

+ T x y r+ =2 2 2 v x r = cos , phng trnh tr thnh r ar =2 2 cos ,

+ Tng ng vi : r a = 2 cos (3)

V tm r= 0nm trn th ca (3).

+ Mt phng php khc xut pht t mt s tnh cht hnh hc c trng cang cong. Trong ng trn xt, ta rng tam gic OPA trong hnh bnphi l tam gic vung. T tam gic OPA vung vi r l cnh bn k vi gc nhn, hin nhin ta c

r a = 2 cos .

Hnh 16.14

V d 17. Tm phng trnh cc ca ng trn vi bn knh a v tm ti imC c to cc l ( )b, , trong b l s dng.

Li gii:+ Ly ( )P r = , l mt im tu trn ng trn,

+ p dng nh l cosin cho tam gic OPC ta c

( )a r b br = + 2 2 2 2 cos .

Cc trng hp c th:

+ ng trn i qua gc to, ta c b a= , khi phng trnh ng trnl:

( )r a = 2 cos

P = (x,y)

a

(a,0)

P = (r, )

r

0 2a A


61/98


Hnh 16.15

+ Tm nm trn trc Ox, ta c = 0 , phng trnh tng ng: r a = 2 cos .

+ Tm nm trn trc Oy, ta c = / 2 , phng trnh tng ng: r a = 2 sin .

V d 18. Tm phng trnh cc ca phn ng cnic vi tm sai e nu tiuim ti gc to v ng chun tng ng l ng thng x p= nm bntri gc to.

Gii. + Tiu im , ng chun, tm sai ca phn cnic l

PFe PF e PD

PD= =hay . . (9)

+ ng cong l elip, parabol hay hyperbol tuthuc theo e e< =1, 1 e>hay 1.

+ Kim tra hnh v ta thy PF r= vPD QR QF FR p r = = + = + cos ,

do t (9): ( )r e p r = + cos .

+ Gii phng trnh theo r, thu c :

epr

e =

.

1 cos(10)

y l phng trnh cc ca conic ta ang xt.

V d 19. Tm phng trnh cc ca ng cnic vi tm sai 13 , tiu im ti gcto v ng chun x= 4.

Li gii. + Ta thay e p= =13 , 4 vo phng trnh (10) ta

c :( )

r

= =

13

13

4 41 cos 3 cos

.

P = (r, )

a

C = (b, )

rb

0

Q

D

x = -p

R

p


62/98


+ ng cong ny l elip.

epr

e =

+.

1 cos

V d 20. Ccng xonc ca Acsimet: Phng trnh cc ca ng xon cl

a. r a= b.a

r r a

= =hay ,

trong a l mt hng s dng.

Hnh 16.22

Gii:a. Do r t l thun vi , bt u t gc, t 0 tng theo ngc chiu kim ng ht v tr ban u dc theo trc cc.Nu < 0 th ng xon c vn nh vy.

b. Dosin

y r sin a a

= = khi 0 nn y = a l tim cn ngang ca ng

cong.+ = 0 th r khng xc nh.+ tng n v cng th gim ti 0.iu ny cho thy ng cong cun quanh gc theo ngc chiu KDH v hn

vng tht dn khi tng v hn.

Bi tp v nh: Trang 515( bi 1 - 9), 519 (bi 1 - 10)c trc cc mc: 7.2, 16.5, 7.3, 7.4, 7.5, 16.4, 7.6, chun b cho Bi s 7

ng dng ca tch phn

e

P

y = a

H nh 16.21

r


63/98


Bi s 7

NG DNG CA TCH PHN XC NH

1. Din tch hnh phng

a. Din tch hnh thang cong

Xt hnh thang cong c gii hn bi :( ) 0

0

y f x

a x b

y

=

.

+ Din tch mi hnh ch nht nh : dA = ydx = f(x) dx :c gi l vi phn din tch.

+ Din tch A ca ton min l tng (lin tc) cc thnh phn din tch dA, khi xtng tan bta c:

( )

b

a

A dA ydx f x dx= = = .

b. Din tch gia hai ng cong

Tnh din tch min gii hn bi( )

( )

y f x

y g x

a x b

=

=

+ Vi phn din tch l dA =[f(x) g(x)]dx

+ Din tch ton min l: ( ) ( )

b

a

A dA f x g x dx= = Tnh din tch bng tch phn xcnh:Bc 1: V min cn tnh din tch, xc nh cc ng bin ca min v tm

to giao im ca chng.Bc 2: Chn vi phn din tch theo:

+ hoc di thng ng vi chiu rng dx+ hoc di nm ngang vi chiu rng dy.

Bc 3: Tnh ra vi phn din tch dA, biu thdA theo bin xhoc y.Bc 4: Ly tch phn dA theo cc cn ca xhoc y.

V d 1. Tnh din tch min phng c gii hn bi ccng

y = x2v y = 4.

Cch 1:Dng di thng ng: x bin thin t - 2 n 2

+ Chiu di ca di l 4 x2


64/98


+ Din tch ca di l: dA = (4 x2)dx.

+ Vy din tch ton min l:

( )3

32

3

88

3

88

344

2

2

32

2

2 =

+

=

=

x

xdxx (vdt).

Cch 2 :Dng vi phn din tch nm ngang: khi y bin thin t 0 n 4

+ Chiu di ca di l y ( y ), nn dA = 2 y dy

+ Din tch ton min l:3

322

4

0

= dyy (vdt)

V d 2. Tm din tch min phng gii hn bi cc ng y = 3 x2v y = x + 1.

+Tm giao im ca cc ng :

+ Chiu di ca gii thng ng :

+ Din tch min phng :

=1

42

(vdt)

c. Din tch hnh phng trong ta cc.

Bi ton : Tm din tch A ca mt min b chn bi mt ng cong( )r f= v hai na ng thng = v = .

Phn tvi phn din tch : dA l din tch ca qut mng vi bn knh r v

gc tm l d : 21

2dA r d =

3 y = x +1

dx

( 1, 2 )x

x 1

y

( -2, -1 )

-2

dy

y = 3- x.x

d

=2

1 r2d


65/98


Din tchA l: 21

2 A dA r d

= =

V d 3. S dng tch phn tm din tch hnh trn 2 cos .r a=

Gii. + ng trn c bin thin t / 2 n / 2 .

+ Do tnh i xng ca hnh trn nn:/2

2

0

1

2 A r d

= =

2a= (vdt).

V d 4. Tm din tch min c gii hn bi ng lemniscate 2 22 cos 2r a= .

Gii. + T tnh i xng, ta tnh din tch ca gc phn t th nht ri nhn vi 4:/4 2

0

14

2 A r d

= =

22 .a=

Ch :S dng tnh i xng tnh din tch s lm cho bi ton n gin hn vtrnh c sai xt, nht l khi ly cn ca .

V d 5. Tm din tch phn trong ca ng trn r = 6a cos v phn ngoi cang hnh tim ( )2 1 cos .r a= +

2a

= /4

a2


66/98


Gii.

+ To giao im: ( )1

2 1 cos 6 cos cos2 3

a a

+ = = =

+ Phn t din tch l:

( ) ( )

22 2 2 2 2 2 2 2

1 2 1 2

1 1 1 136 cos 4 1 cos 2 8cos 1 2 cos .

2 2 2 2dA r d r d r r d a a d a d = = = + =

+ Do tnh i xng:

( ) ( )

[ ]

/ 3 / 32 2 2

0 0

/ 32 2

0

2 2 8cos 1 2 cos 4 4 1 cos 2 1 2 cos

4 3 2sin 2 2sin 4

A a d a d

a a

= = +

= + =

2. Tnh th tch vt th trn xoay.

a.Tnh th tch theo phng phpa :

y = f (x)

dx

a b

x

y = f(x)

ba

x

dx

Min phng gii hn bi y = f(x), x = a, x = b quay quanh trc Ox s to nnmt vt th ba chiu, c gi l vt th trn xoay.

Vi mi di rt mng thng ng vi dy dx v y nm trn trc x to nnmt a mng hnh ng xu vi bn knh y = f(x). V a l hnh tr nn vi phn thtchdV :

dV = y2 dx = . [ f (x)]2 . dx

2a

4a 6a

= /3


67/98


Khi xbin thin t a n b ta c th tch: ( )22

b

a

V dV y dx f x dx = = =

.

V d 6. Tnh th tch hnh cu to bi na ng trn y a x=

2 2

quay quanhng knh ca n.

Gii :+ Vi phn th tch l: 2 2 2dV y dx a x dx = =

+ Th tch hnh cu :

( ) ( )3

2 2 2 2 2 3

0 0

42 2

3 3

aa a

a

xV a x dx a x dx a x a

= = = =

V d 7. Tnh th tch hnh nn chiu cao h, bn knh y r.

Gii :+ Hnh nn l vt th trn xoay c to thnh do quaymt tam gic vung.

+ Cnh huyn ca tam gic vung : y =h

rx,

+ Vi phn th tch l dV=

+ Th tch cn tm l :

213

V r h= = .

i vi vt th c hnh dng khc: khi vi phn th tch c th khng phi la trn.

+ Ta ct vt th bi mt phng vung gc vi mt trc th tit din l mt tam gic,mt hnh vung hoc mt hnh no m ta c th tnh c din tch tit din .

+ Khi vi phn th tch dVl th tch tit din vi dy lt ct mng, do cth tnh c th tch vt th theo phng php dch chuyn lt ct.

V d 8. Tnh th tch ci nm c ct ra t hnh tr vi bnknh y abng mt phng i qua mt ng knh y tr v tovi y tr mt gc 450.

Gii :+ Ct vt th bi mt phng vung gc vi ng knh,tit din l mt tam gic vung cn.

+ Th tch lt ct l:

Hnh 7.8


68/98


dyyayadV2222

2

1= ,

+ Th tch ca vt th l :

( ) 30

32

0

22

3

2

32

12 a

yyadyyaV

aa

=

== (vtt).

Ch : Xt mt di quay quanh mt trc nhng cch trc mt khong no thc mt a vi l hng trong, ging nh ci bn ra. Vi phn th tch l th tchca a tri tri th tch l hng:

( )2 21 2dV y y= Th tch ca n l :

( )2 21 2b

a

V dV y y dx= = ,

trong y1, y2 ln lt l bn knh ngoi v bn knh trong ca bn ra.

b. Tnh th tch bng phng php v:

Bi ton: Tnh th tchvt th trn xoay do min phng( ) 0

0

y f x

x b

= >

quay mt

vng xung quanh Ox.

Di mng c mt cnh dx v mt cnh y quay quanh trc Oy s to ra mt v

hnh tr mng.+ Vi phn th tch dVca v ny l dV = 2xydx.+ Khi bn knh xca v ny tng tx = 0n x = b, c th thy rng th tch ca

vt th l tch phn ca vi phn th tch dV:

0

2 2 . ( )b

V dV xydx x f x dx = = =


69/98


V d 9. Tnh th tch do hnh phng c gii hn bi2

22

y x

y x

=

= trong gc phn t

th nht quay quanh trc Oy.

Gii:+ Chiu cao ca v l: y =(2 x2) x2= 2 2x2,

+ Thnh phn th tch:

dV = 2xy dx = 2x(2 2x2) dx = 4(x x3)dx

+ V honh giao im ca cc ng l x = 1,nn:

( )1

2 413

0 0

4 42 4

x xV x x dx

= = =

V d 10. Mt vt thc to thnh bng cch khoanmt l hnh trng knh a c trc xuyn tm hnh cu bnknh a. Tnh th tch vt th ny.

Gii :

+ Hnh cu c to thnh khi cho na ng trn

y a x= 2 2 quay quanh ng knh ca n (tc trc Oy),bng phng php v ta c :

dV =

+ x bin thin :

332

V a= =

.Ch : C th dng theo phng php bn ra nhng trong trng hp nyphng php v tin hn nhiu.

3. Tnh di cung phnga. di cung trong h to vung gc

Bi ton :Tnh di ca mt ng cong l phn th ( ) : ( )C y f x==== ni tim A n im B.


70/98


Gi s l di bin thin t A n im di ng trnng cong.Gi ds l vi phn cung; dx, dy l s gia tng ng

theo x v y. Xem ds nhn mc phn cung cong nhl mt on thng.

Tnh l Pitago, ta c ds2= dx2+ dy2, t22

2 2 2

2

1 1dy dy

ds dx dy dx dxdx dx

= + = + = += + = + = += + = + = += + = + = +

di ton bng cong AB l tch phn ca ccvi phn cung ds khi ds bin thin dc theo cung, t An B:

di cung AB bng dxdx

dyds

b

a

+=

2

1

Ch :Nu xem x l hm ca hm y. Khi :

dydy

dx

dydy

dx

dydxds

2

2

2

222

11

+=

+=+=

v di ca cung AB l:

2

1

d

c

dxl ds dy

dy

= = +

.

V d 11: Tm di ca ng congx

yx

= +4

2

34 4

vi x 1 2

Gii: +dy dy

x x xdx x dx x x

= + = + = +

2 23 3 3

3 3 3

1 1 11 1

4 4 4.

+ Ta c di cung :dy x

s dx x dx dx x x

= + = + = = 222 2 4

33 2

1 1 1

1 1 1231

4 4 8 32

(vd).

Bd

cA

s

ds

dy

dx

a x b


71/98


b. di cung trong ta cc: Xt mt ng cong c phng trnh cc l( )r f= , gi s l di cung o dc theo ng cong t mt im xc nh theo

mt hng xc nh. Do2 2 2

ds dx dy= +

+ Nhng cosx r= v siny r= nn :

222 2 2 2 2 2

2

dr dr ds r d dr r d r d

dd

= + = + = +

.

+ di cung ca mt ng cong cc vi :

2

2 .dr

s ds r d d

= = +

V d 12. Tm tng chiu di ca ng cardioid ( )1 cosr a= .

Gii. + Ta c sindr ad

=

nn

( ) ( )dr

r a cos a sin a cos a sin d

+ = + = =

222 2 2 2 2 2 21 2 1 4

2

+ V1

sin 02

vi 0 2 , nndr

ds r asin d

= + =

22 2

2

( )22

00

2 sin 4 cos 4 4 8 .2 2

s ds a d a a a a

= = = = =

c. di cung tham s:

di cung: vi p/t: 1 2( )

,( )

x x t t t t

y y t

====

====ta c 2 2'( ) '( )ds x t y t dt = += += += + , do di

cung l:

2

1

2 2'( ) '( )

t

t

s ds x t y t dt = = += = += = += = + .

4. Din tch mt trn xoay


72/98


a. Din tch mt trn xoay trong ta vung gc

Bi ton: Tnh din tch mt trn xoay to bi ng cong trn( ) : ( ),C y f x a x b= = = = nm pha trn trc honh khi quanh xung quanh trc Ox.

+ Nu cc vi phn cung dsquay quanh trc Ox, n s to ra vi phn din tch dAc hnh dng mt di ruy bng.

+ Nu khong cch t trung im dsti trc Oxl yth vi phn din tch s l :2

2 2 . 1dy

dA yds y dxdx

= = +

+ Din tch mt trn xoay l tng cc vi phn din tch dA :2

2 2 . 1b

a

dy A dA yds y dx

dx

= = = +

, trong y = f(x).

Nu ng cong quanh quanh trc Oy, bng cch tng t, din tch mt trnxoay c to ra l :

2 A xds= .

Cng thc tng qut:A = 2 (bn knh mt trn xoay). ds

V d 13. Tm din tch mt cu bn knh bng a.

Gii: + Mt cu c xem l mt trn xoay do na ng trn 22 xay = quayquanh Ox.

+ V ( )22

2/122

xa

xxa

dx

d

dx

dy

== nn

2 22 2 2

2 2

0 0 0

2 2 2 1 4 1 4 4

a a ady x

A yds y dx a x dx adx adx a x

= = + = + = == = + = + = == = + = + = == = + = + = =

(vdt).

Nu xem y l bin ly tch phn:

+ Ta coi ng cong trong gc phn t th nht: 2 2 x a y= = = =


73/98


+ Khi : (((( ))))1/ 2

2 2

2 2dx d ya - y

dy dy a y

= == == == =

+ Din tch mt cn tm l (coi yl bin ly tch phn) :2

2

2 20 0

2 2 2 1 4 4

a adx ydy

A yds y dy a ady a y

= = + = == = + = == = + = == = + = =

vdt

b. Din tch mt trn xoay trong ta cc:

Phn t din tch mt trn xoay trong ta cc :Cung ds sinh ra mt phn t din tch mt: 2 ,dA yds= trong

2 2 2sin ,v y r ds r d dr = = +

do : 2 2 2 4 2 2 22 sin 2 sin .dA r d dr r d r dr = + = +

Khi din tch mt l : S dA====

V d 14. Tm din tch ca mt trn xoay sinh bi ng lemniscate2 22 cos 2r a= quay quanh trc x

Gii. : + T p/trnh ng cong ta c:22 sin 2 ,rdr a d =

Do vy:

( )4 2 2 2 4 2 4 2 2

4 2

4 cos 2 4 sin 2

4

r d r dr a a d

a d

+ = +

=

v 24 sindA a d = .+ Din tch

( )/ 4/ 4

2 2 2 2

0 0

22 4 sin 8 cos 8 1 4 2 2

2 A a d a a a

= = = =

(vdt)

Bi tp v nh: Trang 212(1 - 4), trang 219( bi 1 - 18), trang 222 (1 - 13), Trang226 (bi), trang 229(), trang 233(), trang 534(bi 5 - 17), trang 540(bi 1 - 12).Cc bi Cc bi 1-5(Tr.209),1-4(tr.213), 9,10(tr.216), tr.219,1-14(tr.539), 16-25(tr.

544), tr.220; tr. 226; tr.229; 10-19(tr.534),tr.233,5-6(tr. 534); tr.347.c trc ccmc:14.1, 14.2, 14.3, 14.4, 14.5, 14.6, 14.7 chun b cho Bi s 8: Chui s

r

y

ds


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Bi s 8

CHUI S.

I. S LC V DY S

a.nh ngha: Xt hm s*:

( )

f

n f n

t { } : ( )n n x x f n= , tc l vi mi s nguyn dng n, c tng ng mt s

xn, khi ta ni rng hnh thnh mt dy s:x1, x2, , xn, .,

v thng vit chng di dng {xn}, xn l phn t thn(cn gi l s hngtng qut).Ch :Dy s c thc cho di mt trong cc cch:

+ M t+ Cho cng thc s hng tng qut+ Cho di dng cng thc quy np.

V d 1: Xt chui s {xn} bit phn t thn:( ) 1

na x = , cc s hng : 1, 1, 1; 2/)1(1)( nnxb = , cc s hng :

1,0,1,0;

nxc n /1)( = , cc s hng: ,...4

1,

3

1,

2

1,1 nnxd n /)1()( = , cc s hng :

,...4

3,

3

2,

2

1,0

nxen

n /)1()(1= , cc s hng l: ...;

4

1,

3

1,

2

1,1

=nxf)( 12

1...

4

1

2

11

++++

n

nxg n

1...

4

1

3

1

2

11)( +++++=

n

nn

xh

+=

11)(

Mt dy s trong tt c cc phn tu bng nhau, c gi l dyhng.Mt dy s {xn} c gi l bchnnu tn ti hai s A v B sao cho Axn

Bvi mi n, khi Ac gi l cn di v B c gi l cn trn cady s.Trong V d 1:

+ Cc dy (a) n (f) l dy s b chn,

+ Dy (g) khng b chn ( 1 1 1 1 13 4 4 4 2

+ > + = , 1 1 1 1 1 1 1 1 15 6 7 8 8 8 8 8 2

+ + + > + + + = .

+ Dy s (h) cng b chn, nhng n khng d dng kim tra.


75/98


Mt dy s{ }nx c gi l dy tng nu

x1x2x3 xnxn+1 .

b. nh ngha: Mt dy s { }nx c cho l c gii hn L nu vi mi s

dng tn ti mt s nguyn dng n0 tho mn :| xn- L|


76/98


2

0

1 ...n

n

x x x

=

= + + +

Gii :

+ Tng ring th n ca chui ny c vit di cng thc sau :

12

n

1s 1 ...

1

nn x

x x xx

+= + + + + =

, nu x 1.

+ Nu | x| < 1, chng ta c1

1ns

x

, do vy : 2

11 ... ...

1

n x x x

x+ + + + + =

+ Nu 1x chui phn k.

V d 2. Tnh tng1

1 1 1 1...

( 1) 1.2 2.3 3.4n n n

=

= + + ++

+ Ta thy rng: 1 1 1( 1) 1n n n n

= + +

+ Do tng ring thnnh sau :1 1 1 1 1 1 1

... 11 2 2 3 1 1

nsn n

= + + + =

+ +

+ T: lim 1nn

s+

= . Hay1

11

( 1)n n n

=

=+

.

2. Tnh cht v cc php ton

i. Nu :nn k

a s

=

=

v

nn k

b t

=

=

khi :

( )n nn k

a b s t

= = v n

n k

a s

=

= vi l hng s.

ii. S hi t hay phn k ca mt chui s s khng thay i nu ta bimt s hu hn cc s hng ca chui s.

Phn d: Gi s chui nn ka s

== (hi t), khi

1 2

1

....n n n k k n

R a a a+

+ += +

= + + = c gi l phn d v khi lim 0nn

R+

= .

3. iu kin cn ca s hi t.Nu chui

1 nna

= hi t th ta c lim 0nn a = . iu ngc li khng ng.

Tuy nhin nu lim 0n

na

th chui

nn ka

= s phn k.

V d 3: 12nn k

= hi t v ta nhn thy1

lim lim 02

n nn na

= = .


77/98


+ Nhng chui1

1n n

= phn k (s CM mc sau) mc d1

lim lim 0nn n

an

= = .

+1nn

= phn k v lim limnn na n = = + .

III. KHO ST S HI T CA CHUI S.

1. Chui s dng : Xt chui s dng :

1 21

..., 0,...nn n

n

a na a a a

=

> = + + + .

a. Tiu chun bchnTa c ssn+ an+1 = sn+1i vi mi n, v do cc phn t ca sncu thnhmt dy s tng. iu ny suy ra chui { }ns cc tng ring hi t khi v ch

khicc phn t ca sn b chn trn.

V d 4. Xt chui siu ho :1

1 1 11 ...

2 3n n

=

= + + + Gi m l mt s nguyn

dng v chn n> 2m+1. Khi

( )

1

1

1

1 1 1 11 ...

2 3 4 2

1 1 1 1 1 1 11 ... ... ...

2 3 4 5 8 2 1 2

1 1 1 1 12. 4. ... 2 . 1

2 4 8 2 2

n m

m m

m

m

s

m

+

+

+

> + + + + +

= + + + + + + + + + +

+

> + + + + = +

iu ny chng t rng cc phn tsnkhng b chn v do chui phn k.

Mt s chui - hi t hoc phn k c th to ra t chui iu ho, bngcch xo mt s phn t da trn mt quy lut nht nh.

+ V d nu b tt c cc phn t ngoi tr cc phn t c m bng 2, th

phn cn

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