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MATH141–FinalExam(ReviewofExams1&2)–SampleTest1–DetailedSolutions

1.C.Usetheformulaforthederivativeoftheinverse.

1'(4)'( (4))1'(4)'(7)1'(4)5

gf g

gf

g

=

=

=−

2.A.Rememberthat 11

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

ff f

−−

′ = .Sofirst,wewillfind 1(1)f − bysetting ( ) 1f x =

1

1

(2 sin )

(2 sin )

1

1

ln1 ln0 2 sin0

x x

x x

e

ex x

x

−

−

+

+

−

=

== +=

Nowweknowthat 1(1)f − =0.Sotheformulabecomes: 1 1( ) (1)'(0)

ff

− ′ =

Nowwemustfind '(0)f ,bytakingthederivativeof ( )f x andpluggingin0.

( )

1(2 sin )

2

0

1'( ) 21

'(0) 2 1'(0) 3

x xf x ex

f ef

−+ ⎛ ⎞= +⎜ ⎟

−⎝ ⎠= +=

1 1( ) (1)3

f − ′ =

3.E.Takethederivative.Don’tforgetthechainrule.

1 * sin(5 )*5cos(5 )

sin(5 )5cos(5 )5 tan(5 )

xx

xxx

−

= −

= −

4.D.Usesubstitution.Letu=lnx.

2

2

1( )

xdux u

u du−

∫

∫

ln1

u x

du dxx

dx xdu

=

=

=

1

2

11

1ln1 1

ln ln1 1212

u

u

x

e e

−

=−−=

−=

− −⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= − +

=

5.B.Uselogarithmicdifferentiationtotakethederivative.Thenplugine.

[ ]

2

2

2

2

2

2

2

(ln )

ln ln(ln )ln ln(ln )

1 1 1 ln(ln )2ln

2 ln(ln )ln

2 ln(ln ) (ln )ln

2 ln(ln ) (ln )ln

2 ln(1) (1)1

2 (0)

x

x

x

e

e

y x

y xy x x

dy x x xy dx x xdy x x x ydx xdy x x x xdx x

dy e e e edx edy e edxdy e edx

=

==

⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎡ ⎤= +⎢ ⎥⎣ ⎦⎡ ⎤= +⎢ ⎥⎣ ⎦

⎡ ⎤= +⎢ ⎥⎣ ⎦⎡ ⎤= +⎢ ⎥⎣ ⎦

= + (1) e=

6.A.Usetherulesofexponentialstotakethederivative.

( )2 2

2 2 (2 )(ln 2)x xd xdx

=

7.A.Useu-substituionandletu=1/x.

1/

2

22 ( )

x

u

u

e dxxe x duxe du

−

−

∫

∫∫

1

2

2

1u x

du dxx

dx x du

−=−=

= −

1/

u

x

e Ce C

= − += − +

8.E.Thisisalsoaninversetrigintegral.Substituteu=sinx.

2

2

1

1

cos1 sin

11tantan (sin )

d

duuu C

C

θ θθ

θ

−

−

+

=+

= += +

∫

∫ sincos

udu d

θθ θ

==

9.C.Youhavetorecognizethisasaninversetrigintergral.

1

2sin

525dx x Cx

− ⎛ ⎞= +⎜ ⎟⎝ ⎠−∫

10.A.Useintegrationbypartstosolvethisintegral.Letu x= .

( )

( )

x

x x x

x x x

x x x

x x x

xe dx uv vdu

xe dx x e e dx

xe dx xe e dx

xe dx xe e C

xe dx xe e C

−

− − −

− − −

− − −

− − −

= −

= − − −

= − +

= − + − +

= − − +

∫ ∫∫ ∫∫ ∫∫∫

u xdu dx==

x

x

v edv e dx

−

−

= −=

11.A.Sinceseciseven,thenu=tanx.

( )

[ ]

/44 2

0

2 2 2

2 2

2 2

2 2

2 2

2 4

3 5

/43 5

0

sec tan

sec tan sec

sec

sec

(1 tan )

(1 )

1 13 51 1tan tan3 51 1(1) (1) 03 58 /15

x x dx

x x x dx

x u du

x u du

x u du

u u du

u u du

u u

x x

π

π

+

+

+

+

⎤+ ⎥⎦⎡ ⎤+ −⎢ ⎥⎣ ⎦

∫

∫∫∫∫∫∫

12.D.Sincesinisodd,letu=cosx.

3 6

2 6

sin cos

sin cos sin

x x dx

x x xdx∫∫

cossin

u xdu xdx== −

2 6

2 6

2 6

6 8

7 9

9 7

sin

(1 cos )

(1 )

( )

7 91 1cos cos9 7

x u du

x u du

u u du

u u du

u u C

x x C

−

− −

− −

− −

⎛ ⎞− − +⎜ ⎟⎝ ⎠

+ +

∫∫∫∫

13.C.Useausubstitutionandletu=x+3.

2

3x dxx +∫

3u xdu dx= +=

2x duu∫ 3x u= −

2

2

2

2

( 3)

6 9

96

6 9ln | |2( 3) 6( 3) 9 ln | 3 |2

u duu

u u duu

u duu

u u u C

x x x C

−

− +

⎛ ⎞− +⎜ ⎟⎝ ⎠

− + +

+ − + + + +

∫

∫

∫

Notethatyoumayalsoseetheanswerasthefollowingiftheycombinethe-18withthe+C.

2( 3) 6 9ln | 3 |2

x x x C+ − + + +

14.B.Useatrigsubstitutiontosolvethisintegral.

3sec3sec tan

xdx d

θθ θ θ

==

x 2 9x −

3

2

4

2

4

2

3

3

2

3

2

(3sec ) 93sec tan

(3sec )

9(sec 1) 3sec tan81sec

9(tan ) 3 tan81sec

3tan 3tan81sec

1 tan9 sec1 sin cos9

d

d

d

d

d

d

θ θ θ θθ

θ θ θ θθ

θ θ θθ

θ θ θθ

θ θθ

θ θ θ

−

−

∫

∫

∫

∫

∫

∫

Now,makeausubstitution.

sincos

udu d

θθ θ

==

2

3

3

1919 31 sin27

u du

u C

Cθ

+

+

∫

Tofind sinθ ,lookbacktheoriginaltriangle.

2 3/2

3

1 ( 9)27

xx−

15.C.Thisisanimproperintegralquestion.

0 1

2 21 0

1

2 20 01

1

0 01

0 0

0 0

1 1

1 1lim lim

1 1lim lim

1 1 1 1lim lim1 1

1 1lim 1 lim 1

t

t tt

t

t tt

t t

t t

dx dxx x

dx dxx x

x x

t t

t t

−

→ →−

→ →−

→ →

→ →

+

⎡ ⎤⎡ ⎤+ ⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤− −⎛ ⎞ ⎛ ⎞+⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ − − ⎤ ⎡ − − ⎤⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ − ⎤⎛ ⎞ ⎛+ + − +⎜ ⎟⎢ ⎥⎝ ⎠ ⎝⎣ ⎦

∫ ∫

∫ ∫

⎡ ⎤⎞⎜ ⎟⎢ ⎥⎠⎣ ⎦

Whenyouplugin0,youwillget1/0,whichisinfinity,thereforethisintegraldoesnotexist

(diverges).

16.A.Thisisanimproperintegralquestion.

05

05

0

lim

lim5

lim5 5

1 105 5

x

tt

x

tt

t

t

e dx

e

e e

→−∞

→−∞

→−∞

⎡ ⎤−⎢ ⎥

⎣ ⎦

− =

∫

17.E.Thisisanimproperintegralquestion.

2

0

2 20

0

2 20

01 1

0

1 1 1 1

25

25 25

lim lim25 25

1 1lim tan lim tan5 5 5 5

1 1 1 1lim tan 0 tan lim tan tan 05 5 5 5

t

t tt

t

t tt

t t

dxx

dx dxx x

dx dxx x

x x

t t

∞

−∞∞

−∞

→−∞ →∞

− −

→−∞ →∞

− − − −

→−∞ →∞

+

++ +

++ +

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞+⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣

∫

∫ ∫

∫ ∫

1 10 05 2 5 2

10 10210

5

π π

π π

π

π

⎤⎥⎦

⎡ − ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

+

18.A.Thisisageometricseries.Makesuretogetitinthecorrectformbeforeusingtheformula.

1

11

2 1

11

1

1

343 34

394

n

nn

n

nn

n

n

+∞

−=

−∞

−=

−∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

∑

∑

9, 3 / 4a r= =

Sum=9

1 3 / 4−

9 361/ 4

=

19.E.Thisisageometricseries.Dosomealgebraicmanipulationtogetitinthecorrectform.

1

11

11

11

6 10

106100(10)6(6)

n n

nn

nn

n

nn

∞− +

=

+∞

=

−∞

−=

∑

∑

∑

100 / 6, 10 / 6a r= =

Sincer>1,thisseriesdivergesbygeometricseries.

20.D.

3𝑝 > 1

𝑝 > 1/3

21.D.Usetheratiotest.

1lim

3 45 8lim

3 4 3lim 15 8 5

n

nn

n

nn

n

aan ana

nn

+

→∞

→∞

→∞

+−

+ = <−

22.B.Forthisquestionyoumustknowyourrulesforthedivergencetestandthecomparisontest.

(i)False (ii)True (iii)False

23.A.Usetheratiotestforeach

(i)Converges.

(ii)Diverges.1 !

n

n

nn

∞

=∑

1( 1) !lim( 1)!

( 1)( 1) !lim( 1) !

( 1)lim

1lim

1lim 1

1

n

nn

n

nn

n

nn

n

n

n

n

n nn n

n n nn n n

nn

nn

ne

+

→∞

→∞

→∞

→∞

→∞

++

+ ++

+

+⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞+⎜ ⎟⎝ ⎠>

24.D.Thisisaproblemongeometricseries.

1

1 12 2 2

nn

nn n

x x x −∞ ∞

= =

⎛ ⎞= ⎜ ⎟⎝ ⎠∑ ∑ 𝑟 = 𝑥/2

Theserieswillconvergewhen 𝑟 < 1

!!< 1

−1 < !!< 1

2 2x− < <

25.B.Thisisanerrorquestion.Startbyfindingthefirstfewtermsoftheseries.Wearelookingforsomethingthatislessthan.01=1/100.Notethatyoucanignorethenegativebecausewearelookingforabsolutevalueoferror.

1 1a =

2

3

4

5

12161241120

a

a

a

a

=

=

=

=

Soweonlyneedthefirst4terms.

26.Thisisapartialfractiondecompositionquestion.

2 2 2 2

2 2 2

3 2 3 2

3 2

1( 1) 1

1 ( )( 1) ( 1) ( )11 ( ) ( )

A B Cx Dx x x x xA x x B x Cx D xAx Ax Bx B Cx DxA C x B D x Ax B

+= + ++ +

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

Equationthecoefficientsgivesus:

00

01

A CB DAB

+ =+ ===

So, 0, 1, 0, 1A B C D= = = = −

2 2

22

11

1

0 1 0 11

11

tan11 tan

x dxx x x

x dxx

x x C

x Cx

−

−−

−

−⎛ ⎞+ +⎜ ⎟+⎝ ⎠⎛ ⎞−⎜ ⎟+⎝ ⎠

− +−

− − +

∫

∫

27.Useatrigsubstitutiontoevaluatetheintegral.

3sin3cos

xdx d

θθ θ

==

3 x

29 x−

2

2

2

2

2

2

21

21

99sin (3cos )9(1 sin )

27sin cos3cos

9sin

19 (1 cos(2 ))2

9 1 sin(2 )2 29 1 2sin cos2 2

9 9sin2 3 3 3

9 9sin2 3 9

x dxx

d

d

d

d

C

C

x x x C

x x x

θ θ θθ

θ θ θθ

θ θ

θ θ

θ θ

θ θ θ

−

−

−

−

−

⎡ ⎤− +⎢ ⎥⎣ ⎦⎡ ⎤− +⎢ ⎥⎣ ⎦⎡ ⎤⎛ ⎞−⎛ ⎞ ⎛ ⎞⎢ ⎥− +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦

⎛ ⎞−⎛ ⎞ − ⎜ ⎟⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠

∫

∫

∫∫∫

C⎡ ⎤⎢ ⎥ +⎟⎢ ⎥⎣ ⎦

28.Useintegrationbyparts.

uv vdu− ∫

2 1 sin(2 )2

2 cos(2 )

u x v x

du xdx dv x dx

= =

= =

2

2

1 1sin(2 ) sin(2 )22 21 sin(2 ) sin(2 )2

x x x xdx

x x x x dx

−

−

∫

∫

Useintegrationbypartsagaintoevaluatetheintegral.

1 cos(2 )2

sin(2 )

u x v x

du dx dv x dx

= = −

= =

2

2

2

2

1 sin(2 )21 1 1sin(2 ) cos(2 ) cos(2 )2 2 21 1 1sin(2 ) cos(2 ) cos(2 )2 2 21 1 1sin(2 ) cos(2 ) sin(2 )2 2 4

x x uv vdu

x x x x x dx

x x x x x dx

x x x x x C

⎡ ⎤− −⎣ ⎦

⎡ ⎤− − − −⎢ ⎥⎣ ⎦⎡ ⎤− − +⎢ ⎥⎣ ⎦

+ − +

∫

∫

∫

29.Usetrigsubstitutiontoevaluatetheintegral.

2

2

2 1tan2 sec

1 sec2

xdx d

dx d

θθ θ

θ θ

==

=

24 1x + 2x

1

2

2

2

2

2

1(2 ) 11 sec2tan 1

1 sec2 sec1 sec21 ln sec tan21 ln 4 1 22

dxx

d

d

d

C

x x C

θ θ

θθ θθ

θ θ

θ θ

+

+

+ +

+ + +

∫

∫

∫

∫

30.Evaluatethefollowinglimits:

a)3(ln )lim

x

xx→∞

Answer:_____0_____

Thebottomgrowsfasterthanthetop.

b)0

1lim cscx

xx→

⎛ ⎞−⎜ ⎟⎝ ⎠ Answer:______0____

0

0 0

0

1 1limsinsin 1 cos 0lim limsin cos sin 0

sin 0lim 0sin cos cos 2

x

x x

x

x xx x xx x x x x

xx x x x

→

→ →

→

⎛ ⎞− = ∞−∞⎜ ⎟⎝ ⎠− −⎛ ⎞ ⎛ ⎞→ =⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠⎛ ⎞→ = =⎜ ⎟− + +⎝ ⎠

c) 2lim 2x

x x x→∞

− − Answer:_____-1_____

MultiplybytheconjugatethenuseL.R.

d) ( )1

0lim x xx

e x→

+ Answer:______e2____

( ) ( )( )

1

0 0

0

2

0

1lim lim ln

lnlim

1lim ( 1) 2

x xxx x

x

x

xxx

e x e xx

e xx

e ee x

→ →

→

→

+ → +

+

+ = →+

31.Theseareallsequencessotake limn→∞

a)55sinn

⎧ ⎫⎛ ⎞⎨ ⎬⎜ ⎟⎝ ⎠⎩ ⎭

convergesto____0_____/diverges

b)2

2

2 1( 1)3

n nn n

⎧ ⎫+−⎨ ⎬+⎩ ⎭ convergesto_________/diverges

ThelimitDNE

c)41

n

n⎧ ⎫+⎨ ⎬⎩ ⎭

convergesto____e4_____/diverges

d)ln(6 )ln( )nn

⎧ ⎫⎨ ⎬⎩ ⎭

convergesto___1______/diverges

e)2

9nne

⎧ ⎫⎨ ⎬⎩ ⎭

convergesto_________/diverges

Thetopgrowsfastersothelimit=∞

32.a)2

21

sin( 1)nn

nn

∞

=

−∑ .Startbytestingtheabsolutevalueofthisseries.

2 2

2 21 1

sin sin( 1)nn n

n nn n

∞ ∞

= =

− =∑ ∑

Usethecomparisontestandcompareto 21

1n n

∞

=∑ whichconvergesbyp-series.

2

2 2

2 2 2

2

1 sin

sin1 sin

nn nn n n

n

≥

≥≥

So2

21

sinn

nn

∞

=∑ convergesbythecomparisontest,so

2

21

sin( 1)nn

nn

∞

=

−∑ isabsolutelyconvergent.

b)1

( 1)ln

n

n n

∞

=

−∑ .Startbytestingtheabsolutevalueoftheserieswhichis:1

1lnn n

∞

=∑

Usethecomparisontestandcompareto1

1n n

∞

=∑ ,whichdivergesbyp-series.

1 1

lnlnn nn n

≤

≤

So1

1lnn n

∞

=∑ divergesbythecomparisontest.

Wenowknowthattheseriesisnotabsolutelyconvergent.UsetheAlternatingSeriesTesttodetermineiftheseriesisconditionallyconvergentordivergent.

1.1lim 0lnn n→∞

= 2.

1 1ln( 1) lnln ln( 1)n nn n

<+< +

Bothconditionsofthealternatingseriestesthavebeensatisfiedsothisseriesisconditionallyconvergent

c)2

21

2 1( 1)3 1

n

n

nn

∞

=

+−−∑

Thisseriesdivergesbythetestfordivergencesince lim 0n→∞

≠

d)3

1

5( 1)!

nn

n

nn

∞

=

−∑

Startbyusingtheratiotest.

1 1 3

3

1 3

3

3

3

( 1) 5 ( 1) !lim( 1)! ( 1) 5

5 ( 1) !lim5 ( 1)!

5( 1)lim 0 1( 1)

n n

n nn

n

nn

n

n nn n

n nn n

nn n

+ +

→∞

+

→∞

→∞

− ++ −

++

+ = <+

Sothisseriesisabsolutelyconvergent.

e)2

21

3 1( 1)2 1

nn

n

nn

∞

=

⎛ ⎞+− ⎜ ⎟+⎝ ⎠∑

Usetheroottest.

2

2

3 1 3lim 12 1 2

n

nn

nn→∞

⎛ ⎞+ = >⎜ ⎟+⎝ ⎠

Sothisseriesdiverges

33.a)2

1

( ) nn

n

en

∞

=∑ .Usetheroottest.

2

2

lim

lim 0 1

n

nn

n

en

en

→∞

→∞

⎛ ⎞⎜ ⎟⎝ ⎠

= <

Sothisseriesconvergesbytheroottest.

b) 2

1

3 !n

nn

nπ

∞

=∑ .Usetheratiotest.

2

2

2

2

1

( 1)

1

2 1

2 1

3 ( 1)!lim3 !

3 ( 1) !lim3 !

3( 1)lim 0 1

n n

nnn

n n

nn nn

nn

nn

n nn

n

ππ

ππ

π

+

+→∞

+

+ +→∞

+→∞

+

+

+ = <

Sothisseriesconvergesbytheratiotest

c)13n

n

∞

=∑

Thisseriesdivergesbythetestfordivergence.

1/lim3 1n

n→∞=

d)

1

1

n

n

en

∞

=∑

ThisseriesdivergesbyLCTwith1/n

e)1

15 1n

n

∞

= −∑

ThisseriesconvergesbyLCTwith1

15

n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑ whichisaconvergentgeometricseries.

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MATH141–FinalExam–SampleTest2–DetailedSolutions

1.B.Rememberthat 11

1( ) (2 )'( (2 ))

ff f

ππ

−−

′ = .

Sofirst,wewillfind 1(2 )f π− bysetting ( ) 2f x π= .

2 2 sinx xxπ

π= −

=

Now,wenowthat 1(2 )f π π− = .Sotheformulabecomes:

1 1( ) (2 )'( )

ff

ππ

− ′ =

Tofind '( )f π ,justtakethederivativeof ( )f x andpluginπ .

'( ) 2 cos'( ) 2 cos'( ) 2 ( 1) 3

f x xff

π ππ

= −− −= − − =

So 1 1( ) (2 )3

f π− ′ =

2.E.Startwiththeformula 11

1( ) (4)'( (4))

ff f

−−

′ =

Sofirst,wewillfind 1(4)f − bysetting ( ) 4f x = .

3 2

3 2

4 2 51 21

x xx x

x

= − +− = −=

Nowweknowthat 1(4) 1f − = ,sotheformulabecomes:

1 1( ) (4)'(1)

ff

− ′ =

Tofind '(1)f ,justtakethederivativeof ( )f x andplugin1.

2'( ) 3 4

'(1) 1f x x xf

= −= −

So 1 1( ) (4) 11

f − ′ = = −−

3.C.Useasubstitutionandlet 2 3 2u x x= + + .

2

21

4 63 2

2(2 3)2 3

x dxx xx duu x

++ ++

+

∫

∫

2 3 2(2 3)

2 3

u x xdu x dx

dudxx

= + += +

=+

[ ] [ ]

22

1

2

2

2 ln

2ln( 3 2)

2ln(12) 2ln 62(ln12 ln 6)2ln(12 / 6)2 ln 2ln 2 ln 4

duuu

x x+ +

−−

=

∫

4.A.Sinceyouhaveafunctionraisedtoafunction,uselogarithmicdifferentiation.

sin

sin

sin

ln lnln (sin )(ln )1 1' (sin ) ln (cos )

sin' ln (cos )

sin' ln (cos )

x

x

x

y xy xy x x

y x x xy x

xy x x yxxy x x xx

===

⎡ ⎤= +⎢ ⎥⎣ ⎦⎡ ⎤= +⎢ ⎥⎣ ⎦⎡ ⎤= +⎢ ⎥⎣ ⎦

5.B.Usesubstitutiontotaketheintegral.Letu=1+lnx.

51

64(1 ln )

e

dtt t+∫

1 ln1

u x

du dxx

dx xdu

= +

=

=

51

5

4

4

41

4 4

4 4

64( )

64

644

16

16(1 ln )

16 16(1 ln ) (1 ln1)

16 16(2) (1)1 16 15

e

e

tdut u

u du

u

u

t

e

−

−

−

−

−+

⎡ ⎤ ⎡ ⎤− −−⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤− −−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦− + =

∫

∫

6.B.Usetherulesfortakingderivativesoflogs.

29

22

( ) log ( 1)1 1'( ) (2 )1 ln 9

1 1'(0) (2)1 1 ln 9

'(0) 1/ ln 9

x

xx

f x e

f x ee

f

f

= +

=+

=+

=

7.C.Usesubstitution.Let 2xu e= +

ln 4

0 2

x

x

e dxe +∫

2x

x

x

u edu e dx

dudxe

= +=

=

[ ] [ ]

ln 4

0

ln 4

0

ln 4 0

1

ln

ln( 2)

ln( 2) ln( 2)

ln(6) ln(3)ln 2

x

x

x

e duu e

duuu

e

e e

+

⎡ ⎤ ⎡ ⎤+ − +⎣ ⎦ ⎣ ⎦−

∫

∫

8.D.Makeasubstitutionandlet xu e= ,thenyoushouldbeabletorecognizethisasaninversetrigintegral.

ln(1/2)

20 1

x

x

ee−∫

2 2x x

x

x

u e u edu e dx

dudxe

= ==

=

2

2

1

ln1/21

01 1

111

sin ( )

sin ( )

sin (1/ 2) sin (1)

6 226

3

x

x

x

e dueu

duuu

e

π π

π

π

−

−

− −

−

−

−

−

−

−

∫

∫

9.A.Usetheidentity 2 2tan sec 1x x= − .

( )2sec 1

tan

x dx

x x C

−

− +∫

10.C.Useintegrationbyparts.

2

1ln

1 1

u x vx

du dx dv dxx x

−= =

= =

2

1

2

1

1 1 1ln

ln

ln1

ln 1

ln 2 1 ln1 12 2 1

ln 2 1 12 2

1 ln 22 2

uv vdu

x dxx x x

x x dxxx xxxx x

−

−

−

⎛ ⎞ ⎛ ⎞⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

− +

− +−

− −

⎡ ⎤ ⎡ ⎤− − − − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦− − +

−

∫

∫

∫

11.E.Usesubstitution.Letu=x–5.

5x dxx −∫

5 5u x x udu dx= − = +=

5

51

5ln | |5 5ln | 5 |

x duuu duu

duu

u u Cx x C

+

⎛ ⎞+⎜ ⎟⎝ ⎠+ ++ + + +

∫

∫

∫

12.A.Completethesquareonthebottom.

2

2

2

6 13

6 9 13 9( 3) 4

x x

x xx

+ +

+ + + −+ +

2

1( 3) 4

dxx + +∫

2

3 2 tan2sec

xdx d

θθ θ

+ ==

2

2

2

2

2

2sec4 tan 42sec4(tan 1)

2sec2secsec

d

d

d

d

θ θθθ θθ

θ θθ

θ θ

+

+

∫

∫

∫∫

13.D.Thisisanimproperintegral.

( )

( )

( )( )

( )( ) ( )( )

( )( ) ( )( )

3

3

2

2 2

2 2

1ln

1limln

1lim2 ln

1 1lim2 ln 2 ln

1 1lim2 ln 2 ln

1 102 2

e

t

te

t

t

e

t

t

dxx x

dxx x

x

t e

t e

∞

→∞

→∞

→∞

→∞

⎡ ⎤+ ⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞−⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤⎛ ⎞− −⎢ ⎥⎜ ⎟−⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦⎡ ⎤⎛ ⎞− −⎢ ⎥⎜ ⎟−⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

+ =

∫

∫

14.E.Thisisanimproperintegralquestion.

( )

( )

( ) ( )

25

5

3lim4

3lim4

3 3lim4 5 4

30 31

t

t

t

dxx

x

x

∞

→∞

∞

→∞

→∞

−

−−

⎡ ⎤− −−⎢ ⎥− −⎣ ⎦−− =

∫

15.C.Thisisageometricseries.Makesuretogetitinthecorrectformbeforeusingtheformula.

( )

1

1 1

1

11 1

1 1

1 1

1 261 26 6

1 2 26 6 6 6

1 1 2 2 1 1 76 6 6 6 5 2 10

n

nn

n

n nn n

n

nn n

n n

n n

∞

=

∞ ∞

= =

−∞ ∞

−= =

− −∞ ∞

= =

+

+

⎛ ⎞⎛ ⎞+ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞+ = + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

∑

∑ ∑

∑ ∑

∑ ∑

16.E.Thisisanothergeometricseriesquestion.

1

1

1

1

3

13 3

n

nn

n

n

π

π

−∞

=

−∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

∑

Theseriesdivergessince!!> 1

17.B.

(i)False,since( )1/21/2

1 15n n

≥+

,andthecomparisontestfordivergencerequirestheopposite.

(ii)True,since1 1, , lim lim5 5

nn n x x

n

a na bbn n n→∞ →∞

= = =+ +

=1.

18.D.

( )( )

1 cos( )1.

cos( )1 0 !

cos

2.lim limcos !1x x

nn nn n n

FALSE

nn n undefined FALSE

n

π

π

π

π→∞ →∞

≤

≤≤

= =

3.Forthealternatingseriestest,notthatthenumeratorwillalwaysbe1or-1.So1

nb n=

1

1lim 0

1 111

0 1 !

n

n n

n

b b

n nn n

TRUE

→∞

+

=

≤

≤+≤ +≤

So,1and2arefalse,and3istrue.

19.C.Applydivergencetest

2

2

1lim 05 1 5x

nn→∞

= ≠+

20.Thisapartialfractionquestion.

2

5 103 4 4 1

5 10 ( 1) ( 4)

x A Bx x x x

x A x B x

− = +− − − +

− = + + −

Plugginginx=-1givesus:

15 53

BB− = −=

Plugginginx=4givesus:

10 5

2A

A==

2 34 1

2ln | 4 | 3ln | 1|

dx dxx xx x C

+− +

− + + +

∫ ∫

21.Useintegrationbyparts.

sin(2 )xI e x dx= ∫

sin(2 )2cos(2 )

x

x

u x v edu x dv e dx= == =

sin(2 ) 2 cos(2 )

sin(2 ) 2 cos(2 )

x x

x x

uv vdu

e x e x

e x e x

−

−

−

∫∫∫

Useintegrationbypartsagain…

cos(2 )2sin(2 )

x

x

u x v edu x dv e dx= == − =

[ ]

sin(2 ) 2

sin(2 ) 2 cos(2 ) 2sin(2 )

sin(2 ) 2 cos(2 ) 4 sin(2 )

sin(2 ) 2 cos(2 ) 45 sin(2 ) 2 cos(2 )

1 sin(2 ) 2 cos(2 )51 sin(2 ) 2cos(2 )5

x

x x x

x x x

x x

x x

x x

x

e x uv vdu

e x e x x e dx

e x e x e x dx

I e x e x II e x e x C

I e x e x C

I e x x C

⎡ ⎤− −⎣ ⎦⎡ ⎤− − −⎣ ⎦

− −

= − −= − +

= − +

= − +

∫∫∫

22.Let 5sinx θ= 5 x

5cosdx dθ θ=

225 x−

2

2

2

2

2

2

2

2

2

2

2

2

2

21

25

25 25sin 5cos25sin

25(1 sin ) 5cos25sin

25(cos ) 5cos25sin

5cos 5cos25sin

cossin

cot

(csc 1)

cot

25 sin5

x dxx

d

d

d

d

d

d

d

C

x x Cx

θ θ θθ

θ θ θθ

θ θ θθ

θ θ θθ

θ θθ

θ θ

θ θ

θ θ

−

−

−

−

−

− − +

− ⎛ ⎞− − +⎜ ⎟⎝ ⎠

∫

∫

∫

∫

∫

∫

∫

∫

23.Youhavetocompletethesquareinordertoevaluatethisintegral.

2

2

4 4 4( 2) 4x xx− + −− −

( )

2 3/2

3/22

1( 4 )

1

( 2) 4

dxx x

dxx

−

− −

∫

∫

So 2 2secx θ− = x-2

2sec tandx dθ θ θ= 2

2( 2) 4x− −

2 3/2

2 3/2

3

2

2

2

2

2sec tan(4sec 4)

2sec tan(4 tan )

2sec tan8 tan

1 sec4 tan

1 1 cos4 cos sin

1 cos4 sin

d

d

d

d

d

d

θ θ θθ

θ θ θθ

θ θ θθ

θ θθ

θ θθ θ

θ θθ

−∫

∫

∫

∫

∫

∫

1 cos 14 sin sin

1 cot csc4

1 1 2 2csc4 4 2 8

d

d

x xC C C

θ θθ θ

θ θ θ

θ − −− + = − + = − +

∫

∫

24.Evaluatethefollowinglimits:

a) 2

5lim5

x

x

e xx→∞

+ − Answer:______∞____

b)0

1lim sinxx

x→

⎛ ⎞⎜ ⎟⎝ ⎠

Answer:_______1___

Move𝑥tothebottomas1/xanduseL.R.

c)0

tan( )limln(1 )x

xx

π→ +

Answer:____𝜋______

d) ( )1

0lim sin x xx

x e→

+ Answer:_____e2_____

( ) ( )( )

1

0 0

0

2

0

1lim sin lim ln sin

ln sinlim

1lim (cos ) 2sin

x xxx x

x

x

xxx

x e x ex

x ex

x e ex e

→ →

→

→

+ → +

+

+ = →+

25.Theseareallsequencessotake limn→∞

.Theanswersarehighlightedinred.

a)( ) 423

n

n

+⎧ ⎫−⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭

convergesto____0_____/diverges

b) cos2n⎧ ⎫

⎨ ⎬⎩ ⎭

convergesto_________/diverges

c){ }2 nn e− convergesto___0______/diverges

d){ }ln(2 3) ln( 4)n n+ − − convergesto_____ln2____/diverges

26.

a)2

1/31

( 1)n

n n

+∞

=

−∑ .Startbytestingtheabsolutevalueofthisseries.

2

1/3 1/31 1

( 1) 1n

n nn n

+∞ ∞

= =

− =∑ ∑

Bythep-test,theseriesisdivergent.

Nowlookattheoriginalseries2

1/31

( 1)n

n n

+∞

=

−∑ .Thisisanalternatingserieswhere 1/3

1nb n=

lim 0nnb

→∞= and 1n nb b+ ≤

SothisconvergesbyAST.

Therefore,2

1/31

( 1)n

n n

+∞

=

−∑ isconditionallyconvergent.

b)1! n

nn e

∞−

=∑ .Usetheratiotest:

1

1

( 1)!lim!

( 1) !lim!

( 1)lim

n

nn

n

nn

n

n ee n

n n en e ene

+→∞

→∞

→∞

+ ×

+

+ = ∞

So1! n

nn e

∞−

=∑ divergesbytheratiotest.

c)2

1 11 1n n n

∞

=

⎡ ⎤−⎢ ⎥− +⎣ ⎦∑

Thisseriesisabsoluteconvergent.

Itisatelescopingseries.Notethatallthetermsarepositive,sosincetheseriesconverges,itisalsoabsolutelyconvergent.

d)( )

11

53

n

nn

n∞

−=

−∑ .Usetheratiotest.

1 1

1 1

1

1 1

( 5) ( 1) 3lim3 ( 5)

5( 1)3lim3 3

5( 1) 5lim 13 3

n n

n nn

n

nn

n

nn

nn

nn

+ −

+ −→∞

−

−→∞

→∞

− + ×−

+

+ = >

Sothisseriesisdivergent.

e)1

1( 1)2

n

n

nn n

∞

=

+−+∑

Thisseriesdivergesbythetestfordivergence.

27.a)1

1 cos3nn

n∞

=

+∑ .Usethecomparisontest.

Compareto1 1

2 123 3

n

nn n

∞ ∞

= =

⎛ ⎞= ⎜ ⎟⎝ ⎠∑ ∑ ßconvergesbygeometricseries.

2 1 cos3 32(3) 3 (1 cos )2 1 cos1 cos

n n

n n

n

nn

n

+≥

≥ +≥ +≥

So1

1 cos3nn

n∞

=

+∑ convergesbycomparisontestwith1

23nn

∞

=∑

b)( )22

1lnn n n

∞

=∑ .Usetheintegraltest.

( ) ( )

( )

2 22 2

22

1 1ln ln

1limln

n

t

t

dxn n x x

x x

∞∞

=

→∞

→∑ ∫

∫

Useu-substitutiontotaketheintegral.

2

1lim

1limln

1 1 1limln ln 2 ln 2

t

t

t

t

u

x

t

→∞

→∞

→∞

−⎡ ⎤⎢ ⎥⎣ ⎦

−⎡ ⎤⎢ ⎥⎣ ⎦−⎡ ⎤+ =⎢ ⎥⎣ ⎦

Sothisseriesconvergesbytheintegraltest.

c) 21

11 sin

5n

nn

∞

=

⎛ ⎞+ ⎜ ⎟⎝ ⎠+∑

Thisseriesconvergesbythecomparisontest.

2 2

11 sin2

5n

n n

⎛ ⎞+ ⎜ ⎟⎝ ⎠ ≤+

Since 21

2n n

∞

=∑

converges, 21

11 sin

5n

nn

∞

=

⎛ ⎞+ ⎜ ⎟⎝ ⎠+∑

converges.

d)1

54

n

nn

nn

∞

=

++∑

ThisseriesdivergesbyLCTwith1

54

n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑

e)2

1 2n

nn

∞

= +∑

Divergesbythetestfordivergence.