MATH141 – Final Exam (Review of Exams 1&2) – Sample...
Transcript of MATH141 – Final Exam (Review of Exams 1&2) – Sample...
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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.