Antiderivatives and Indefinite Integration 5

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Thisquestionhasseveralpartsthatmustbecompletedsequentially.Ifyouskipapartofthequestion,youwillnotreceiveanypointsfortheskippedpart,andyouwillnotbeabletocomebacktotheskippedpart.

TutorialExerciseFindthegeneralsolutionofthedifferentialequationandchecktheresultbydifferentiation.

Part1of5

Whensolvingadifferentialequation, itisconvenienttowriteitintheequivalentdifferentialform Tofindthegeneral

solution,we integrate integrate bothsides.

Thus,

Part2of5Integratebothsidesandusethepowerruleontherightside.

Part3of5

Thus,thegeneralsolutionof is

Part4of5

Tocheckfortheresult,wehavetoprovethat

Part5of5Thus,the derivative derivative of istheoriginalfunctionYouhavenowcompletedtheMasterIt.

1. 13/13points|PreviousAnswersLarsonET55.1.005.MI.SA.

AntiderivativesandIndefiniteIntegration5.1(Homework)KhaledAlmarri[2015FA]MATH265DL1,sectionDL1,Fall2015Instructor:GinaRomano

WebAssign

=8t3dydt

=8t3,dydt

dy=8t3dt.

= dy dy

$$8t3

dt

y = 8 +C

= 2 2 t +C

t4 4

4 4

4 4

=8t3dydt

y=$$2t4+C

.

2t4+C =8t3.ddt

2t4+C = 2 4 4 t

= 8 8 t

ddt

3 3

3 3

2t4+C 8t3.


TutorialExerciseFindthegeneralsolutionofthedifferentialequationandchecktheresultbydifferentiation.

Part1of5Rewritethedifferentialinequivalentform Tofindthegeneralsolution, integrate integrate bothsides.Thus,

Part2of5Usethepowerruleontherightsidetoevaluatetheintegral.

y =

y =

Rewriteyinthemoreconvenientform,

Part3of5

Thus,thegeneralsolutionof is

Part4of5

Tochecktheresult,differentiateyouranswertoprovethat

Evaluatetheleftsideoftheequation.

Part5of5Thusthe derviative derviative ofyisequaltotheoriginalfunction.YouhavenowcompletedtheMasterIt.


=5x6dydx

dy=5x6dx.

=dy 5 5 x dx.6 6

5 +Cx5 5

5 5

x5+C

y= +C. 1 1

x5 5

=5x6dydx

y=$$1x5+C

.

+C =5x6.ddx

1x5

+C = (1)( 5 5 )x

= 5x6

ddx

1x5

6 6

Completethetable.(UseCfortheconstantofintegration.)

OriginalIntegral Rewrite Integrate Simplify

+C

3. 3/3points|PreviousAnswersLarsonET55.1.012.

Findtheindefiniteintegralandchecktheresultbydifferentiation.(UseCfortheconstantofintegration.)

x66+8x+C



14x4+C



14x(12)+2x(32)3+C


dxx(x3+5) dxx4

+5x

x55+5x22+C

2x3+25

x210

(x5+8)dx

dx1x5

dxx+7

x


TutorialExerciseFindtheindefiniteintegralandchecktheresultbydifferentiation.

Part1of4Expandtheintegrand

=

=

Part2of4Nowobtaintheintegral.(UseCfortheconstantofintegration.)

=

=

$$5t520t3+36t+C

Part3of4

Tochecktheresult,provethat


Part4of4

Thus,the derivative derivative of isequaltotheoriginalfunction, Thisconfirmsthattheresultis

correct.YouhavenowcompletedtheMasterIt.


(5t26)2dt

(5t26)2.

(5t26)2 (5t26)(5t2 6)

25 25 t4 60 60 t +362 2

dt(5t26)2 dt25 25 t4 60 60 t +362 2

5t520t3+36t+C = 5t26 .ddt

2

5t520t3+36t+C

= 5( 5 5 )t 20( 3 3 )t2+ 36 36

= 25t 60t +36

= ( 5 5 t 6)2

ddt

4 4

4 4 2 2

2 2

5t520t3+36t+C 5t26 .2



Part1of3

Tofindtheindefiniteintegral,findtheantiderivativeofthegivenintegrand

Usethepowerruleofintegrationtointegratethefirstterm,andthebasicruleofintegrationoftrigonometricquantitiestointegratethesecondterm.(UseCfortheconstantofintegration.)

$$44+tan()+C

Part2of3Tocheckfortheresult,youhavetoprovethat


Part3of3Thus,theleftsideoftheequationis equalto equalto therightside.YouhavenowcompletedtheMasterIt.



3tan()+8cos()+C


(3+sec2)d

.3+sec2 d

3+sec2 d

4+tan+C =3+sec2.dd

14

4+tan+C = ( 4 4 ) +sec2

=3+sec2

dd

14

14

3 3

3(sec())28sin() d



Part1of4First,solvethedenominatoroftheintegrand.Recallthat

Part2of4Therefore,rewritetheintegrand.

Rewritetherightsideasaproductoftwofractionsandevaluatetheintegral.(UseCfortheconstantofintegration.)

Part3of4

Tocheckfortheresult,provethat


10.10/10points|PreviousAnswersLarsonET55.1.040.MI.SA.

dxcosx1cos2x

sin2x+cos2x= 1 1 .

=dxcosx1cos2x

dx

cosx$$sin2(x)

=

=

=

$$csc(x)+C

dx1sinx

cosx$$sin(x)

($$csc(x)

)(cotx)dx

cscx+C = .ddx

cosx1cos2x

Part4of4Thus,theleftsideoftheequationis equalto equalto therightside.YouhavenowcompletedtheMasterIt.

Findtheequationofy,giventhederivativeandtheindicatedpointonthecurve.y=$$x22x+2

dy

dx=2(x1)

11.1/1points|PreviousAnswersLarsonET55.1.054.

cscx+C =

cscx

$$cot(x)

=

=

=

ddx

1$$sin(x)

cosxsinx

cosx$$sin2(x)

cosx1

$$cos2(x)


TutorialExerciseSolvethedifferentialequation.

Part1of4Tosolvethedifferentialequation, integrate integrate bothsideswithrespectto x x .

Part2of4Obtaintheintegralonbothsides.

Part3of4Substitutethevaluefor whenx=0intheexpressionforthefunction

ObtainthevalueofC.

Part4of4Thus,thesolutionof satisfyingthecondition is

YouhavenowcompletedtheMasterIt.


f'(x)=7x,f(0)=6

=

f(x) = 7/2 3.5 x +C

f'(x)dx 7 7 xdx

2 2

f(0), f(x).

f(0)=6=3.5( 0 0 ) +C2 2

C= 6 6

f'(x)=7x f(0)=6

f(x)=$$3.5x2+6


TutorialExerciseSolvethedifferentialequation.

Part1of4Tosolvethedifferentialequation, integrate integrate bothsideswithrespectto s s .

Part2of4Obtaintheintegralonbothsides.

Part3of4Substitutethevaluefor when intheexpressionforthefunction

ObtainthevalueofC.

Part4of4Thus,thesolutionof satisfyingthecondition is

YouhavenowcompletedtheMasterIt.


f'(s)=16s16s3,f(3)=2

=

= 8s 4s +C

f'(s)ds ( 16 16 s16s )ds3 3

2 2 4 4

f(3) s=3 f(s).

f(3) = 2= 8( 3 3 ) 4( 3 3 ) +C

2= 72 324 324 +C

2 2 4 4

C= 254 254

f'(s)=16s16s3 f(3)=2

f(s)=$$8s24s4+254

.


TutorialExerciseUse astheaccelerationduetogravity.(Neglectairresistance.)

Withwhatinitialvelocitymustanobjectbethrownupward(fromaheightof3meters)toreachamaximumheightof450meters?

Part1of5Anobjectataninitialposition,s0metersabovetheground,isthrownupwithaninitialvelocityofv0meterspersecond.Theaccelerationacting

downwards(duetogravity)is Thedisplacementoftheobject, asafunctionoftimetisgivenby

Here meters,andthemaximumheight= 450 450 meters.

Part2of5Substitutes0=3intothefunction

Whentheobjectisatthegivenmaximumheightwehave

Part3of5Now, differentiate differentiate withrespectto$$t

,toobtainthevelocity asafunctionoftimet,

Whentheobjectreachesthetopmostpoint,itcomestorest(beforestartingtofalldown).Hence,it'svelocity, atthatpointoftimeis 0 0 m/sec.

Substitutethisvaluefor intheexpressionfor andsimplifytoobtaintintermsofv0.

Part4of5Substitutetheexpressionfortobtainedinthelaststepintheexpression toobtainthevalueofv0.

Simplifytheexpression.

Part5of5Theinitialvelocitywithwhichtheobjectisthrownuproundedtoonedecimalplaceis m/sec.YouhavenowcompletedtheMasterIt.


a(t)=9.8m/sec2

9.8m/sec2. f(t)

f(t)=4.9t2+v0t+s0.

s0= 3 3

f(t)=4.9t2+v0t+s0.

f(t) =4.9t2+v0t+ 3 3

4.9t2+v0t= 447 447 .

f(t)

v(t)

v(t)= 9.8 9.8 t+v0.

v(t),

v(t) v(t)

t= v09.8 9.8

4.9t2+v0t=447

4.9 +v0 =447v0

9.8 9.82

v09.8 9.8

4.9v02+9.8v02 = (9.8)2 447 447

v02 = 8761.20 8761.20 (roundedtotwodecimalplaces)

v0= 93.6 93.6

Consideraparticlemovingalongthexaxiswherex(t)isthepositionoftheparticleattimet,x'(t)isitsvelocity,andx''(t)isitsacceleration.

(a)Findthevelocityandaccelerationoftheparticle.

(b)Findtheopentintervalsonwhichtheparticleismovingtotheright.

(c)Findthevelocityoftheparticlewhentheaccelerationis0.51

15.3/4points|PreviousAnswersLarsonET55.1.087.MI.

x(t)=t312t2+21t3,0t10

v(t) =3t224t+21

a(t) =6t24

0

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Antiderivatives and Indefinite Integration 5