Contentsjktartir.people.ysu.edu/1572/exam.pdf · 2019-04-19 · 1 Chapter 1: Math 1572 Summer 2002 Section 1.1: Review Solutions to Review Problems 1. For each of the following, ﬁnd

Contents

Chapter 1: Math 1572 Summer 2002 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Section 1.1: Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1


Section 2.1: Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Section 2.2: Quiz 1572 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Section 2.3: Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Section 2.4: Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Section 2.5: Final . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Chapter 3: Fall 2005 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Section 3.1: Exam 1 Math 1572 Fall 2005 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Chapter 4: Math 1572 Fall 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Section 4.1: Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Section 4.2: Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Section 4.3: Exam 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

Chapter 5: Math 1572 Spring 2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Section 5.1: Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Section 5.2: Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

Section 5.3: Exam 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89


Section 6.1: Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

Section 6.2: Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

Section 6.3: final . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

i

ii

Chapter 7: Math 1572 Fall 2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

Section 7.1: Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

Section 7.2: Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

Section 7.3: Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Section 7.4: Exam 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

Section 7.5: Final . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

Section 7.6: Final . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

Chapter 8: Math 1572 Spring 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

Section 8.1: Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

Section 8.2: Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

Section 8.3: Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

Section 8.4: Exam 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

Section 8.5: Final Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

Chapter 9: Spring 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

Section 9.1: Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

Section 9.2: Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

Section 9.3: Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

Section 9.4: Exam 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

Section 9.5: Final Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

Chapter 10: Spring 2019 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

Section 10.1: Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

Section 10.2: Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

iii

Section 10.3: Exam 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

1

Chapter 1: Math 1572 Summer 2002

Section 1.1: Review

Solutions to Review Problems

1. For each of the following, find the equation of the line that that is tangent to the graphof f

-1at the given point.

a. f(x) = x3 + 2x+ 3; x = -30

f ′(x) = 3x2 + 2

f-1(-30) = -3

f ′(-3) = 29

[

f-1]′(-30) = 1

29

y + 3 = 129(x+ 30)

b. f(x) = tanx; x = 1

f ′(x) = sec2 x

f-1(1) = π

4

f ′(π4) = 2

[

f-1]′(1) = 1

2

y − π4= 1

2(x− 1)

2. Calculate the following limits.

a. limx→0

sin x

xL′H= lim

x→0

cos x

1= 1

b. limx→0

ex

ln x= 0

c. limx→∞

ln x

exL′H= lim

x→∞

1

xex= 0

d. limx→∞

(√x2 + 1− x

)

limx→∞

(√x2 + 1− x

)

= limx→∞

(√x2 + 1− x

)

·√x2 + 1 + x√x2 + 1 + x

= limx→∞

1√x2 + 1 + x

= 0

e. limx→0

(

sin x)x

First, consider

limx→0

(

ln sin x)x

= limx→0

ln sin x1x

L′H= lim

x→0-x2 cosx

sin x

= limx→0

-x cosxx

sin x

= 0.

Therefore, limx→0

(

sin x)x

= e0 = 1.

f. limx→0

ln(sin x) = -∞

2

g. limn→∞

n!

nn

limn→∞

n!

nn

= limn→∞

n(n− 1)(n− 2) · · ·2 · 1n · n · · ·n

= limn→∞

(

n

n· n− 1

n· n− 2

n· · · 2

n· 1n

)

≤ limn→∞

1

n

= 0

h. limn→∞

en

n!

Since limn→∞

en+1

(n + 1)!· n!en

= limn→∞

e

n + 1= 0, the series

∞∑

n=0

en

n!converges by the ratio test.

Therefore, limn→∞

en

n!= 0 by the divergence test.

3. Differentiate.

a. k(x) = esinx

k′(x) = esinx(cosx)

b. f(x) = ln |x3 + x2 − 1|

f ′(x) =3x2 + 2x

x3 + x2 − 1

c. g(x) = tan-1(sin x)

g′(x) =cos x

1 + sin2 x

d. f(x) = sin (ex)

f ′(x) = ex cos (ex)

e. y = xsinx

y = xsinx

ln y = ln xsinx

ln y = sin x ln x

y′

y= cosx ln x+ sin x · 1

x

y′

y=

x cosx ln x+ sin x

x

y′ =x cosx ln x+ sin x

x· y

y′ =x

sinx

(x cosx ln x+ sin x)

x

y′ = xsin x−1


f. h(x) = xex

y = xex

ln y = lnxex

ln y = ex ln x

y′

y= ex lnx+

ex

x

3

y′

y=

xex ln x+ ex

x

y′ =xex ln x+ ex

x· y

y′ =x

ex

(xex ln x+ ex)

x

y′ = xex−1(xex lnx+ ex)

g. f(x) = eex

f ′(x) = exeex

4

4. Integrate.

a.

∫ √x2 + 1dx

✟✟✟✟✟✟✟✟✟✟

θ

√x2 + 1

1

x

x = tan θ

dx = sec2 θ dθ

∫ √x2 + 1dx =

∫ √tan2 θ + 1 sec2 θ dθ =

∫

sec3 θ dθ

Use integration by parts letting u = sec x, du = sec x tanx dx, v = tan x, and dv = sec2 x dx.

∫

sec2 x · sec xdx = sec x tanx−∫

sec x · tan2 xdx

∫

sec3 xdx = sec x tanx−∫

sec x(sec2 x− 1)dx

∫


(

sec3 x− sec x)

dx

∫


sec3 xdx+

∫

sec xdx

2

∫

sec3 xdx = sec x tan x+

∫

sec xdx

2

∫

sec3 xdx = sec x tan x+ ln | sec x+ tan x|+ C

∫

sec3 xdx = 12(sec x tanx+ ln | sec x+ tan x|) + C

5

b.

∫ 1

0

ex sin xdx

Use integration by parts letting u = sin x, du = cosx dx, v = ex, and dv = ex dx.

∫ 1

0

ex sin xdx = ex sin x

∣

∣

∣

∣

1

0

−∫ 1

0

ex cos xdx

Use integration by parts again letting u = cosx, du = - sin x dx, v = ex, and dv = ex dx.

∫ 1

0


∣

∣

∣

∣

1

0

−∫ 1

0

ex cos xdx

∫ 1

0

ex sin xdx = e sin 1−(

ex cos x

∣

∣

∣

∣

1

0

−∫ 1

0

ex(- sin x)dx

)

∫ 1

0

ex sin xdx = e sin 1− e cos 1 + 1−∫ 1

0

ex sin xdx

2

∫ 1

0

ex sin xdx = e sin 1− e cos 1 + 1

∫ 1

0

ex sin xdx = 12(e sin 1− e cos 1 + 1)

c.

∫

x2 + 2x+ 3

x3 + x2 + x+ 1dx

∫

x2 + 2x+ 3

x3 + x2 + x+ 1dx

=

∫

x2 + 2x+ 3

(x2 + 1)(x+ 1)dx

=

∫(

2

x2 + 1+

1

x+ 1

)

dx

= 2 tan-1 x+ ln |x+ 1|+ C

6

d.

∫

sin2 θ cos2 θ dθ

=

∫

(1− cos2 θ) cos2 θ dθ

=

∫

[1− 12(cos 2θ + 1)] · 1

2(cos 2θ + 1)dθ

= 12

∫

(

-12cos 2θ + 1

2

)

(cos 2θ + 1)dθ

= -14

∫

(cos 2θ − 1)(cos 2θ + 1)dθ

= -14

∫

(

cos2 2θ − 1)

dθ

= -14

∫

(

12(cos 4θ + 1)− 1

)

dθ

= -14

∫

(

12cos 4θ − 1

2

)

dθ

= -18

∫

(

cos 4θ − 1)

dθ

= -18

(

14sin 4θ − θ

)

+ C

= - 132sin 4θ + 1

8θ + C

e.

∫

sin20 x cos3 xdx

=

∫

sin20 x · cos2 x · cosxdx

=

∫

sin20 x(1− sin2 x) cosxdx

=

∫

(sin20 x− sin22 x) cosxdx

Let u = sin x and du = cosx dx.

∫


=

∫

(

u20 − u22)

du

= 121u21 − 1

23u23 + C

= 121sin21 x− 1

23sin23 x+ C

f.

∫

2x√x2 − 4

dx

Let u = x2 − 4 and du = 2x dx.

∫

2x√x2 − 4

dx

=

∫

1√udu

= 2√u+ C

= 2√x2 − 4 + C

7

g.

∫

1√x2 − 4

dx

✟✟✟✟✟✟✟✟✟✟

θ

x √x2 − 4

2

x = 2 sec θ

dx = 2 sec θ tan θ dθ

∫

1√x2 − 4

dx

=

∫

2 sec θ tan θ√4 sec2 θ − 4

dθ

=

∫

sec θ dθ

= ln | sec θ + tan θ|+ C

= ln

∣

∣

∣

∣

x

2+

√x2 − 4

2

∣

∣

∣

∣

+ C

h.

∫ π

2

-π2

x4 sin xdx

Since f(x) = x4 sin x is an odd function,

∫ π

2

-π2

x4 sin xdx = 0. Alternatively, use integration

by parts 4 times.

5. Find the solution of the differential equation that satisfies the given conditions.

a.dy

dx=

3

2y; x ≥ 2, y(2) = 1

dy

dx=

3

2y

2y dy = 3 dx

∫

2y dy =

∫

3dx

y2 = 3x+ C

1 = 6 + C

C = -5

y2 = 3x− 5

b.dy

dx=

-y2 − 2

2xy; x > 0, y(1) = 2

dy

dx=

-y2 − 2

2xy

2y

y2 + 2dy = -

1

xdx

∫

2y

y2 + 2dy =

∫

-1

xdx

ln |y2 + 2| = - ln |x|+ C

ln 6 = - ln 1 + C

C = ln 6

ln |y2 + 2| = - ln |x|+ ln 6

y2 + 2 = 6x

xy2 + 2x = 6

8

6. Let C be the graph of y = 19

√x(x− 27) from x = 1 to x = 9. Also, let R be the region

bounded by the curves x = 1, x = 9, C, and the x-axis. Finally, let S be the solid formedby revolving R about the x-axis.

y = 19

√x(x− 27)

y = 19

√x3 − 3

√x

y′ = 16

√x− 3

2√x

[y′]2 =(√

x

6− 3

2√x

)2

[y′]2 = x36

− 12+ 9

4x

[y′]2 + 1 = x36

+ 12+ 9

4x

[y′]2 + 1 =x2 + 18x+ 81

36x

[y′]2 + 1 =(x+ 9)2

36x

√

[y′]2 + 1 =

√

(x+ 9)2

36x

√

[y′]2 + 1 =x+ 9

6√x

√

[y′]2 + 1 =√x

6+ 3

2√x

a. Find the arc length of C.

∫ 9

1

(√x

6+ 3

2√x

)

dx =(

19

√x3 + 3

√x)

∣

∣

∣

∣

9

1

= 12− 289= 80

9

b. Find the area of R.

-

∫ 9

1

(

19

√x3 − 3

√x)

dx =(

2√x3 − 2

45

√x5)

∣

∣

∣

∣

9

1

= 54− 545−(

2− 245

)

= 2165

− 8845

= 185645

c. Find the surface area of S.

-2π

∫ 9

1

19

√x(x− 27)

(

x+ 9

6√x

)

dx

= - π27

∫ 9

1

(x− 27)(x+ 9)dx

= - π27

∫ 9

1

(

x2 − 18x− 243)

dx

= - π27

(

13x3 − 9x2 − 243

2x)

∣

∣

∣

∣

9

1

= - π27(-3159

2+ 781

6)

= - π27

· -43483

= 4348π81

d. Find the volume of S.

π

∫ 9

1

(

19

√x3 − 3

√x)2

dx = π

∫ 9

1

(

181x3 − 2

3x2 + 9x

)

dx

9

= π(

1324

x4 − 29x3 + 9

2x2)

∣

∣

∣

∣

9

1

= π[

814− 162 + 729

2−(

1324

− 29+ 9

2

)]

= 17696π81

7. Let C be the graph of y =√x(x − 1

3) from x = 1 to x = 9. Also, let R be the region


y =√x(x− 1

3) from

y =√x3 − 1

3

√x

y′ = 32

√x− 1

6√x

[y′]2 =(

3√x

2− 1

6√x

)2

[y′]2 = 9x4− 1

2+ 1

36x

[y′]2 + 1 = 9x4+ 1

2+ 1

36x

[y′]2 + 1 =81x2 + 18x+ 1

36x

[y′]2 + 1 =(9x+ 1)2

36x

√

[y′]2 + 1 =

√

(9 + 1)2

36x

√

[y′]2 + 1 =9x+ 1

6√x

√

[y′]2 + 1 = 3√x

2+ 1

6√x


∫ 9

1

(

3√x

2+ 1

6√x

)

dx =(√

x3 + 13

√x)

∣

∣

∣

∣

9

1

= 28− 43= 80

3


∫ 9

1

(√x3 − 1

3

√x)

dx =(

25

√x5 − 2

9

√x3)

∣

∣

∣

∣

9

1

= 4865

− 6−(

25− 2

9

)

= 4565

− 845

= 409645


2π

∫ 9

1

√x(x− 1

3)

(

9x+ 1

6√x

)

dx

= π3

∫ 9

1

(x− 13)(9x+ 1)dx

= π3

∫ 9

1

(

9x2 − 2x− 13

)

dx

= π3

(

3x3 − x2 − 13x)

∣

∣

∣

∣

9

1

= π3(2103− 5

3)

= π3· 6304

3

= 6304π9


10

π

∫ 9

1

(√x3 − 1

3

√x)2

dx

= π

∫ 9

1

(

x3 − 23x2 + 1

9x)

dx

= π(

14x4 − 2

9x3 + 1

18x2)

∣

∣

∣

∣

9

1

= π[

65614

− 162 + 92−(

14− 2

9+ 1

18

)]

= 4448π3

8. Let C be the graph of y = x2

8− ln x from x = 1 to x = 4. Also, let R be the region

bounded by the curves C, x = 1, and x = 4. Finally, let S be the solid formed by revolvingR about the x-axis.





9. Let C be the plane curve defined by the parametric equations x = et cos t and y = et sin tfor 0 ≤ t ≤ π.

x = et cos t

dx

dt= et cos t− et sin t

y = et sin t

dy

dt= et sin t+ et cos t

dy

dx=

dy

dtdxdt

=et sin t+ et cos t

et cos t− et sin t=

sin t+ cos t

cos t− sin t

(

dx

dt

)2

= e2t(cos t− sin t)2 = e2t(cos2 t+ sin2 t− 2 cos t sin t) = e2t(1− 2 cos t sin t)

(

dy

dt

)2

= e2t(cos t + sin t)2 = e2t(cos2 t+ sin2 t + 2 cos t sin t) = e2t(1 + 2 cos t sin t)

(

dx

dt

)2

+

(

dy

dt

)2

= 2e2t

√

(

dx

dt

)2

+

(

dy

dt

)2

=√2et

a. Find the equation of the line that is tangent to C at the point where t = π4.

11

t = π4

x =

√2e

π

4

2

y =

√2e

π

4

2

dxdt

= 0

dy

dxis undefined

The tangent line is vertical.

x =

√2e

π

4

2

b. Find the length of C.

∫ π

0

√2et dt =

√2et∣

∣

∣

∣

π

0

=√2 (eπ − 1)

c. Find the surface area of the solid formed by rotating C about the x-axis.

2π

∫ π

0

√2et · et sin tdt = 2π

∫ π

0

√2e2t sin tdt = 2

√2 π

∫ π

0

e2t sin tdt

Use integration by parts letting u = sin t, du = cos t dt, v = 12e2t, and dv = e2t dt.

2√2 π

∫ π

0

e2t sin tdt = 2√2π

(

12e2t sin t

∣

∣

∣

∣

π

0

−∫ π

0

12e2t cos tdt

)

2√2 π

∫ π

0

e2t sin tdt = -√2 π

∫ π

0

e2t cos tdt

Use integration by parts again letting u = cos t, du = - sin t dt, v = 12e2t, and dv = e2t dt.

2√2 π

∫ π

0


(

12e2t cos t

∣

∣

∣

∣

π

0

−∫ π

0

-12e2t sin tdt

)

2√2 π

∫ π

0


(

-12e2π − 1

2−∫ π

0

-12e2t sin tdt

)

2√2 π

∫ π

0

e2t sin tdt =

√2πe2π

2+

√2π

2−

√2π

2

∫ π

0

e2t sin tdt

2√2 π

∫ π

0

e2t sin tdt =

√2πe2π

2+

√2π

2− 1

4

(

2√2π

∫ π

0

e2t sin tdt

)

5

4

(

2√2π

∫ π

0

e2t sin tdt

)

=

√2πe2π

2+

√2π

2

12

2√2 π

∫ π

0

e2t sin tdt =4

5

(√2πe2π

2+

√2π

2

)

2√2 π

∫ π

0

e2t sin tdt =2√2πe2π + 2

√2π

5

10. Let C be the plane curve defined by the parametric equations x = cos3 θ and y = sin3 θfor 0 ≤ θ ≤ π

2.

x = cos3 θ

dx

dθ= 3 cos2 θ · (- sin θ)

y = sin3 θ

dy

dθ= 3 sin2 θ · cos θ

dy

dx=

dy

dθdxdθ

=3 sin2 θ · cos θ-3 cos2 θ · sin θ = - tan θ

(

dx

dθ

)2

= 9 cos4 θ · sin2 θ

(

dy

dθ

)2

= 9 sin4 θ · cos2 θ

(

dx

dθ

)2

+

(

dy

dθ

)2


√

(

dx

dθ

)2

+

(

dy

dθ

)2

= 3 sin θ · cos θ

a. Find the equation of the line that is tangent to C at the point where θ = π6.

θ = π6

x = 3√3

8

y = 18

dy

dx= - 1√

3

y − 18= - 1√

3

(

x− 3√3

8

)


∫ π

2

0

3 sin θ · cos θ dθ = 32sin2 θ

∣

∣

∣

∣

π

2

0

= 32


2π

∫ π

2

0

(

sin3 θ)

3 sin θ · cos θ dθ = 6π

∫ π

2

0

sin4 θ · cos θ dθ = 6π5sin5

∣

∣

∣

∣

π

2

0

= 6π5

11. For each of the following, sketch the curves with the given polar equations and find thearea inside the first curve and outside the second curve.

13

a.

i. r = 1

ii. r = sin 2θ

π − 4

∫ π

2

0

12sin2 2θ dθ

= π − 2

∫ π

2

0

12(1− cos 4θ)dθ

= π −(

θ − 14sin 4θ

)

∣

∣

∣

∣

π

2

0

= π2

✲✛

✻

❄

b.

i. r = 2 sin θ

ii. r = 1

2 sin θ = 1

sin θ = 12

θ = π6

2

∫ π

2

π

6

12

(

4 sin2 θ − 1)

dθ

=

∫ π

2

π

6

(

2(1− cos 2θ)− 1)

dθ

=

∫ π

2

π

6

(

1− 2 cos 2θ)

dθ

=(

θ − sin 2θ)

∣

∣

∣

∣

π

2

π

6

= π2−(

π6−

√32

)

= π3+

√32

= 2π+3√3

6

✲✛

✻

❄

14

12. For each of the following, sketch the curve and find an integral that represents thelength of the given polar curve. Do not integrate.

a. r = 4 cos θ

drdθ

= -4 sin θ

r2 = 16 cos2 θ

(

drdθ

)2= 16 sin2 θ

r2 +(

drdθ

)2= 16 cos2 θ + 16 sin2 θ

√

r2 +(

drdθ

)2=

√16 cos2 θ + 16 sin2 θ

√

r2 +(

drdθ

)2= 4

∫ π

0

4dθ

= 4θ

∣

∣

∣

∣

π

0

= 4π

✲✛

✻

❄

b. r = 3 cos 2θ

drdθ

= -6 sin 2θ

r2 = 9 cos2 2θ

(

drdθ

)2= 36 sin2 2θ

r2 +(

drdθ

)2= 9 cos2 2θ + 36 sin2 2θ

√

r2 +(

drdθ

)2=

√9 cos2 2θ + 36 sin2 2θ

√

r2 +(

drdθ

)2=

√9 + 27 sin2 2θ

√

r2 +(

drdθ

)2= 3

√1 + 3 sin2 2θ

∫ 2π

0

3√1 + 3 sin2 2θ dθ

✲✛

✻

❄

15

c. r = 2 cos 3θ

drdθ

= -6 sin 3θ

r2 = 4 cos2 3θ

(

drdθ

)2= 36 sin2 3θ

r2 +(

drdθ

)2= 4 cos2 3θ + 36 sin2 3θ

√

r2 +(

drdθ

)2=

√4 cos2 3θ + 36 sin2 3θ

√

r2 +(

drdθ

)2=

√4 + 32 sin2 3θ

√

r2 +(

drdθ

)2= 2

√1 + 8 sin2 3θ

∫ π

0

2√1 + 8 sin2 3θ dθ

✲✛

✻

❄

d. r = cos θ + 1

drdθ

= - sin θ

r2 = cos2 θ + 2 cos θ + 1

(

drdθ

)2= sin2 θ

r2 +(

drdθ

)2= cos2 θ + 2 cos θ + 1 + sin2 θ

r2 +(

drdθ

)2= 2 cos θ + 2

√

r2 +(

drdθ

)2=

√2 cos θ + 2

∫ 2π

0

√2 cos θ + 2dθ

✲✛

✻

❄

16

13. For each of the following, determine whether the series converges absolutely, convergesconditionally, or diverges.

a.∞∑

n=0

en

n!

Since limn→∞

en+1

(n+ 1)!· n!en

= limn→∞

e

n + 1= 0, the series converges by the ratio test.

b.

∞∑

n=1

( n

n + 1

)n2

Since limn→∞

n

√

( n

n+ 1

)n2

= limn→∞

( n

n+ 1

)n

= 1e, the series converges by the root test.

c.∞∑

n=0

( n

n+ 1

)n

Since limn→∞

( n

n + 1

)n

= 1e, the series diverges by the divergence test.

d.∞∑

n=1

(-1)n

nn

Note that for n ≥ 2,1

nn≤ 1

2n. Since

∞∑

n=1

1

2nis a convergent geometric series,

∞∑

n=1

(-1)n

nn

converges absolutely by the comparison test.

e.∞∑

n=1

1√n · 2

√n

∫ ∞

1

1√x · 2

√xdx

= limt→∞

∫ t

1

1√x · 2

√xdx

= limt→∞

∫ t

1

1√x·(

1

2

)

√x

dx

= limt→∞

(

2

ln 12

·(

1

2

)

√x ∣

∣

∣

∣

t

1

)

= limt→∞

(

2

ln 12

·(

1

2

)

√t

− 1

ln 12

)

= 0

Therefore,∞∑

n=1

1√n · 2

√nconverges by the integral test.

17

f.∞∑

n=0

n2

n!

Since limn→∞

(n + 1)2

(n + 1)!· n!n2

= limn→∞

(n+ 1)2

n+ 1· 1

n2= lim

n→∞

n2 + 2n+ 1

n3 + n20, the series converges by

the ratio test.

g.

∞∑

n=4

(-1)n

n2 − 5n+ 6

Since

∞∑

n=4

1

n2 − 5n+ 6=

∞∑

n=4

(

1

n− 3− 1

n− 2

)

= 1− 12+ 1

2− 1

3+ 1

3− 1

4+ 1

4· · · , the series

∞∑

n=4

(-1)n

n2 − 5n+ 6is absolutely convergent.

14. For each of the following power series, find the interval of convergence and radius ofconvergence.

a.∞∑

n=1

2nxn

n2

Since limn→∞

∣

∣

∣

∣

2n+1xn+1

(n + 1)2· n2

2nxn

∣

∣

∣

∣

= |2x| < 1 if and only if |x| < 12, the radius of convergence

is 12. Note that if x = 1

2,

∞∑

n=1

2nxn

n2=

∞∑

n=1

1

n2which is a convergent p-series. If x = -1

2,

∞∑

n=1

2nxn

n2=

∞∑

n=1

(-1)

n2which is absolutely convergent. Therefore, the interval of convergence

is[

-12, 12

]

.

b.

∞∑

n=0

5n(x− 2)n

nn

Since limn→∞

n

√

∣

∣

∣

∣

5n(x− 2)n

nn

∣

∣

∣

∣

= limn→∞

5|x− 2|n

= 0 for all x, the interval of convergence is

(-∞,∞) and the radius of convergence is ∞.

c.∞∑

n=0

n2xn

2n

18

Since limn→∞

∣

∣

∣

∣

(n+ 1)2xn+1

2n+1· 2n

xnn2

∣

∣

∣

∣

=∣

∣

∣

x

2

∣

∣

∣< 1 if and only if |x| < 2, the radius of convergence

is 2. Note that if x = 2, then∞∑

n=0

n2xn

2n=

∞∑

n=0

n2 which diverges and Note that if x = -2,

then∞∑

n=0

n2xn

2n=

∞∑

n=0

-n2 which diverges. Therefore, the interval of convergence is (-2, 2).

d.∞∑

n=1

(-1)n(x+ π)n

2n

Note that if x = -π − 1, then

∞∑

n=1

(-1)n(x+ π)n

2n=

∞∑

n=1

1

2nwhich diverges. If x = -π + 1,

then

∞∑

n=1

(-1)n(x+ π)n

2n=

∞∑

n=1

(-1)n

2nwhich converges. Therefore, the interval of convergence

is (-π − 1, -π + 1] and the radius of convergence is 1.

15. Let g(x) = x tan-1 x. Express g as a power series. What is the radius of convergence?

Since tan-1 x =∞∑

n=0

(-1)nx2n+1

2n+ 1, x tan-1 x =

∞∑

n=0

(-1)nx2n+2

2n+ 1.

The radius of convergence is 1.

16. Let f(x) =√xex. Find the Maclaurin series of f . What is the radius of convergence?

17. Let f(x) = sin x. Find the Taylor series of f centered at π2. What is the radius of

convergence?

f(x) = sin x

f ′(x) = cosx

f ′′(x) = - sin x

f ′′′(x) = - cos x

f (4)(x) = sin x

...

f(

π2

)

= 1

f ′ (π2

)

= 0

f ′′ (π2

)

= -1

f ′′′ (π2

)

= 0

f (4)(

π2

)

= 1

...

∞∑

n=0

(-1)n(

x− π2

)

(2n)!

19

18. Let g(x) = cosx. Find the Taylor series of f centered at π4. What is the radius of

convergence?

g(x) = cos x

g′(x) = - sin x

g′′(x) = - cosx

g′′′(x) = sin x

g(4)(x) = sin x

...

g(

π4

)

=√22

g′(

π4

)

= -√22

g′′(

π4

)

= -√22

g′′′(

π4

)

=√22

g(4)(

π4

)

=√22

...

√2

2 · 0! −√2

2 · 1!(

x− π4

)

−√2

2 · 2!(

x− π4

)2+

√2

2 · 3!(

x− π4

)3+

√2

2 · 4!(

x− π4

)4 −√2

2 · 5!(

x− π4

)5

−√2

2 · 6!(

x− π4

)6+

√2

2 · 7!(

x− π4

)7 · · ·

20


Section 2.1: Quiz

Quiz 107-01-02

(1) 19. Differentiate the function f(x) =√x3 − 2x2 + x+ 11.

f(x) = (x3 − 2x2 + x+ 11)12

f ′(x) = 12(x3 − 2x2 + x+ 11)

- 12(3x2 − 4x+ 1)

20. Integrate.

(1) a.

∫ 2

1

(

x2 − x+ 1)

dx =(

13x3 − 1

2x2 + x

)

∣

∣

∣

∣

2

1

= 83− 2 + 2−

(

13− 1

2+ 1)

= 116

(1) b.

∫

(2x3 + 4) (x4 + 8x− 2)6dx = 1

14(x4 + 8x− 2)

7+ C

Quiz 207-02-02

(1) 21. Let h(x) = x+√x and find h

-1(6).

See the student solutions manual.

(1) 22. Let f(x) =√5x+ 2 and find f

-1(x).


(1) 23. Let f(x) = tan πx2+ x2 + 3 on (-1, 1) and find

(

f-1)′(3).


Quiz 307-03-02

24. Calculate each of the following.

(1) a. log3127

= -3

21

(1) b. log27 3 = 13


(1) a. sin-1 12= π

6

(1) b. sin(

cos-1 23

)

x2 + y2 = 1

49+ y2 = 1

y2 = 59

y =√53

Quiz 407-05-02

26. Differentiate.

(1) a. f(x) = e3x3−4x2+1

f ′(x) = e3x3−4x2+1(9x2 − 8x)

(1) b. f(x) = sin-1 (ex)

f ′(x) =1√

1− e2x· ex =

ex√1− e2x

27. For each of the following, find the equation of the line that is tangent to the graph ofthe given function at the specified point.

(1) a. f(x) = ln(x− 1)2; x =√e + 1

f ′(x) = 2x−1

f(√e + 1) = ln e = 1

f ′(√e + 1) = 2√

e

y − 1 = 2√e(x−√

e− 1)

Quiz 507-08-02

(1) 28. Let f(x) = tan πx2+ x2 + 3 on (-1, 1) and find

(

f-1)′(3).


(1) 29. Given that Q(t) = Q0

(

12

)t

λ = Q0e-kt, solve for k in terms of λ.

22

Q0

(

12

)t

λ = Q0e-kt

(

12

)t

λ = e-kt

ln(

12

)t

λ = ln e-kt

tλln 1

2= -kt

1λln 1

2= -k

1λ(- ln 2) = -k

1λln 2 = k

k = ln 2λ

Quiz 607-09-02

Integrate.

(1) 30.

∫ 1

0

xex dx

Use integration by parts letting u = x, du = dx, v = ex, and dv = ex dx.

∫ 1

0

xex dx = xex∣

∣

∣

∣

1

0

−∫ 1

0

ex dx = xex∣

∣

∣

∣

1

0

−ex∣

∣

∣

∣

1

0

= e− 0− (e− 1) = 1

(1) 31.

∫

xex2dx = 1

2ex

2+ C

(1) 32.

∫

ln xdx

Use integration by parts letting u = ln x, du = 1xdx, v = x, and dv = dx.

∫

ln xdx = x lnx−∫

x · 1xdx = x ln x−

∫

1dx = x ln x− x+ C

Quiz 707-15-02

33. Integrate.

(1) a.

∫ 1

0

ex sin xdx


∫ 1

0


∣

∣

∣

∣

1

0

−∫ 1

0

ex cos xdx

23


ex sin x

∣

∣

∣

∣

1

0

−∫ 1

0

ex cos xdx = e sin 1−(

ex cosx

∣

∣

∣

∣

1

0

−∫ 1

0

ex(− sin x)dx

)

= e sin 1− e cos 1 + 1−∫ 1

0

ex sin xdx

Therefore, 2

∫ 1

0

ex sin xdx = e sin 1−e cos 1+1 and so

∫ 1

0

ex sin xdx = 12(e sin 1−e cos 1+1).

(1) b.

∫ 3√

32

0

x3

(√4x2 + 9

)3 dx

See page 521 of text.

(1) c.

∫

sin 5x cos 3xdx = 12

∫

(

sin 2x+ sin 8x)

dx = -14cos 2x− 1

16cos 8x+ C

Quiz 807-16-02

34. Integrate.

(1) a.

∫

x2 + 1

x2 − xdx


(1) b.

∫

e2x

e2x + 3ex + 2dx


Quiz 907-17-02

35. Determine whether or not the following series converges or diverges.

(1) a.∞∑

n=1

ln

(

n + 1

n

)

Since limn→∞

ln n+1n

1n

= limx→∞

ln x+1x

1x

L′H= lim

x→∞

xx+1

· x−(x+1)x2

- 1x2

= limx→∞

x

x+ 1= 1 and the series

∞∑

n=1

1n

diverges, the series

∞∑

n=1

ln(

n+1n

)

diverges by the limit comparison test.

(1) 36. State the limit comparison theorem for series.

24

See the text or your class notes.

(1) 37. Prove the limit comparison theorem for series.

See the text or your class notes.

Quiz 1007-19-02

38. Integrate.

(1) a.

∫

1

x ln xdx

Let u = ln x and du = 1xdx.

∫

1

x ln xdx =

∫

1

udu = ln |u|+ C = ln | lnx| + C

(1) b.

∫

ln xdx

Use integration by parts letting u = ln x, du = 1xdx, v = x, and dv = dx.

∫

ln xdx = x lnx−∫

x · 1xdx = x ln x−

∫

1dx = x ln x− x+ C

39. Determine whether or not the following series converges or diverges.

(1) a.∞∑

n=2

1

n2 − 2n+ 1

Note that

∞∑

n=2

1

n2 − 2n + 1=

∞∑

n=2

1

(n− 1)2=

∞∑

n=1

1

n2which is a convergent p-series.

Quiz 1107-22-02

40. Integrate.

(1) a.

∫ π

2

π

6

cot x ln(sin x)dx

Let u = ln(sin x) and du = cotx dx.

25

∫ π

2

π

6

cotx ln(sin x)dx =

∫ 0

ln 12

udu = 12u2

∣

∣

∣

∣

0

ln 12

= -12

(

ln 12

)2

(1) b.

∫

x2 + 1

x2 − xdx


(1) c.

∫

e2x

e2x + 3ex + 2dx


Quiz 1207-23-02

41. For each of the following determine whether the given series converges or diverges.

(1) a.∞∑

n=1

(-1)n+1(

1n

)

Since limn→∞

1

n= 0, the series

∞∑

n=1

(-1)n+1(

1n

)

converges by the alternting series test.

(1) b.∞∑

n=1

1√n3 + 1

Since1√

n3 + 1≤ 1√

n3and

∞∑

n=1

1√n3

is a convergent p-series,

∞∑

n=1

1√n3 + 1

converges by the

comparison test.

(1) c.∞∑

n=0

(

n+ 1

n

)n

Since limn→∞

(

n+ 1

n

)n

= e,

∞∑

n=0

(

n+ 1

n

)n

diverges by the divergence test.

Quiz 1307-24-02

42. Integrate.

(1) a.

∫ 1

0

ex+ex dx

∫ 1

0

ex+ex dx =

∫ 1

0

ex · eex dx

26

Let u = ex and du = ex dx.

∫ 1

0

ex · eex dx =

∫ e

1

eu du = eu∣

∣

∣

∣

e

1

= ee − e

(1) b.

∫

4x2 − x− 1

x3 − xdx =

∫(

1

x+

2

x+ 1+

1

x− 1

)

dx = ln |x|+2 ln |x+1|+ ln |x− 1|+C

43. For each of the following, determine whether the series is absolutely convergent, con-ditionally convergent, or divergent.

(1) a.

∞∑

n=1

(-1)n(

n+ 1

n2

)n

Since limn→∞

n

√

(

n+ 1

n2

)n

= limn→∞

n + 1

n2= 0, the series

∞∑

n=1

(-1)n(

n+ 1

n2

)n

is absolutely con-

vergent by the root test.

(1) b.

∞∑

n=1

2n

2n!

For n ≥ 3,2n

2n!≤ 2n

22n=

1

2n. Since

∞∑

n=1

1


∞∑

n=1

2n

2n!converges

by the comparison test.

(1) c. Find the sum of the series∞∑

n=2

(

1

2

)n

.

Since∞∑

n=0

(

1

2

)n

= 2,∞∑

n=2

(

1

2

)n

= 2− 1− 12= 1

2.

27

Quiz 1407-30-02

(4) 44. For each of the following, determine which curve represents the graph of the givenparametric equation.

a. x = 2 sin t and y = 3 cos t+ t

b. x = 8 cos t− 1 and y = 6 sin t+ 1

c. x = 310t3 − 2t− 4 and y = 4

5t2 − t− 6

d. x = t sin t and y = t cos t

I

✲✛

✻

❄

II

✲✛

✻

❄

III

✲✛

✻

❄

IV

✲✛

✻

❄

See solutions from section 12.1 of the class notes.

Quiz 15

28

07-31-02

(1) 45. Let C be the parametric curve defined by x = et and y = ln t. Find the equationof the line that is tangent to the curve at the point corresponding to t = 2.

dxdt

= et

dy

dt= 1

t

dy

dx=

dy

dtdxdt

=1

tet

t = 2

x = e2

y = ln 2

dy

dx=

1

2e2

y − ln 2 = 12e2

(x− e2)

(1) 46. Find the length of the plane curve given by x = et and y = 4 where 0 ≤ t ≤ ln 3.

Note that the curve being described is the line segment with endpoints (1, 4) and (3, 4).This line segment has length 2.

(1) 47. Let C be the parametric curve defined by x = t2 + 1 and y = 6t where√7 ≤ t ≤√

55. Find the surface area of the solid formed by rotating C about the x-axis.

dxdt

= 2t

dy

dt= 6

∫

√55

√7

2π · 6t√4t2 + 36dt =

∫

√55

√7

12πt√4t2 + 36dt =

∫

√55

√7

24πt√t2 + 9dt

Let u = t2 + 9 and du = 2t dt.

∫

√55

√7

24πt√t2 + 9dt =

∫ 64

16

12π√udu = 8π

√u3

∣

∣

∣

∣

64

16

= 8π(512− 64) = 3584π

Quiz 1608-02-02

48. For each of the following, find a power series representation for the given function andfind the radius of convergence.

(1) a. f(x) =1

x+ 2

See page 780 of the text.

(1) b. g(x) = ln(5− x)


29

Quiz 1708-06-02

49. Integrate.

(1) a.

∫

3x2 − x+ 2

x3 − x2 + x− 1dx

∫

3x2 − x+ 2

x3 − x2 + x− 1dx

=

∫

3x2 − x+ 2

(x− 1)(x2 + 1)dx

=

∫(

2

x− 1+

x

x2 + 1

)

dx

= 2 ln |x− 1|+ 12ln |x2 + 1|+ C

(1) b.

∫

18x2 + 24x

3x3 + 6x2 + 2dx

Let u = 3x3 + 6x2 + 2 anddu = (9x2 + 12x) dx.

∫

18x2 + 24x

3x3 + 6x2 + 2dx

=

∫

2

udu

= 2 ln |u|+ C

= 2 ln |3x3 + 6x2 + 2|+ C

(1) c.

∫ 1

0

(ee)x dx

∫ 1

0

(ee)x dx

=

∫ 1

0

eex dx

= 1eeex∣

∣

∣

∣

1

0

= 1e(ee − 1)

Quiz 1808-07-02

50. Integrate.

(1) a.

∫

3x2 − x+ 2

x3 − x2 + x− 1dx

See quiz 17.

(1) b.

∫ 1

0

(ee)x dx

See quiz 17.

51. Find the sum of each of the following series.

(1) a.∞∑

n=0

3(

12

)n=

3

1− 12

= 6

30

(1) b.∞∑

n=0

( ∞∑

k=0

1

2n+1 · 2k

)

=∞∑

n=0

(

1

2n+1

∞∑

k=0

1

2k

)

=∞∑

n=0

2

2n+1 =∞∑

n=0

1

2n= 2

Section 2.2: Quiz 1572

Quiz 106-30-03

(2) 52. Let f(x) =√3x+ 1. Find the equation of the line that is tangent to the graph

of f at x = 1.

f(x) = (3x+ 1)12

f ′(x) = 12(3x+ 1)

- 12 (3)

f ′(x) =3

2√3x+ 1

f(1) = 2

f ′(1) = 34

y − 2 = 34(x− 1)

(3) 53. Find the area of the region bounded by the curves x = 0, x = 1, y = 0, andy = 2x sin x2.

∫ 1

0

2x sin x2 dx

Let u = x2 and du = 2x dx.

∫ 1

0

2x sin x2 dx

=

∫ 1

0

sin udu

= - cosu

∣

∣

∣

∣

1

0

= - cos 1− -1

= 1− cos 1

Quiz 207-01-03

54. For each of the following, determine whether or not the given function is 1–1. If thefunction is 1–1, find the inverse.

(3) a. f(x) = 5x− 6

Suppose that x1 , x2 ∈ R and f(x1) = f(x2). Then

5x1 − 6 = 5x2 − 6

5x1= 5x

2

x1 = x2 .

31

Therefore, f is 1–1.

Find the inverse of f.

y = 5x− 6

5x = y + 6

x =y + 6

5

f-1(x) =

x+ 6

5

(2) b. f(x) = x2 − 2x− 15

Since f(5) = f(-3) = 0, f is not 1–1.

Quiz 307-02-03

55. Calculate.

(1) a. log3 27 = 3

(1) b. log25 5 = 12

(1) c. log2515= -1

2

(1) d. eln 4 = 4

(1) e. sec-1(√2) = π

4

Quiz 407-07-03

56. Differentiate.

(1) a. f(x) = e√x

f(x) = ex12

f ′(x) = ex12 · 1

2x

- 12

f ′(x) =e√x

2√x

(1) b. g(x) =√ex

g(x) = (ex)12

g(x) = e12 x

g′(x) = e12x · 1

2

g′(x) =

√ex

2

(3) 57. Let f(x) = esinx. Find the equation of the line that is tangent to the graph of fwhen x = 0.

f(x) = esinx f ′(x) = cosx · esinx

32

f(0) = 1

f ′(0) = 1

y − 1 = 1(x− 0)

y = x+ 1

Quiz 507-08-03

58. Differentiate.

(1) a.

f(x) = 5x

f ′(x) = 5x · ln 5

(1) b.

g(x) = log5 x

g′(x) = 1x ln 5

(1) c.

y = e√x2+1

y′ = e√x2+1 · 1

2(x2 + 1)

- 12 (2x)

y′ =xe

√x2+1

√x2 + 1

59. Integrate.

(1) a.

∫

1

x2 + 1dx = tan-1 x+ C

(1) b.

∫

2x

x2 + 1dx

Let u = x2 + 1 and du = 2x dx.

∫

2x

x2 + 1dx

=

∫

1

udu

= ln |u|+ C

= ln |x2 + 1|+ C

Quiz 607-09-03

60. Differentiate.

(2) a. y = x√

x

ln y = ln x√

x

ln y =√x ln x

y′

y= 1

2x

- 12 ln x+√x · 1

x

y′

y= lnx

2√x+ 1√

x

y′ = y(

lnx+22√x

)

y′ = x√

x

(

lnx+22√x

)

33

61. Integrate.

(2) a.

∫

2x

x2 + 1dx


∫

2x

x2 + 1dx

=

∫

1

udu

= ln |u|+ C

= ln |x2 + 1|+ C

(2) 62. Let f(x) = x5 + x3 +1. Find the equation of the line that is tangent to the graph

of f-1at the point (3, 1).

f(x) = x5 + x3 + 1

f ′(x) = 5x4 + 3x2

f ′(1) = 8

[f-1]′(3) = 1

8

y − 1 = 18(x− 3)

34

Quiz 707-11-03


(2) a. limx→0

ln xx

= limx→0

x ln x

= limx→0

lnx1x

L’H= lim

x→0

1x

- 1x2

= limx→0

-x

= 0

64. Integrate.

(2) a.

∫

x sin xdx

Use integration by parts letting u = x, du = dx, v = - cosx, and dv = sin x dx.

∫

x sin xdx = -x cosx−∫

- cosxdx = -x cosx+ sin x+ C

(1) 65. Let f(x) = x5 + x3 +1. Find the equation of the line that is tangent to the graph

of f-1at the point (3, 1).

f(x) = x5 + x3 + 1

f ′(x) = 5x4 + 3x2

f ′(1) = 8

[f-1]′(3) = 1

8

y − 1 = 18(x− 3)

Quiz 807-14-03

66. Integrate.

(2) a.

∫

cotxdx

=

∫

cos x

sin xdx


∫

cosx

sin xdx

35

=

∫

1

udu

= ln |u|+ C

= ln | sinx|+ C

(2) b.

∫ 1

0

xex dx


∫ 1

0

xex dx = xex∣

∣

∣

∣

1

0

−∫ 1

0

ex dx = xex∣

∣

∣

∣

1

0

− ex∣

∣

∣

∣

1

0

= e− 0− (e− 1) = 1

(2) 67.

∫

tan-1 xdx

Use integration by parts letting u = tan-1 x, du = dxx2+1

, v = x, and dv = dx.

∫

tan-1 xdx = x tan-1 x −∫

x

x2 + 1dx = x tan-1 x − 1

2ln |x2 + 1|+ C

Quiz 907-15-03

(6) 68. Let f(x) = ex sin x on the interval [-π, π]. Find the intervals of increase anddecrease, local extrema, intervals of concavity, and inflection points. Use these answers tosketch the graph.

f(x) = ex sin x

f ′(x) = ex sin x+ ex cos x

f ′(x) = ex(sin x+ cosx)

Dec:[

-π, -π4

]

∪[

3π4, π]

Inc:[

-π4, 3π

4

]

Local max:4√e3π√2

at 3π4

Local min: - 1√2 4√eπ

at -π4

f ′(x) = ex(sin x+ cosx)

f ′′(x) = ex(sin x+cosx)+ex(cosx−sin x)

f ′′(x) = 2ex cos x

CD:(

-π, -π2

)

∪(

π2, π)

CU:(

-π2, π2

)

IP:(

-π2, - 1√

eπ

)

,(

π2,√eπ)

f(x) = ex sinx

36

✲✛

✻

❄

π

2-π2

5

-5

Quiz 1007-16-03

(2) 69. Integrate.

∫ eπ

1

sin(ln x)

xdx


∫ eπ

1

sin(ln x)

xdx

=

∫ π

0

sin udu

= - cosu

∣

∣

∣

∣

π

0

= 2

70. Answer each of the following as true or false (write the entire word) and justify youranswer.

(2) a. limn→∞

n

√

n+ 1

n= 0

False.

For all n ∈ N, n

√

n+ 1

n≥ 1. So lim

n→∞n

√

n+ 1

n6= 0.

(2) b. The series∞∑

n=1

1nconverges by the divergence test.

False.

This statement is false for two reasons. The first reason is that this series diverges. Thesecond reason is that the divergence test can never be used to show that a series converges.

37

Quiz 1107-21-03

71. For each of the following, determine whether the given series converges or diverges.

(2) a.

∞∑

n=1

en

n

limn→∞

en

n

= limx→∞

ex

x

L’H= lim

x→∞

ex

1

= ∞

Diverges by the divergence test.

(3) b.

∞∑

n=1

1

n + 1

Hint: Recall that

∞∑

n=1

1

ndiverges.

If

∞∑

n=1

1

n + 1= s, then

∞∑

n=1

1

n= s− 1. This is not possible since

∞∑

n=1

1

ndiverges. Therefore,

∞∑

n=1

1

n+ 1diverges.

Quiz 1207-22-03

(2) 72. Answer the following as true or false (write the entire word) and justify youranswer.

The series

∞∑

n=1

n

endiverges by the divergence test.

False.

limn→∞

n

en

= limx→∞

x

ex

L’H= lim

x→∞

1

ex

= 0

The divergence test fails.

(3) 73. Determine whether the given series converges or diverges.

38

∞∑

n=1

1

n+ 1

Quiz 1307-23-03

74. Integrate.

(3) a.

∫

√x2 − 1

xdx

1

√x2 − 1

x

θ

x = sec θ

dx = sec θ tan θ dθ

∫

√x2 − 1

xdx

=

∫

√sec2 θ − 1

sec θsec θ tan θ dθ

=

∫

tan2 θ dθ

=

∫

(

sec2 θ − 1)

dθ

= tan θ − θ + C

=√x2 − 1− sec-1 x+ C

(3) b.

∫ 1

0

√1− x2 dx

√1− x2

x1

θ

x = sin θ

dx = cos θ dθ

∫ 1

0

√1− x2 dx

=

∫ π

2

0

√

1− sin2 θ cos θ dθ

=

∫ π

2

0

cos2 θ dθ

= 12

∫ π

2

0

(

1 + cos 2θ)

dθ

= 12

(

θ + 12sin 2θ

)

∣

∣

∣

∣

π

2

0

= π4

Quiz 1407-25-03

(5) 75. Integrate.

∫

5x2 + 11x+ 7

x3 + 2x2 + x+ 2dx

39

=

∫(

4x+ 3

x2 + 1+

1

x+ 2

)

dx =

∫(

4x

x2 + 1+

3

x2 + 1+

1

x+ 2

)

dx

= 2 ln |x2 + 1|+ 3 tan-1 x+ ln |x+ 2|+C

Quiz 1507-28-03

(5) 76. Integrate.

∫

5x2 + 11x+ 7

x3 + 2x2 + x+ 2dx

Quiz 1607-29-03

(2) 77. Differentiate and simplify.

f(x) = ln | sec x+ tan x|

f ′(x) =sec x tanx+ sec2 x

sec x+ tan x

f ′(x) =sec x(tanx+ sec x)

sec x+ tanx

f ′(x) = sec x

40

78. For each of the following, find the length of the given curve.

(5) a. y =3

3√x2

2; [0, 3

√3]

y = 32x

23

y′ = x- 13

y′ = 13√x

(y′)2 = 13√x2

∫ 3√3

0

√

1 + 13√x2dx

=

∫ 3√3

0

√

3√x2 + 13√x2

dx

=

∫ 3√3

0

√

3√x2 + 13√x

dx

=

∫ 3√3

0

x- 13

(

x23 + 1

)12

dx

=(

x23 + 1

)32

∣

∣

∣

∣

3√3

0

=

√

(

3√x2 + 1

)3∣

∣

∣

∣

3√3

0

= 8− 1

= 7

(3) b. y = ln(cos x);[

0, π4

]

y = ln(cosx)

y′ = - sinxcos x

y′ = - tan x

(y′)2 = tan2 x

∫ π

4

0

√1 + tan2 xdx

=

∫ π

4

0

√sec2 xdx

=

∫ π

4

0

sec xdx

= ln | sec x+ tan x|∣

∣

∣

∣

π

4

0

= ln |√2 + 1| − ln |1 + 0|

= ln(√2 + 1)

Quiz 1707-30-03

79. For each of the following, determine whether the series converges or diverges.

(2) a.∞∑

n=1

nn

2n

41

Note that for n ≥ 3,nn

2n=(n

2

)n

≥(

3

2

)n

. Since∞∑

n=1

3n

2nis a divergent geometric series,

∞∑

n=1

nn

2ndiverges by the divergence test.

(2) b.∞∑

n=1

110√n11

Since the series is a p-series with p = 1110

> 1, the series converges.

(4) c.∞∑

n=1

lnn

n2

∫ ∞

1

ln x

x2dx

= limt→∞

∫ t

1

ln x

x2dx

Use integration by parts letting u = ln x, du = 1xdx, v = - 1

x, and dv = 1

x2 dx.

limt→∞

∫ t

1

ln x

x2dx

= limt→∞

[

-lnx

x

∣

∣

∣

∣

∞

1

−∫ t

1

-1

x2dx

]

= limt→∞

(

-ln x

x

∣

∣

∣

∣

∞

1

− 1

x

∣

∣

∣

∣

t

1

)

= limt→∞

(

-ln t

t− 0− 1

t+ 1

)

= 1

Therefore, the series converges by the integral test.

Quiz 1808-01-03

(5) 80. Find the length of the given curve.

y =3

3√x2

2; [0, 3

√3]

42

Quiz 1908-04-03

81. Let f(x) =√x+ 1. Also, let R be the region bounded by the curves x = 1, x = 5,

y = 0, and y = f(x) and let S be the 3-dimensional solid formed by revolving R about thex-axis.

(2) a. Find the area of R.

∫ 5

1

√x+ 1dx

= 23

√

(x+ 1)3∣

∣

∣

∣

5

1

= 23

(

6√6− 2

√2)

= 4√6− 4

√2

3

(3) b. Find the volume of S.

π

∫ 5

1

(

x+ 1)

dx

= π(

12x2 + x

)

∣

∣

∣

∣

5

1

= π[

252+ 5−

(

12+ 1)]

= 16π

(4) c. Find the surface area of S.

f(x) =√x+ 1

f ′(x) = 12√x+1

[f ′(x)]2 = 14(x+1)

2π

∫ 5

1

√x+ 1

√

1 + 14(x+1)

dx

= 2π

∫ 5

1

√x+ 1

√

4x+54(x+1)

dx

= π

∫ 5

1

√4x+ 5dx

= π6

√

(4x+ 5)3∣

∣

∣

∣

5

1

= π6(125− 27)

= 49π3

Quiz 2008-05-03

For each of the following, determine whether the series is absolutely convergent, condition-ally convergent, or divergent.

(2) 82.∞∑

n=0

cos nπ

n+ 2

43

∞∑

n=0

cosnπ

n + 2= 1

2− 1

3+ 1

4− 1

5+ · · ·

The alternating harmonic series converges by the alternating series test but fails to beabsolutely convergent by the divergence test.

(2) 83.

∞∑

n=0

n− 1

n + 1

Since limn→∞

n− 1

n+ 1= 1,

∞∑

n=0

n− 1

n+ 1diverges by the divergence test.

(3) 84.

∞∑

n=1

(-1)nn tan 1n

Apply the divergence test. Since limn→∞

(-1)nn tan 1n= 0 if and only if lim

n→∞

∣

∣(-1)nn tan 1n

∣

∣ = 0,

we need only check limn→∞

n tan 1n.

Since limx→∞

x tan 1x= lim

x→∞

tan 1x

1x

L’H= lim

x→∞

- 1x2 sec

2 1x

- 1x2

= limx→∞

sec2 1x= 1, the series diverges by

the divergence test.

Quiz 2108-06-03

(3) 85. Let C be the parametric curve defined by x = et and y = ln t. Find the equationof the line that is tangent to the curve at the point corresponding to t = 2.

dxdt

= et

dy

dt= 1

t

dy

dx=

dy

dtdxdt

=1

tet

t = 2

x = e2

y = ln 2

dy

dx=

1

2e2

y − ln 2 = 12e2

(x− e2)

86. Let x = t2 + 1 and y = t ln t.

(2) a. Find dy

dxdy

dx=

ln t+ 1

2t

44

(2) b. Find d2y

dx2

d2y

dx2

=

(1t)(2t)− (ln t + 1)(2)

(2t)2

2t

=-2 ln t

8t3

= -ln t

4t3

Section 2.3: Review

Solutions to Review Problems

Omit questions 13, 16, and 21 due to typos. You are not required to turn in your photocopy.

87. Calculate.

a. csc-1 2 = π6

b. sin (tan-1 3)

y

x= 3

y = 3x

x2 + y2 = 1

x2 + 9x2 = 1

10x2 = 1

x2 = 110

x = 1√10

y = 3√10

sin (tan-1 3) = 3√10

88. Given that loga b = 1, simplify a− b.

Since loga b = 1, a = b and so a− b− 0.

89. Given that loga b = -1, simplify a · b.

Since loga b = -1, b = 1aand so a · b = 1.

90. Given that loga b =12, simplify a

b.

Since loga b =12, b =

√a and so a

b= a√

a=

√a.

91. Given the graph of f, sketch the graph of f-1.

45

✲✛

✻

❄

✲✛

✻

❄

92. For each of the following, find the equation of the line that that is tangent to the graphof f

-1at the given point.

a. f(x) = x3 + 2x+ 3; x = -30

f ′(x) = 3x2 + 2

f-1(-30) = -3

f ′(-3) = 29

[

f-1]′(-30) = 1

29

y + 3 = 129(x+ 30)

b. f(x) = tanx; x = 1

f ′(x) = sec2 x

f-1(1) = π

4

f ′(π4) = 2

[

f-1]′(1) = 1

2

y − π4= 1

2(x− 1)


a. limx→0

sin x

xL′H= lim

x→0

cos x

1= 1

b. limx→0

ex

ln x= 0

c. limx→∞

ln x

exL′H= lim

x→∞

1

xex= 0

d. limx→∞

(√x2 + 1− x

)

limx→∞

(√x2 + 1− x

)

= limx→∞

(√x2 + 1− x

)

·√x2 + 1 + x√x2 + 1 + x

= limx→∞

1√x2 + 1 + x

e. limx→0

(

sin x)x

First, consider

limx→0

(

ln sin x)x

= limx→0

x ln sin x

46

= limx→0

ln sin x1x

L′H= lim

x→0-x2 cosx

sin x

= limx→0

-x cosxx

sin x

= 0.

Therefore, limx→0

(

sin x)x

= e0 = 1.

f. limx→0

ln(sin x) = -∞

g. limn→∞

n!

nn

limn→∞

n!

nn

= limn→∞

n(n− 1)(n− 2) · · ·2 · 1n · n · · ·n

= limn→∞

(

n

n· n− 1

n· n− 2

n· · · 2

n· 1n

)

≤ limn→∞

1

n

= 0

h. limn→∞

en

n!

Since limn→∞

en+1

(n + 1)!· n!en

= limn→∞

e


∞∑

n=0

en



en


94. Differentiate.

a. k(x) = esinx

k′(x) = esinx(cosx)

b. f(x) = ln |x3 + x2 − 1|

f ′(x) =3x2 + 2x

x3 + x2 − 1

c. g(x) = tan-1(sin x)

g′(x) =cos x

1 + sin2 x

d. f(x) = sin (ex)


e. y = xsinx

y = xsinx

ln y = lnxsin x

ln y = sin x ln x

y′

y= cosx ln x+ sin x · 1

x

y′

y=

x cosx ln x+ sin x

x

y′ =x cosx ln x+ sin x

x· y

y′ =x

sinx


x

y′ = xsin x−1


47

f. h(x) = xex

y = xex

ln y = ln xex

ln y = ex ln x

y′

y= ex ln x+

ex

x

y′

y=

xex ln x+ ex

x

y′ =xex ln x+ ex

x· y

y′ =x

ex

(xex ln x+ ex)

x

y′ = xex−1(xex lnx+ ex)

g. f(x) = eex

f ′(x) = exeex

48

95. Integrate.

a.

∫ √x2 + 1dx

✟✟✟✟✟✟✟✟✟✟

θ

√x2 + 1

1

x

x = tan θ

dx = sec2 θ dθ

∫ √x2 + 1dx =

∫ √tan2 θ + 1 sec2 θ dθ =

∫

sec3 θ dθ


∫


sec x · tan2 xdx

∫



∫


(

sec3 x− sec x)

dx

∫


sec3 xdx+

∫

sec xdx

2

∫


∫

sec xdx

2

∫


∫


49

b.

∫ 1

0

ex sin xdx


∫ 1

0


∣

∣

∣

∣

1

0

−∫ 1

0

ex cos xdx


∫ 1

0


∣

∣

∣

∣

1

0

−∫ 1

0

ex cos xdx

∫ 1

0

ex sin xdx = e sin 1−(

ex cos x

∣

∣

∣

∣

1

0

−∫ 1

0

ex(- sin x)dx

)

∫ 1

0

ex sin xdx = e sin 1− e cos 1 + 1−∫ 1

0

ex sin xdx

2

∫ 1

0

ex sin xdx = e sin 1− e cos 1 + 1

∫ 1

0

ex sin xdx = 12(e sin 1− e cos 1 + 1)

c.

∫

x2 + 2x+ 3

x3 + x2 + x+ 1dx

∫

x2 + 2x+ 3

x3 + x2 + x+ 1dx

=

∫

x2 + 2x+ 3

(x2 + 1)(x+ 1)dx

=

∫(

2

x2 + 1+

1

x+ 1

)

dx

= 2 tan-1 x+ ln |x+ 1|+ C

50

d.

∫

sin2 θ cos2 θ dθ

=

∫

(1− cos2 θ) cos2 θ dθ

=

∫

[1− 12(cos 2θ + 1)] · 1

2(cos 2θ + 1)dθ

= 12

∫

(

-12cos 2θ + 1

2

)

(cos 2θ + 1)dθ

= -14

∫

(cos 2θ − 1)(cos 2θ + 1)dθ

= -14

∫

(

cos2 2θ − 1)

dθ

= -14

∫

(

12(cos 4θ + 1)− 1

)

dθ

= -14

∫

(

12cos 4θ − 1

2

)

dθ

= -18

∫

(

cos 4θ − 1)

dθ

= -18

(

14sin 4θ − θ

)

+ C

= - 132sin 4θ + 1

8θ + C

e.

∫

sin20 x cos3 xdx

=

∫

sin20 x · cos2 x · cosxdx

=

∫


=

∫



∫


=

∫

(

u20 − u22)

du

= 121u21 − 1

23u23 + C

= 121sin21 x− 1

23sin23 x+ C

f.

∫

2x√x2 − 4

dx

Let u = x2 − 4 and du = 2x dx.

∫

2x√x2 − 4

dx

=

∫

1√udu

= 2√u+ C

= 2√x2 − 4 + C

51

g.

∫

1√x2 − 4

dx

✟✟✟✟✟✟✟✟✟✟

θ

x √x2 − 4

2

x = 2 sec θ

dx = 2 sec θ tan θ dθ

∫

1√x2 − 4

dx

=

∫

2 sec θ tan θ√4 sec2 θ − 4

dθ

=

∫

sec θ dθ

= ln | sec θ + tan θ|+ C

= ln

∣

∣

∣

∣

x

2+

√x2 − 4

2

∣

∣

∣

∣

+ C

h.

∫ π

2

-π2

x4 sin xdx

Since f(x) = x4 sin x is an odd function,

∫ π

2

-π2

x4 sin xdx = 0. Alternatively, use integration

by parts 4 times.


a.dy

dx=

3

2y; x ≥ 2, y(2) = 1

dy

dx=

3

2y

2y dy = 3 dx

∫

2y dy =

∫

3dx

y2 = 3x+ C

1 = 6 + C

C = -5

y2 = 3x− 5

b.dy

dx=

-y2 − 2

2xy; x > 0, y(1) = 2

dy

dx=

-y2 − 2

2xy

2y

y2 + 2dy = -

1

xdx

∫

2y

y2 + 2dy =

∫

-1

xdx

ln |y2 + 2| = - ln |x|+ C

ln 6 = - ln 1 + C

C = ln 6

ln |y2 + 2| = - ln |x|+ ln 6

y2 + 2 = 6x

xy2 + 2x = 6

52




y = 19

√x(x− 27)

y = 19

√x3 − 3

√x

y′ = 16

√x− 3

2√x

[y′]2 =(√

x

6− 3

2√x

)2

[y′]2 = x36

− 12+ 9

4x

[y′]2 + 1 = x36

+ 12+ 9

4x

[y′]2 + 1 =x2 + 18x+ 81

36x

[y′]2 + 1 =(x+ 9)2

36x

√

[y′]2 + 1 =

√

(x+ 9)2

36x

√

[y′]2 + 1 =x+ 9

6√x

√

[y′]2 + 1 =√x

6+ 3

2√x


∫ 9

1

(√x

6+ 3

2√x

)

dx =(

19

√x3 + 3

√x)

∣

∣

∣

∣

9

1

= 12− 289= 80

9


-

∫ 9

1

(

19

√x3 − 3

√x)

dx =(

2√x3 − 2

45

√x5)

∣

∣

∣

∣

9

1

= 54− 545−(

2− 245

)

= 2165

− 8845

= 185645


-2π

∫ 9

1

19

√x(x− 27)

(

x+ 9

6√x

)

dx

= - π27

∫ 9

1

(x− 27)(x+ 9)dx

= - π27

∫ 9

1

(

x2 − 18x− 243)

dx

= - π27

(

13x3 − 9x2 − 243

2x)

∣

∣

∣

∣

9

1

= - π27(-3159

2+ 781

6)

= - π27

· -43483

= 4348π81


π

∫ 9

1

(

19

√x3 − 3

√x)2

dx = π

∫ 9

1

(

181x3 − 2

3x2 + 9x

)

dx

53

= π(

1324

x4 − 29x3 + 9

2x2)

∣

∣

∣

∣

9

1

= π[

814− 162 + 729

2−(

1324

− 29+ 9

2

)]

= 17696π81

98. Let C be the graph of y =√x(x − 1

3) from x = 1 to x = 9. Also, let R be the region


y =√x(x− 1

3) from

y =√x3 − 1

3

√x

y′ = 32

√x− 1

6√x

[y′]2 =(

3√x

2− 1

6√x

)2

[y′]2 = 9x4− 1

2+ 1

36x

[y′]2 + 1 = 9x4+ 1

2+ 1

36x

[y′]2 + 1 =81x2 + 18x+ 1

36x

[y′]2 + 1 =(9x+ 1)2

36x

√

[y′]2 + 1 =

√

(9 + 1)2

36x

√

[y′]2 + 1 =9x+ 1

6√x

√

[y′]2 + 1 = 3√x

2+ 1

6√x


∫ 9

1

(

3√x

2+ 1

6√x

)

dx =(√

x3 + 13

√x)

∣

∣

∣

∣

9

1

= 28− 43= 80

3


∫ 9

1

(√x3 − 1

3

√x)

dx =(

25

√x5 − 2

9

√x3)

∣

∣

∣

∣

9

1

= 4865

− 6−(

25− 2

9

)

= 4565

− 845

= 409645


2π

∫ 9

1

√x(x− 1

3)

(

9x+ 1

6√x

)

dx

= π3

∫ 9

1

(x− 13)(9x+ 1)dx

= π3

∫ 9

1

(

9x2 − 2x− 13

)

dx

= π3

(

3x3 − x2 − 13x)

∣

∣

∣

∣

9

1

= π3(2103− 5

3)

= π3· 6304

3

= 6304π9


54

π

∫ 9

1

(√x3 − 1

3

√x)2

dx

= π

∫ 9

1

(

x3 − 23x2 + 1

9x)

dx

= π(

14x4 − 2

9x3 + 1

18x2)

∣

∣

∣

∣

9

1

= π[

65614

− 162 + 92−(

14− 2

9+ 1

18

)]

= 4448π3

99. Let C be the graph of y = x2

8− ln x from x = 1 to x = 4. Also, let R be the region

bounded by the curves C, x = 1, and x = 4. Finally, let S be the solid formed by revolvingR about the x-axis.





100. Let C be the plane curve defined by the parametric equations x = et cos t and y =et sin t for 0 ≤ t ≤ π.

x = et cos t

dx

dt= et cos t− et sin t

y = et sin t

dy

dt= et sin t+ et cos t

dy

dx=

dy

dtdxdt

=et sin t+ et cos t

et cos t− et sin t=

sin t+ cos t

cos t− sin t

(

dx

dt

)2

= e2t(cos t− sin t)2 = e2t(cos2 t+ sin2 t− 2 cos t sin t) = e2t(1− 2 cos t sin t)

(

dy

dt

)2

= e2t(cos t + sin t)2 = e2t(cos2 t+ sin2 t + 2 cos t sin t) = e2t(1 + 2 cos t sin t)

(

dx

dt

)2

+

(

dy

dt

)2

= 2e2t

√

(

dx

dt

)2

+

(

dy

dt

)2

=√2et

a. Find the equation of the line that is tangent to C at the point where t = π4.

55

t = π4

x =

√2e

π

4

2

y =

√2e

π

4

2

dxdt

= 0

dy

dxis undefined

The tangent line is vertical.

x =

√2e

π

4

2


∫ π

0

√2et dt =

√2et∣

∣

∣

∣

π

0

=√2 (eπ − 1)


2π

∫ π

0

√2et · et sin tdt = 2π

∫ π

0

√2e2t sin tdt = 2

√2 π

∫ π

0

e2t sin tdt

Use integration by parts letting u = sin t, du = cos t dt, v = 12e2t, and dv = e2t dt.

2√2 π

∫ π

0

e2t sin tdt = 2√2π

(

12e2t sin t

∣

∣

∣

∣

π

0

−∫ π

0

12e2t cos tdt

)

2√2 π

∫ π

0


∫ π

0

e2t cos tdt

Use integration by parts again letting u = cos t, du = - sin t dt, v = 12e2t, and dv = e2t dt.

2√2 π

∫ π

0


(

12e2t cos t

∣

∣

∣

∣

π

0

−∫ π

0

-12e2t sin tdt

)

2√2 π

∫ π

0


(

-12e2π − 1

2−∫ π

0

-12e2t sin tdt

)

2√2 π

∫ π

0

e2t sin tdt =

√2πe2π

2+

√2π

2−

√2π

2

∫ π

0

e2t sin tdt

2√2 π

∫ π

0

e2t sin tdt =

√2πe2π

2+

√2π

2− 1

4

(

2√2π

∫ π

0

e2t sin tdt

)

5

4

(

2√2π

∫ π

0

e2t sin tdt

)

=

√2πe2π

2+

√2π

2

56

2√2 π

∫ π

0

e2t sin tdt =4

5

(√2πe2π

2+

√2π

2

)

2√2 π

∫ π

0

e2t sin tdt =2√2πe2π + 2

√2π

5


2.

x = cos3 θ

dx


y = sin3 θ

dy


dy

dx=

dy

dθdxdθ


(

dx

dθ

)2


(

dy

dθ

)2


(

dx

dθ

)2

+

(

dy

dθ

)2


√

(

dx

dθ

)2

+

(

dy

dθ

)2



θ = π6

x = 3√3

8

y = 18

dy

dx= - 1√

3

y − 18= - 1√

3

(

x− 3√3

8

)


∫ π

2

0


∣

∣

∣

∣

π

2

0

= 32


2π

∫ π

2

0

(

sin3 θ)


∫ π

2

0


∣

∣

∣

∣

π

2

0

= 6π5

102. For each of the following, sketch the curves with the given polar equations and findthe area inside the first curve and outside the second curve.

57

a.

i. r = 1

ii. r = sin 2θ

π − 4

∫ π

2

0

12sin2 2θ dθ

= π − 2

∫ π

2

0

12(1− cos 4θ)dθ

= π −(

θ − 14sin 4θ

)

∣

∣

∣

∣

π

2

0

= π2

✲✛

✻

❄

b.

i. r = 2 sin θ

ii. r = 1

2 sin θ = 1

sin θ = 12

θ = π6

2

∫ π

2

π

6

12

(

4 sin2 θ − 1)

dθ

=

∫ π

2

π

6

(

2(1− cos 2θ)− 1)

dθ

=

∫ π

2

π

6

(

1− 2 cos 2θ)

dθ

=(

θ − sin 2θ)

∣

∣

∣

∣

π

2

π

6

= π2−(

π6−

√32

)

= π3+

√32

= 2π+3√3

6

✲✛

✻

❄

58

103. For each of the following, sketch the curve and find an integral that represents thelength of the given polar curve. Do not integrate.

a. r = 4 cos θ

drdθ

= -4 sin θ

r2 = 16 cos2 θ

(

drdθ

)2= 16 sin2 θ

r2 +(

drdθ

)2= 16 cos2 θ + 16 sin2 θ

√

r2 +(

drdθ

)2=

√16 cos2 θ + 16 sin2 θ

√

r2 +(

drdθ

)2= 4

∫ π

0

4dθ

= 4θ

∣

∣

∣

∣

π

0

= 4π

✲✛

✻

❄

b. r = 3 cos 2θ

drdθ

= -6 sin 2θ

r2 = 9 cos2 2θ

(

drdθ

)2= 36 sin2 2θ

r2 +(

drdθ

)2= 9 cos2 2θ + 36 sin2 2θ

√

r2 +(

drdθ

)2=

√9 cos2 2θ + 36 sin2 2θ

√

r2 +(

drdθ

)2=

√9 + 27 sin2 2θ

√

r2 +(

drdθ

)2= 3

√1 + 3 sin2 2θ

∫ 2π

0

3√1 + 3 sin2 2θ dθ

✲✛

✻

❄

59

c. r = 2 cos 3θ

drdθ

= -6 sin 3θ

r2 = 4 cos2 3θ

(

drdθ

)2= 36 sin2 3θ

r2 +(

drdθ

)2= 4 cos2 3θ + 36 sin2 3θ

√

r2 +(

drdθ

)2=

√4 cos2 3θ + 36 sin2 3θ

√

r2 +(

drdθ

)2=

√4 + 32 sin2 3θ

√

r2 +(

drdθ

)2= 2

√1 + 8 sin2 3θ

∫ π

0

2√1 + 8 sin2 3θ dθ

✲✛

✻

❄

d. r = cos θ + 1

drdθ

= - sin θ

r2 = cos2 θ + 2 cos θ + 1

(

drdθ

)2= sin2 θ

r2 +(

drdθ

)2= cos2 θ + 2 cos θ + 1 + sin2 θ

r2 +(

drdθ

)2= 2 cos θ + 2

√

r2 +(

drdθ

)2=

√2 cos θ + 2

∫ 2π

0

√2 cos θ + 2dθ

✲✛

✻

❄

60


a.∞∑

n=0

en

n!

Since limn→∞

en+1

(n+ 1)!· n!en

= limn→∞

e

n + 1= 0, the series converges by the ratio test.

b.

∞∑

n=1

( n

n + 1

)n2

Since limn→∞

n

√

( n

n+ 1

)n2

= limn→∞

( n

n+ 1

)n

= 1e, the series converges by the root test.

c.∞∑

n=0

( n

n+ 1

)n

Since limn→∞

( n

n + 1

)n

= 1e, the series diverges by the divergence test.

d.∞∑

n=1

(-1)n

nn

Note that for n ≥ 2,1

nn≤ 1

2n. Since

∞∑

n=1

1


∞∑

n=1

(-1)n

nn

converges absolutely by the comparison test.

e.∞∑

n=1

1√n · 2

√n

∫ ∞

1

1√x · 2

√xdx

= limt→∞

∫ t

1

1√x · 2

√xdx

= limt→∞

∫ t

1

1√x·(

1

2

)

√x

dx

= limt→∞

(

2

ln 12

·(

1

2

)

√x ∣

∣

∣

∣

t

1

)

= limt→∞

(

2

ln 12

·(

1

2

)

√t

− 1

ln 12

)

= 0

Therefore,∞∑

n=1

1√n · 2

√nconverges by the integral test.

61

f.∞∑

n=0

n2

n!

Since limn→∞

(n + 1)2

(n + 1)!· n!n2

= limn→∞

(n+ 1)2

n+ 1· 1

n2= lim

n→∞

n2 + 2n+ 1

n3 + n20, the series converges by

the ratio test.

g.

∞∑

n=4

(-1)n

n2 − 5n+ 6

Since

∞∑

n=4

1

n2 − 5n+ 6=

∞∑

n=4

(

1

n− 3− 1

n− 2

)

= 1− 12+ 1

2− 1

3+ 1

3− 1

4+ 1

4· · · , the series

∞∑

n=4

(-1)n

n2 − 5n+ 6is absolutely convergent.


a.∞∑

n=1

2nxn

n2

Since limn→∞

∣

∣

∣

∣

2n+1xn+1

(n + 1)2· n2

2nxn

∣

∣

∣

∣

= |2x| < 1 if and only if |x| < 12, the radius of convergence

is 12. Note that if x = 1

2,

∞∑

n=1

2nxn

n2=

∞∑

n=1

1

n2which is a convergent p-series. If x = -1

2,

∞∑

n=1

2nxn

n2=

∞∑

n=1

(-1)

n2which is absolutely convergent. Therefore, the interval of convergence

is[

-12, 12

]

.

b.

∞∑

n=0

5n(x− 2)n

nn

Since limn→∞

n

√

∣

∣

∣

∣

5n(x− 2)n

nn

∣

∣

∣

∣

= limn→∞

5|x− 2|n

= 0 for all x, the interval of convergence is

(-∞,∞) and the radius of convergence is ∞.

c.∞∑

n=0

n2xn

2n

62

Since limn→∞

∣

∣

∣

∣

(n+ 1)2xn+1

2n+1· 2n

xnn2

∣

∣

∣

∣

=∣

∣

∣

x

2

∣

∣

∣< 1 if and only if |x| < 2, the radius of convergence

is 2. Note that if x = 2, then∞∑

n=0

n2xn

2n=

∞∑

n=0

n2 which diverges and Note that if x = -2,

then∞∑

n=0

n2xn

2n=

∞∑

n=0

-n2 which diverges. Therefore, the interval of convergence is (-2, 2).

d.∞∑

n=1

(-1)n(x+ π)n

2n

Note that if x = -π − 1, then

∞∑

n=1

(-1)n(x+ π)n

2n=

∞∑

n=1

1

2nwhich diverges. If x = -π + 1,

then

∞∑

n=1

(-1)n(x+ π)n

2n=

∞∑

n=1

(-1)n

2nwhich converges. Therefore, the interval of convergence

is (-π − 1, -π + 1] and the radius of convergence is 1.

106. Let g(x) = x tan-1 x. Express g as a power series. What is the radius of convergence?

Since tan-1 x =∞∑

n=0

(-1)nx2n+1

2n+ 1, x tan-1 x =

∞∑

n=0

(-1)nx2n+2

2n+ 1.

The radius of convergence is 1.

107. Let f(x) =√xex.

108. Let f(x) = sin x. Find the Taylor series of f centered at π2. What is the radius of

convergence?

f(x) = sin x

f ′(x) = cosx

f ′′(x) = - sin x

f ′′′(x) = - cos x

f (4)(x) = sin x

...

f(

π2

)

= 1

f ′ (π2

)

= 0

f ′′ (π2

)

= -1

f ′′′ (π2

)

= 0

f (4)(

π2

)

= 1

...

∞∑

n=0

(-1)n(

x− π2

)

(2n)!

63

109. Let g(x) = cosx. Find the Taylor series of f centered at π4. What is the radius of

convergence?

g(x) = cos x

g′(x) = - sin x

g′′(x) = - cosx

g′′′(x) = sin x

g(4)(x) = sin x

...

g(

π4

)

=√22

g′(

π4

)

= -√22

g′′(

π4

)

= -√22

g′′′(

π4

)

=√22

g(4)(

π4

)

=√22

...

√2

2 · 0! −√2

2 · 1!(

x− π4

)

−√2

2 · 2!(

x− π4

)2+

√2

2 · 3!(

x− π4

)3+

√2

2 · 4!(

x− π4

)4 −√2

2 · 5!(

x− π4

)5

−√2

2 · 6!(

x− π4

)6+

√2

2 · 7!(

x− π4

)7 · · ·

Section 2.4: Exam 1

Midterm Exam 1572 Summer 2003

110. Differentiate.

(3) a. f(x) = ex4+x2−4

f ′(x) = (4x3 + 2x)ex4+x2−4

(3) b. g(x) = ln | cosx|

g′(x) =- sin x

cosx

g′(x) = - tanx

(3) c. h(x) = tan-1 ex

h′(x) =ex

1 + e2x

(5) d. y = xex

ln y = lnxex

ln y = ex ln x

ddx

ln y = ddx

(

ex ln x)

y′

y= ex ln x+ ex · 1

x

y′

y=

ex(x ln x+ 1)

x

y′ =yex(x ln x+ 1)

x

y′ =xexex(x ln x+ 1)

x

64

y′ = xex−1ex(x ln x+ 1)

(3) 111. State the definition of a one-to-one function.

112. Let f(x) = x7 + x5 + 2x.

(2) a. Find f ′(x).

f ′(x) = 7x6 + 5x4 + 2

(4) b. Use the derivative of f to prove that f is one-to-one.

Proof : Since f ′(x) > 0 for all x ∈ R, f is strictly increasing and therefore, one-to-one.

(4) c. Find the equation of the line that is tangent to the graph of f-1(x) at the point

where x = 4.

f-1(4) = 1

f ′(x) = 7x6 + 5x4 + 2

f ′(1) = 14

(

f-1)′(4) = 1

14

y − 1 = 114(x− 4)

65


(3) a. limx→∞

ex

x

L’H= lim

x→∞

ex

1= ∞

(6) b. limx→0

xx

= limx→0

elnxx

= limx→0

ex lnx

= elim

x→0x lnx

limx→0

x ln x

= limx→0

ln x1x

L’H= lim

x→0

1x

- 1x2

= limx→0

-x2

= 0

elim

x→0x lnx

= e0

= 1

114. Integrate.

(4) a.

∫ eπ

1

sin(lnx)

xdx


∫ eπ

1

sin(ln x)

xdx

=

∫ π

0

sin udu

= - cosu

∣

∣

∣

∣

π

0

= 2

(5) b.

∫ π

3

π

4

tan2 x csc xdx

=

∫ π

3

π

4

sin x

cos2 xdx

=

∫ π

3

π

4

tan x sec xdx

= sec x

∣

∣

∣

∣

π

3

π

4

= 2−√2

(4) c.

∫

sin 2x cos 3xdx

=

∫

12(sin(-x) + sin 5x)dx

=

∫

(

12sin(-x) + 1

2sin 5x

)

dx

= -12cos(-x)− 1

10cos 5x+ C

= -12cosx− 1

10cos 5x+ C

(5) d.

∫

ex

1 + e2xdx

Let u = ex and du = ex dx.

∫

ex

1 + e2xdx

66

=

∫

1

1 + u2du

= tan-1 u+ C

= tan-1 ex + C

(4) e.

∫

x4 ln xdx

Use integration by parts letting u = ln x, du = 1xdx, v = 1

5x5, and dv = x4 dx.

∫

x4 lnxdx = 15x5 ln x−

∫

15x4 dx = 1

5x5 ln x− 1

25x5 + C


(4) a.

∞∑

n=1

3n3 + 2n− 1

6n3 + n2

limn→∞

3n3 + 2n− 1

6n3 + n2=

1

2


(5) b.

∞∑

n=1

1

n2 + n

=∞∑

n=1

1

n(n + 1)

=∞∑

n=1

(

1

n− 1

n+ 1

)

= limn→∞

(

1− 1n+1

)

= 1

(5) c.∞∑

n=1

n

√

n+ 1

n

For all n ∈ N, n

√

n + 1

n≥ 1.

So limn→∞

n

√

n+ 1

n6= 0.


(3) 116. If possible, give an example of a sequence {xn}∞n=1 such that lim

n→∞x

n= 0 and

∞∑

n=1

xndiverges. If this is not possible, explain why it is not.

Section 2.5: Final

Final Exam Math 1572 Summer 2002

117. Calculate.

a. tan-1 1 = π4

b. sec(

sin-1√32

)

= sec π3= 2

c. log9 3 = 12

d. log319= -2

67

118. Find the equation of the line that that is tangent to the graph of f-1at the given

point.

f(x) = 2x3 + x+ 3; x = -15

f ′(x) = 6x2 + 1

f-1(-15) = -2

f ′(-2) = 25

[

f-1]′(-15) = 1

25

y + 2 = 125(x+ 15)


a. limx→∞

x2

exL’H

= limx→∞

2x

exL’H

= limx→∞

2

ex= 0 b. lim

x→0

cosx

x= ∞

c. limn→∞

en

n!

Since limn→∞

en+1

(n + 1)!· n!en

= limn→∞

e


∞∑

n=0

en



en


120. Differentiate.

a. f(x) = ex7+x2

f ′(x) = ex7+x2

(7x6 + 2x)

b. g(x) = ln | sin x|

g′(x) =cosx

sin x= cot x

c. h(x) = tan-1(x2 + 4x+ 1)

h′(x) =2x+ 4

(x2 + 4x+ 1)2 + 1

d. f(x) = sec(sin x)

f ′(x) = sec(sin x) · tan(sin x) · cos x

e. y = x(x2+1)

y = x(x2+1)

ln y = lnx(x2+1)

ln y = (x2 + 1) lnx

y′

y= 2x ln x+

x2 + 1

x

y′

y=

2x2 ln x+ x2 + 1

x

y′ =2x2 ln x+ x2 + 1

x· y

y′ =2x2 ln x+ x2 + 1

x· x(

x2+1)

y′ = xx2

(2x2 ln x+ x2 + 1)

68

121. Integrate.

a.

∫

2x+ 1

x2 + 1dx =

∫(

2x

x2 + 1+

1

x2 + 1

)

dx = ln |x2 + 1|+ tan-1 x+ C

b.

∫ π

2

0

x sin xdx

Use integration by parts letting u = x, du = dx, v = - cosx, and dv = sin x dx.

∫ π

2

0

x sin xdx

= -x cos x

∣

∣

∣

∣

π

2

0

−∫ π

2

0

- cosxdx

= -x cosx

∣

∣

∣

∣

π

2

0

+ sinx

∣

∣

∣

∣

π

2

0

= 1

c.

∫

sin3 xdx

∫

sin3 xdx

=

∫

sin2 x · sin xdx

=

∫

(1− cos2 x) sin xdx

Let u = cosx and du = - sin x dx.

∫

(1− cos2 x) sin xdx

=

∫

-(1− u2)du

=

∫

(

u2 − 1)

du

= 13u3 − u+ C

= 13cos3 x− cos x+ C

d.

∫

4

x2 − 2x− 3dx =

∫(

1

x− 3− 1

x+ 1

)

dx = ln |x− 3| − ln |x+ 1|+ C

e.

∫

sec3 xdx =

∫

sec2 x · sec xdx


∫


sec x · tan2 xdx

∫



∫


(

sec3 x− sec x)

dx

69

∫


sec3 xdx+

∫

sec xdx

2

∫


∫

sec xdx

2

∫


∫





y = 13

√x(x− 3)

y = 13

√x3 −√

x

y′ = 12

√x− 1

2√x

[y′]2 =(√

x

2− 1

2√x

)2

[y′]2 = x4− 1

2+ 1

4x

[y′]2 + 1 = x4+ 1

2+ 1

4x

[y′]2 + 1 =x2 + 2x+ 1

4x

[y′]2 + 1 =(x+ 1)2

4x

√

[y′]2 + 1 =

√

(x+ 1)2

4x

√

[y′]2 + 1 =x+ 1

2√x

√

[y′]2 + 1 =√x

2+ 1

2√x


∫ 9

3

(√x

2+ 1

2√x

)

dx =(

13

√x3 +

√x)

∣

∣

∣

∣

9

3

= 12− 2√3


∫ 9

3

(

13

√x3 −√

x)

dx =(

215

√x5 − 2

3

√x3)

∣

∣

∣

∣

9

3

= 1625

− 18−(

65

√3− 2

√3)

= 725+ 4

5

√3 = 72+4

√3

5


2π

∫ 9

3

13

√x(x− 3)

(

x+ 1

2√x

)

dx = π3

∫ 9

3

(x− 3)(x+ 1)dx

70

= π3

∫ 9

3

(

x2 − 2x− 3)

dx

= π3

(

13x3 − x2 − 3x

)

∣

∣

∣

∣

9

3

= π3(135 + 9)

= π3(135 + 9)

= 48π


π

∫ 9

3

(

13

√x3 −√

x)2

dx

= π

∫ 9

3

(

19x3 − 2

3x2 + x

)

dx

= π(

136x4 − 2

9x3 + 1

2x2)

∣

∣

∣

∣

9

3

= π[

7294

− 162 + 812−(

94− 6 + 9

2

)]

= 60π

123. Let C be the plane curve defined by the parametric equations x = 23t3 and y = t2 + 1

for 0 ≤ t ≤ 2√2.

x = 23t3

dx

dt= 2t2

y = t2 + 1

dy

dt= 2t

dy

dx=

dy

dtdxdt

=2t

2t2=

1

t

(

dx

dt

)2

= 4t4

(

dy

dt

)2

= 4t2

(

dx

dt

)2

+

(

dy

dt

)2

= 4t4 + 4t2

√

(

dx

dt

)2

+

(

dy

dt

)2

=√4t4 + 4t2

√

(

dx

dt

)2

+

(

dy

dt

)2

= 2t√t2 + 1

a. Find the equation of the line that is tangent to C at the point where t =√3.

t =√3

x = 2√3

y = 4

dy

dx= 1√

3

y − 4 = 1√3(x− 2

√3)


71

∫ 2√2

0

2t√t2 + 1dt = 2

3

√

(t2 + 1)3∣

∣

∣

∣

2√2

0

= 18− 23= 52

3


2π

∫ 2√2

0

(t2 + 1) · 2t√t2 + 1dt

= 2π

∫ 2√2

0

2t√

(t2 + 1)3 dt

= 2π(

25

√

(t2 + 1)5)

∣

∣

∣

∣

2√2

0

= 45π(√

(t2 + 1)5)

∣

∣

∣

∣

2√2

0

= 45π(√

(t2 + 1)5)

∣

∣

∣

∣

2√2

0

= 45π(243− 1)

= 968π5


2.

x = cos3 θ

dx


y = sin3 θ

dy


dy

dx=

dy

dtdxdt


(

dx

dθ

)2


(

dy

dθ

)2


(

dx

dθ

)2

+

(

dy

dθ

)2


√

(

dx

dθ

)2

+

(

dy

dθ

)2



θ = π6

x = 3√3

8

y = 18

dy

dx= - 1√

3

y − 18= - 1√

3

(

x− 3√3

8

)


∫ π

2

0


∣

∣

∣

∣

π

2

0

= 32

72


2π

∫ π

2

0

(

sin3 θ)


∫ π

2

0


∣

∣

∣

∣

π

2

0

= 6π5

125. For each of the following, sketch the curve with the given polar equation and find thearea enclosed by the curve.

a. r = 8 sin θ

16π

✲✛

✻

❄

73

b. r = 4 cos 2θ

8

∫ π

4

0

12(4 cos 2θ)2 dθ

= 64

∫ π

4

0

cos2 2θ dθ

= 32

∫ π

4

0

(

1 + cos 4θ)

dθ

= 32(

θ + 14sin 4θ

)

∣

∣

∣

∣

π

4

0

= 8π

✲✛

✻

❄

126. Find the length of the polar curve r = eθ with 0 ≤ θ ≤ 2.

drdθ

= eθ

drdθ

= eθ

r2 +(

drdθ

)2= 2e2θ

√

r2 +(

drdθ

)2=

√2eθ

∫ 2

0

√2eθ dθ =

√2eθ∣

∣

∣

∣

2

0

=√2(e2 − 1)


a.

∞∑

n=1

n!

n2

Sincen!

n2≥ n(n− 1)(n− 2)

n2for n ≥ 3 and lim

n→∞

n(n− 1)(n− 2)

n2= ∞, lim

n→∞

n!

n2= ∞.

Therefore,∞∑

n=1

n!

n2diverges by the divergence test.

b.∞∑

n=1

(

sinn

n

)n

Since limn→∞

n

√

∣

∣

∣

∣

(

sinn

n

)n∣∣

∣

∣

= limn→∞

| sinn|n

= 0,∞∑

n=1

(

sinn

n

)n

converges absolutely by the root

test.

74

c.∞∑

n=1

√n

n2 + 2n + 5

Since

√n

n2 + 2n+ 5≤

√n

n2=

1

n32

and∞∑

n=1

1

n32

is a convergent p-series,∞∑

n=1

√n

n2 + 2n + 5con-

verges by the comparison test.

d.∞∑

n=1

(-1)n

3√n

The series∞∑

n=1

(-1)n

3√n

converges by the alternating series test but fails to be absolutely

convergent since∞∑

n=1

13√n

is a divergent p-series.

e.∞∑

n=1

(

n+ 1

n

)n

Since limn→∞

∞∑

n=1

(

n+ 1

n

)n

= e > 0,

∞∑

n=1

(

n + 1

n

)n



a.∞∑

n=0

n!xn

en

Since limn→∞

∣

∣

∣

∣

(n+ 1)!xn+1

en+1· en

n!xn

∣

∣

∣

∣

= limn→∞

(n+ 1)|xn|e

= ∞ for all x 6= 0, the interval of con-

vergence is {0} and the radius of convergence is 0.

b.

∞∑

n=1

(-1)n(x− 2)n

n

Since

∞∑

n=1

(-1)n

nconverges and

∞∑

n=1

1

ndiverges, the interval of convergence must be (1, 3]

and so the radius of convergence is 1.

129. Let g(x) = ex.

a. Give the Maclaurin series for g and find the radius of convergence.

75

g(x) = ex

g′(x) = ex

g(n)(x) = ex for all n

g(0) = 1

g′(0) = 1

g(n)(0) = 1 for all n

Thus, the Maclaurin series representation for ex is 1 + x+ x2

2!+ x3

3!+ · · · =

∞∑

n=0

xn

n!.

Since limn→∞

∣

∣

∣

∣

xn+1

(n + 1)!· n!xn

∣

∣

∣

∣

= limn→∞

|x|n + 1

= 0 for all x, the interval of convergence is (-∞,∞)

and the radius of convergence is ∞.

b. Use your answer from the previous part to express e2 as an infinite series.

e2 =∞∑

n=0

2n

n!

76

Chapter 3: Fall 2005

Section 3.1: Exam 1 Math 1572 Fall 2005

Exam 1 Math 1572 Fall 2005

130. Let f(x) = x55 + 11x43 + 12x.

(3) a. Show that f is 1–1.

Proof : Since f ′(x) = 55x54+473x42+12 = ≥ 12 > 0, f is strictly increasing which impliesthat f is 1–1.

(3) b. f-1(-24) = -1

131. Differentiate.

(4) a. f(x) = esinx

f ′(x) = cosxesinx

(4) b. g(x) = ln(x3 − 21x+ 1)

g′(x) =3x2 − 21

x3 − 21x+ 1

(4) c. h(x) = sin-1

(cosx)

h′(x) =- sin x√1− cos2 x

=- sin x√sin2 x

=- sin x

| sin x|

(4) d. y = xlnx

ln y = ln xlnx

ln y = ln x ln x

ln y = (ln x)2

y′

y= 2 lnx

x

y′ = y · 2 lnxx

y′ = xlnx · 2 lnxx

y′ = 2xlnx−1 ln x

(Page 484: 39) (3) 132. A ladder 10 ft long leans against a vertical wall. If the bottom ofthe ladder slides away from the base of the wall at a speed of 2 ft/s, how fast is the anglebetween the ladder and the wall changing when the bottom of the ladder is 6 ft from thebase of the wall?



(4) a. limx→∞

x

exL’H= lim

x→∞

1

ex= 0 (4) b. lim

x→-∞

x

ex= lim

x→∞-xex = -∞

77

134. Integrate.

(4) a.

∫

(sec2 x) etanx dx = etan x + C

(4) b.

∫ 4

2

x3 lnxdx

Use integration by parts letting u = ln x,du = 1

xdx, v = 1


∫ 4

2

x3 ln xdx

= 14x4 lnx

∣

∣

∣

∣

4

2

−∫ 4

2

14x3 dx

= 14x4 lnx

∣

∣

∣

∣

4

2

− 116x4

∣

∣

∣

∣

4

2

= 64 ln 4− 4 ln 2− (16− 1)

= 64 ln 4− 4 ln 2− 15

= 64 ln 22 − 4 ln 2− 15

= 128 ln 2− 4 ln 2− 15

= 124 ln 2− 15

(4) c.

∫

cos2 xdx

=

∫

12(1 + cos 2x)dx

=

∫

(

12+ 1

2cos 2x

)

dx

= 12x+ 1

4sin 2x+ C

(4) d.

∫ π

4

0

tan3 θ sec2 θ dθ

= 14tan4 θ

∣

∣

∣

∣

π

4

0

= 14

(4) e.

∫

tan-1 xdx

Use integration by parts letting u =tan-1 x, du = dx

1+x2 , v = x, and dv = dx.

∫

tan-1 xdx

= x tan-1−∫

x

1 + x2dx

= x tan-1−12ln(1 + x2) + C

Total Points: 310

78

Chapter 4: Math 1572 Fall 2006

Section 4.1: Exam 1



(4) a. limx→0

sin-1

x

x

L’H= lim

x→0

1√1− x2

= 1

(4) b. limx→0

x ln x

= limx→0

ln x1x

L’H= lim

x→0

1x

- 1x2

= limx→0

-x

= 0

136. Differentiate.

(4) a. y = ln | sin x|

y′ =cosx

sin x= cot x

(4) b. f(x) = ln xx

f(x) = x ln x

f ′(x) = ln x+ xx

f ′(x) = ln x+ 1

(4) c. f(x) = sin-1

ex

f ′(x) =ex√

1− e2x

(4) d. f(x) = esin-1

x

f ′(x) =esin

-1x

√1− x2

137. Integrate.

(4) a.

∫ π

4

0

tan xdx

- ln | cosx|∣

∣

∣

∣

π

4

0

= - ln√22

= ln√2

(4) b.

∫

3x+ 13

x2 + 9x+ 20dx

=

∫(

1

x+ 4+

2

x+ 5

)

dx

= ln |x+ 4|+ 2 ln |x+ 5|+ C

(4) c.

∫

2x3

x2 + 1dx

79

=

∫

2xx2

x2 + 1dx


∫

2xx2

x2 + 1dx

=

∫

u− 1

udx

=

∫

(

1− 1u

)

dx

= u− ln |u|+ C

= x2 + 1− ln |x2 + 1|+ C

(4) d.

∫ √1− x2 dx

(4) e.

∫ e

1

ln x2

xdx =

∫ e

1

2 lnx

xdx


∫ e

1

2 lnx

xdx

=

∫ 1

0

2udu

= u2

∣

∣

∣

∣

1

0

= 1

(4) f.

∫

x sec-1xdx

Use integration by parts letting u = sec-1x, du = 1

x√x2−1

dx, v = 12x2, and dv = x dx.

∫

x sec-1xdx = 1

2x2 sec

-1x−

∫

x

2√x2 − 1

dx = 12x2 sec

-1x− 1

2

√x2 − 1 + C

Section 4.2: Exam 2


(Page 588: 5) (5) 138. Find the length of the curve defined by f(x) = 6√x3 + 1 from

x = 0 to x = 1.

(Page 595: 7) (5) 139. Find the surface area of the solid of revolution created by rotatingthe following curve about the x-axis.

y =√x; from x = 4 to x = 9.

140. For each of the following, determine whether the series is absolutely convergent,conditionally convergent, or divergent.

(5) a.∞∑

n=0

1

4n

80

This is a geometric series with r = 4.

∞∑

n=0

1

4n=

4

3

(5) b.∞∑

n=1

n!

nn

For n ≥ 3,n!

nn=

1 · 2 · 3 · · ·nn · n · n · · ·n ≤ 2

n2. Since

∞∑

n=1

1

n2is a convergent p-series,

∞∑

n=1

n!

nnconverges


(5) c.∞∑

n=0

en

n!

Since limn→∞

en+1

(n+ 1)!· n!en

= limn→∞

e


∞∑

n=0

en


(5) d.

∞∑

n=1

nn

(n2 + 1)n

Since limn→∞

n

√

nn

(n2 + 1)n= lim

n→∞

n

(n2 + 1)= 0, the series

∞∑

n=1

nn

(n2 + 1)nconverges by the root

test.

81

(5) e.∞∑

n=1

(

n + 1

n

)n

Since limn→∞

(

n+ 1

n

)n

= e > 0,∞∑

n=1

(

n + 1

n

)n


(5) f.∞∑

n=1

(-1)n√n

First, note that

∞∑

n=1

(-1)n√n

converges by the alternating series test. The series is not abso-

lutely convergent since∞∑

n=1

1√n

is a divergent p-series.

(Page 784: 31) (5) g.

∞∑

n=1

5n

4n + 3n

(5) h.

∞∑

n=1

sin n+ 2

n

Sincesin n+ 2

n≥ 1

nand

∞∑

n=1

1

nis a divergent p-series,

∞∑

n=1

sinn+ 2

ndiverges.

Section 4.3: Exam 3


141. Eliminate the parameter to find a Cartesian equation of the curve and sketch thecurve.

(6) a. x = et, y = t, t ∈ R

t = ln x

y = ln x

y = lnx

✲✛

✻

❄

82


2.

x = cos3 θ

dx


y = sin3 θ

dy


dy

dx=

dy

dtdxdt


(

dx

dθ

)2


(

dy

dθ

)2


(

dx

dθ

)2

+

(

dy

dθ

)2


√

(

dx

dθ

)2

+

(

dy

dθ

)2


(6) a. Find the equation of the line that is tangent to C at the point where θ = π6.

θ = π6

x = 3√3

8

y = 18

dy

dx= - 1√

3

y − 18= - 1√

3

(

x− 3√3

8

)

(6) b. Find the length of C.

∫ π

2

0


∣

∣

∣

∣

π

2

0

= 32

(6) c. Find the surface area of the solid formed by rotating C about the x-axis.

2π

∫ π

2

0

(

sin3 θ)


∫ π

2

0


∣

∣

∣

∣

π

2

0

= 6π5

143. Let C be the plane curve with parametric equations x = 2 cos θ and y = 4 sin θ for0 ≤ θ ≤ π.

(6) a. Find the area of the region enclosed by C.

dx

dθ= -2 sin θ

dy

dθ= 4 cos θ

∣

∣

∣

∣

∫ π

0

2 cos θ · 4 cos θ dθ∣

∣

∣

∣

=

∣

∣

∣

∣

∫ π

0

8 cos2 θ dθ

∣

∣

∣

∣

83

=

∣

∣

∣

∣

∫ π

0

4(1 + cos 2θ)dθ

∣

∣

∣

∣

=(

4θ + 2 sin 2θ)

∣

∣

∣

∣

π

0

= 4π

144. Consider the polar curve r = 4 cos 2θ.

(6) a. Sketch the curve.

r = 4 cos 2θ

✲✛

✻

❄

(6) b. Calculate the area enclosed bythe curve.

r = 4 cos 2θ

8

∫ π

4

0

12(4 cos 2θ)2 dθ

= 64

∫ π

4

0

cos2 2θ dθ

= 32

∫ π

4

0

(

1 + cos 4θ)

dθ

= 32(

θ + 14sin 4θ

)

∣

∣

∣

∣

π

4

0

= 8π

(6) c. Give the integral whose value is the length of the curve. Do not evaluate the integral.


(Page 643: 11) (6) a.dy

dx= y2 + 1; y(1) = 0

(6) 146. A tank contains 500 gallons of brine which contains .5 pounds of salt per gallon.Pure water flows into the tank at a rate of 10 gallons per minute. The solution is keptthoroughly mixed and drains from the tank at a rate of 10 gallons per minute. Express theamount of salt in the tank as a function of time.

Let Q = Q(t) be the amount of salt in the tank at time t.

dQ

dt= -

Q

50

dQ

Q= -

dt

50

∫

dQ

Q=

∫

-dt

50

ln |Q| = - t50

+ C

Since Q(0) = 250, C = ln 250.

84

ln |Q| = t50

+ ln 250

Q = et50+ln 250

Q = 250e- t50

85

Chapter 5: Math 1572 Spring 2007

Section 5.1: Exam 1

Exam 1 Math 1572 Spring 2007

(4) 147.

Given the graph of f, sketch the graph of f-1.

y = f(x)

✲✛

✻

❄

y = f-1

(x)

✲✛

✻

❄

(4) 148. Suppose that g(x) = f-1(x), f(2) = 7, and f ′(2) = 5. Find g′(7).

g′(7) = 15

(Page 431: 15) (4) 149. Find the domain and range of f(x) =1

ex + 1.

150. Calculate.

(Page 483: 7) (4) a. tan[sin-1(23)] (4) b. eln 6 = 6

151. Differentiate.

(4) a. g(x) = ln(x3 + x2 − 4)

g′(x) =3x2 + 2x

x3 + x2 − 4

(4) b. f(x) = etan−1 x

f ′(x) =etan

−1 x

x2 + 1

(Page 450: 43) (4) c. y = xx

86

(4) 152. A radioactive substance decays from 50 grams to 45 grams in one year. Findthe half-life.

Q(t) = Q0

12

t

λ

45 = 50 · 12

1λ

4550

= 12

1λ

ln 910

= ln 12

1λ

ln 910

= 1λln 1

2

λ =ln 1

2

ln 910

153. Determine whether each of the following sequences converges or diverges.

(4) a.

{

n2

n!

}∞

n=1

limn→∞

n2

n!

≤ limn→∞

n2

n(n− 1)(n− 2)

= limn→∞

n2

n3 − 3n2 + 2n

= 0

(4) b.{( n

n+ 1

)n}∞

n=1

limn→∞

( n

n + 1

)n

= 1e


(4) a.

∞∑

n=1

n+ 1

n2

Sincen+ 1

n2=

1

n+

1

n2≥ 1

nand

∞∑

n=1

1

ndiverges,

∞∑

n=1

n+ 1

n2diverges by the comparison test.

(4) b.

∞∑

n=1

1

nn

Since1

nn≤ 1

2nand

∞∑

n=1

1

2nconverges,

∞∑

n=1

1

nnconverges by the comparison test.

Section 5.2: Exam 2


87

155. Integrate.

(5) a.

∫ 1

0

x2ex3dx = 1

3ex

3

∣

∣

∣

∣

1

0

= 13e− 1

3

(5) b.

∫

1

x√x2 + 1

dx

1

x

θ

√ x2 + 1

x = tan θ

dx = sec2 θ dθ

∫

1

x√x2 + 1

dx

=

∫

cot θ cos θ sec2 θ dθ

=

∫

csc θ dθ

= ln | csc θ − cot θ|+ C

= ln∣

∣

∣

√x2+1x

− 1x

∣

∣

∣+ C

(5) c.

∫

4

x2 − 2x− 3dx

=

∫(

1

x− 3− 1

x+ 1

)

dx

= ln |x− 3| − ln |x+ 1|+ C

(5) d.

∫ π

2

0

sin9 x cos3 xdx

=

∫ π

2

0

sin9 x cos2 x cos xdx

=

∫ π

2

0


=

∫ π

2

0

(sin9 x− sin11 x) cos xdx

=(

110sin10 x− 1

12sin12 x

)

∣

∣

∣

∣

π

2

0

= 110

− 112

= 160

(5) e.

∫

x√x+ 1dx

Let u = x+ 1 and du = dx.

∫

x√x+ 1dx

=

∫

u12 (u− 1)du

=

∫

(

u32 − u

12

)

du

= 25u

52 − 2

3u

32 + C

= 25(x+ 1)

52 − 2

3(x+ 1)

32 + C

(5) f.

∫

x sec x tan xdx

Use integration by parts letting u = x, du = dx, v = sec x and dv = sec x tanx dx.

∫

x sec x tanxdx = x sec x−∫

sec xdx = x sec x− ln | sec x+ tanx| + C

88

156. For each of the following, determine whether the series is absolutely convergent,conditionally convergent, or divergent.

(5) a.∞∑

n=1

(

n + 1

n

)n

Since limn→∞

(

n+ 1

n

)n

= e,∞∑

n=1

(

n+ 1

n

)n


Note that the root test is inconclusive.

(5) b.∞∑

n=2

(-1)n

n√lnn

Since limn→∞

1

n√lnn

= 0,∞∑

n=2

(-1)n

n√lnn

converges by the alternating series test. Also, since

∫ ∞

2

1

x√ln x

dx= limt→∞

∫ t

2

1

x√ln x

dx= limt→∞

2√ln x

∣

∣

∣

∣

t

2

= limt→∞

(

2√ln t− 2 ln 2

)

=∞,∞∑

n=2

1

n√lnn

diverges by the integral test. Therefore,∞∑

n=2

(-1)n

n√lnn

is conditionally convergent.

(5) c.∞∑

n=1

(n + 1)3n

n!

Since limn→∞

∣

∣

∣

∣

(n + 2)3n+1

(n+ 1)!· n!

(n+ 1)3n

∣

∣

∣

∣

= limn→∞

3(n+ 2)

(n+ 1)(n+ 1)= lim

n→∞

3n+ 6

n2 + 2n+ 1= 0, the se-

ries is absolutely convergent by the ratio test.

(5) d.

∞∑

n=1

sin n

n2

Since

∣

∣

∣

∣

sinn

n2

∣

∣

∣

∣

≤ 1

n2and

∞∑

n=1

1

n2is a convergent p-series,

∞∑

n=1

sinn

n2is absolutely convergent


(5) e.∞∑

n=1

tan-1 n

n

Sincetan-1 n

n≥ π

4nand

∞∑

n=1

π

4nis a divergent p-series,

∞∑

n=1

tan-1 n

ndiverges by the compar-

ison test.

89

Section 5.3: Exam 3


(5) 157. Find the length of the following curve.

y = 12

√x(x− 2); 16 ≤ x ≤ 48

y′ = 14x

- 12 (x− 2) + 12

√x

y′ = (x−2)4√x+ 1

2√x

y′ = x−2+24√x

y′ =√x

4

√

(y′)2 + 1 =√

x16

+ 1

∫ 48

16

√

x16

+ 1dx

=(

323

√

( x16

+ 1)3)

∣

∣

∣

∣

48

16

= 2563

− 32√2

3

158. Let R be the region bounded by the curves y =√x, x = 3

4, x = 2, and the x-axis.

Also, let S be the solid formed by rotating R about the x-axis.


∫ 2

34

√xdx

= 23x

32

∣

∣

∣

∣

2

34

= 4√2

3−

√34

= 16√2−3

√3

12

(5) b. Find the surface area of S.

y′ = 12√x

∫ 2

34

2π√x√

14x

+ 1dx

=

∫ 2

34

2π√x

√

4x+ 1

4xdx

=

∫ 2

34

π√4x+ 1dx

= π6(4x+ 1)

32

∣

∣

∣

∣

2

34

= π6(27− 8)

= 19π6

(5) c. Find the center of mass of R.

x

= 1216

√2−3

√3

∫ 2

34

x√xdx

= 1216

√2−3

√3

(

25

√x5)

∣

∣

∣

∣

2

34

= 1216

√2−3

√3

(

25

√x5)

∣

∣

∣

∣

2

34

90

= 1216

√2−3

√3· 25

(

4√2− 9

√3

32

)

y

= 1216

√2−3

√3

∫ 2

34

12xdx

= 1216

√2−3

√3

(

14x2)

∣

∣

∣

∣

2

34

= 1216

√2−3

√3

(

1− 964

)

(Page 605: 1) (5) 159. An aquarium 5 ft long and 2 ft wide, 3 ft deep is full of water.Find the hydrostatic force on one end of the aquarium.

(5) 160. Eliminate the parameter to find a Cartesian equation of the curve and sketchthe curve.

x = et,

y = t,

t ∈ R

t = ln x

y = ln x

y = lnx

✲✛

✻

❄


2.

x = cos3 θ

dx


y = sin3 θ

dy


dy

dx=

dy

dtdxdt


(

dx

dθ

)2


(

dy

dθ

)2


(

dx

dθ

)2

+

(

dy

dθ

)2


√

(

dx

dθ

)2

+

(

dy

dθ

)2


91

(5) a. Find the equation of the line that is tangent to C at the point where θ = π6.

θ = π6

x = 3√3

8

y = 18

dy

dx= - 1√

3

y − 18= - 1√

3

(

x− 3√3

8

)


∫ π

2

0


∣

∣

∣

∣

π

2

0

= 32


2π

∫ π

2

0

(

sin3 θ)


∫ π

2

0


∣

∣

∣

∣

π

2

0

= 6π5

(5) 162. A tank contains 1000 gallons of brine which contains .25 pounds of salt pergallon. Brine which contains .5 pounds of salt per gallon flows into the tank at a rate of 10gallons per minute. The solution is kept thoroughly mixed and drains from the tank at arate of 10 gallons per minute. Express the amount of salt in the tank as a function of time.


dQ

dt= 5− Q

100

dQ

dt= 500−Q

100

dQ

500−Q= dt

100

∫

dQ

500−Q=

∫

dt100

- ln |500−Q| = t100

+ C

Since Q(0) = 250, C = - ln 250.

- ln |500−Q| = t100

− ln 250

- ln(500−Q) = t100

− ln 250

ln(500−Q) = - t100

+ ln 250

500−Q = e- t100+ln 250

Q = 500− e- t100+ln 250

Q = 500− 250e- t25

92


Section 6.1: Exam 1

Exam 1 Math 1572 Summer 2007

163. Let f(x) = ex + 5.

(2) a. Show that f is one-to-one.

Proof : Since f ′(x) = ex > 0, f is strictly increasing and hence 1–1.

(2) b. Find (f-1)′(6).

(f-1)(6) = 0

f ′(0) = 1

(f-1)′(6) =

1

f ′(f -1(6))= 1

(2) c. Find f-1(x).

f-1(x) = ln(x− 5)

(2) d. Find (f-1)′(x). Hint: Use this answer to check your answer in part b.

(f-1)′(x) =

1

x− 5

(2) e. Sketch the graph of f and f-1on the same axes.

164. Solve for x.

(3) a. log416 = x

x = 2

(3) b. 3x = 5

ln 3x = ln 5

x ln 3 = ln 5

x = ln 5ln 3

(3) c. log5x = 2

x = 25

(3) d. sec x =√2

x = π4

(3) e. tan(sin-1 4

5) = x

93

Let θ = sin-1 4

5. Then sin θ = 4

5, cos θ = 3

5, and tan θ = 4

3.

94

165. Differentiate.

(3) a. h(t) = esin t

h′(t) = esin t · cos t

(3) b. f(x) =√ln x

f(x) = (ln x)12

f ′(x) = 12(ln x)

- 12 · 1x= 1

2x√lnx

(3) c. g(x) = log x

g′(x) = 1(ln 10)(ln x)

(Page 450: 45) (3) d. y = xsinx

(Page 431: 15) (3) 166. Find the domain and range of f(x) =1

1 + ex.

(3) 167. Find the half-life of a radioactive substance that decays from 50 mg to 40 mg intwo years.

Q = 50(

12

)

t

λ

50(

12

)

2λ

= 40

(

12

)

2λ

= 45

ln(

12

)

2λ

= ln 45

2λln 1

2= ln 4

5

λ =2 ln 1

2

ln 45

168. Integrate.

(4) a.

∫ π

0

x sin xdx (Page 516: 27) (4) b.

∫

cosx ln(sin x)dx

Section 6.2: Exam 2

Exam 2 Math 1572 Summer 2007

169. Integrate.

(5) a.

∫

3x2 − x+ 2

(x+ 1)(x2 + 1)dx

=

∫(

3

x+ 1− 1

x2 + 1

)

dx

= 3 ln |x+ 1| − tan-1 x+ C

(5) b.

∫

cos θ sec2(sin θ)dθ

= tan(sin θ) + C

(5) c.

∫ 1

0

1√x2 + 1

dx

1

x

θ

√ x2 + 1

95

x = tan θ

dx = sec2 θ dθ

∫ 1

0

1√x2 + 1

dx

=

∫ π

4

0

cos θ sec2 θ dθ

=

∫ π

4

0

sec θ dθ

= ln | sec θ + tan θ|∣

∣

∣

∣

π

4

0

= ln(√2 + 1)

(5) d.

∫

x√x+ 1

dx

Let u =√x+ 1.

Then x = u2 − 1 and dx = 2u du.

∫

x√x+ 1

dx

=

∫

2(u2 − 1)du

= 23u3 − 2u+ C

= 23

√

(x+ 1)3 − 2√x+ 1 + C

96

170. Let C be the graph of y = 14x2 − ln

√x from x = 1 to x = 4. Also, let R be the region


y = 14x2 − ln

√x

y′ = 12x− 1

2x

(y′)2 = 14x2 − 1

2+ 1

4x2

(y′)2 + 1 = 14x2 + 1

2+ 1

4x2

(y′)2 + 1 =x4 + 2x2 + 1

4x2

(y′)2 + 1 =(x2 + 1)2

4x2

√

(y′)2 + 1 =x2 + 1

2x

√

(y′)2 + 1 = 12x2 + 1

2x

(5) a. Find the length of C.

∫ 4

1

(

12x+ 1

2x

)

dx

=(

14x2 + 1

2lnx)

∣

∣

∣

∣

4

1

= 4 + 12ln 4− (1

4+ 0)

= 154+ ln 2

(5) b. Find the surface area of S.

∫ 4

1

2π(x2 − ln√x)(1

2x+ 1

2x)dx

=

∫ 4

1

π(x2 − ln√x)(x+ 1

x)dx

= π

∫ 4

1

(

x3 + x− x ln√x− ln

√x

x

)

dx

= π

∫ 4

1

(

x3 + x− 12x ln x− lnx

2x

)

dx

= π

∫ 4

1

(

x3 + x− lnx2x

− 12x ln x

)

dx

= π2

∫ 4

1

(

2x3 + 2x− lnxx

− x ln x)

dx

= π2

(

12x4 + x2 − 1

2(ln x)2 − 1

2x2 ln x+ 1

4x2)

∣

∣

∣

∣

4

1

= π2

(

12x4 + 5

4x2 − 1

2(ln x)2 − 1

2x2 ln x

)

∣

∣

∣

∣

4

1

= π(

14x4 + 5

8x2 − 1

4(ln x)2 − 1

4x2 ln x

)

∣

∣

∣

∣

4

1

= π(

64 + 10− 14(ln 4)2 − 4 ln 4− 1

4− 5

8

)

= π(

74− 78− 1

4(ln 4)2 − 4 ln 4

)

97


(5) a.{ n

en

}∞

n=1

limn→∞

n

en

L’H= lim

n→∞

1

en

= 0

(5) b.

{

(

n + 1

n

)2n}∞

n=1

limn→∞

(

n + 1

n

)2n

= limn→∞

[(

1 + 1n

)n]2

= e2


(5) a.

∞∑

n=1

√

n + 1

n

Since limn→∞

√

n + 1

n= 1, the series diverges by the divergence test.

(5) b.

∞∑

n=1

1

n5n

Since1

n5n≤ 1

5nand

∞∑

n=1

1

5nis a convergent geometric series, the series

∞∑

n=1

1

n5nconverges


(5) c.

∞∑

n=1

lnn

n

Sincelnn

n≥ 1

nfor n ≥ 3 and

∞∑

n=1

1

ndiverges, the series

∞∑

n=1

lnn

ndiverges by the comparison

test.

Section 6.3: final

Final Exam Math 1572 Summer 2007

173. Differentiate.

(3) a. h(x) = sin-1

x (3) b. g(x) = (ex)3

(3) c. f(x) = ln(tan-1x)

98

174. Integrate.

(3) a.

∫

x ln xdx (3) b.

∫

x− 8

x2 − x− 2dx (3) c.

∫

x2

√x− 1

dx

99

175. Let C be the plane curve with parametric equations x = cos t and y = 2 sin t where0 ≤ t ≤ 2π.

(3) a. Sketch the graph of C.

(3) b. Find the equation of the line that is tangent to the curve at the point correspondingto t = π

6.

(3) c. Find the area of the region enclosed by C.

(3) d. Find the surface area of the solid formed by rotating C about the x-axis. Hint: Useonly the part of C defined by 0 ≤ t ≤ π.

100

176. (3) a. Sketch the graph of the polar curve r = 2 cos 3θ.

(3) b. Find the area of the region enclosed by the graph of the polar curve r = 2 cos 3θ.

101


(3) a.∞∑

n=0

1

2n(3) b.

∞∑

n=2

(-1)n+1

n lnn

(3) c.

∞∑

n=1

( n

n+ 1

)n2

(3) d.

∞∑

n=1

esinn

n2

102

(3) 178. Find the radius of convergence and the interval of convergence of the power

series∞∑

n=1

(x+ 1)n

n.

(3) 179. Express1

1− xas a power series.

(3) 180. What function is represented by the power series

∞∑

n=1

nxn−1. Hint: Calculate

∫

[ ∞∑

n=1

nxn−1

]

dx.

(3) 181. Find the Maclaurin Series for f(x) = ex.

Points: 791

Chapter 7: Math 1572 Fall 2007

Section 7.1: Quizzes

Quiz 1

Name:

Directions: Show all of your work and justify all of your answers.

Name:

E-mail (optional):

Directions: Show all of your work and justify all of your answers. An answer withoutjustification will receive a zero.

1. Determine whether or not each of the following is 1–1.

(Page 420: 9) (1) a. f(x) = 12(x+ 5) (Page 420: 13) (1) b. h(x) = x4 + 5

(1) 2. Find a formula for the inverse of the function f(x) =√10− 3x.

Quiz 2

Name:


Name:

E-mail (optional):


1. Calculate.

(Page 439: 5) (1) a. log5125

(Page 439: 7) (1) b. log12 3 + log12 48

(Page 439: 31) (1) 2. Solve the following for x.

5x−3 = 10

Quiz 3

Name:


Name:

E-mail (optional):


1. Calculate.

(Page 483: 3) (1) a. tan-1 √

3 (Page 483: 3) (1) b. sin-1

- 1√2

2. Differentiate.

(Page 431: 33) (1) a. f(x) = e1x (Page 431: 39) (1) b. y = ee

x

Quiz 4

Name:


Name:

E-mail (optional):


1. Differentiate.

(Page 449: 3) (1) a. f(θ) = ln(cos θ) (Page 449: 27) (1) b. y = log10 x

(Page 449: 33) (1) 2. Find f ′(e) where f(x) =x

ln x.

(Page 449: 35) (1) 3. Find f ′(e) where f(x) = ln(ln x).

Quiz 5

Name:


Name:

E-mail (optional):


1. Differentiate.

(Page 450: 43) (1) a. y = xx (Page 484: 23) (1) b. y = tan-1 √

x

(Page 484: 49) (1) 2. A ladder 10 ft long leans against a vertical wall. If the bottom ofthe ladder slides away from the base of the wall at a speed of 2 ft/s, how fast is the anglebetween the ladder and the wall changing when the bottom of the ladder is 6 ft from thebase of the wall?

Quiz 6

Name:


Name:

E-mail (optional):


1. Integrate.

(Page 530: 5) (1) a.

∫ 2

√2

1

t3√t2 − 1

dt (Page 530: 11) (1) b.

∫ √1− 4x2 dx

Quiz 7

Name:


Name:

E-mail (optional):


1. Integrate.

(Page 540: 15) (1) a.

∫ 1

0

2x+ 3

(x+ 1)2dx (Page 540: 29) (1) b.

∫

x+ 4

x2 + 2x+ 5dx

Quiz 8

Name:


Name:

E-mail (optional):


1. Integrate.

(Page 546: 15) (1) a.

∫ 12

0

x√1− x2

dx (Page 546: 37) (1) b.

∫

cos2 θ tan2 θ dθ

Quiz 9

Name:


Name:

E-mail (optional):


1. Integrate.

(Page 546: 29) (1) a.

∫ 5

0

3w − 1

w + 2dw (Page 546: 43) (1) b.

∫

ex√1 + ex dx

Quiz 10

Name:


Name:

E-mail (optional):


1. Integrate.

(Page 546: 41) (1) a.

∫

θ tan2 θ dθ (Page 546: 23) (1) b.

∫ 1

0

(1 +√x)8 dx

Quiz 11

Name:


Name:

E-mail (optional):


1. For each of the following, determine whether the sequence converges or diverges.

(1) a. an=

2n

3n+1 (1) b. an=

n2

en(1) c. a

n=

n!

2n

Quiz 12

Name:


Name:

E-mail (optional):


1. Determine whether each of the following series is convergent or divergent.

(Page 756: 21) (1) a.∞∑

n=1

n

n + 5(Page 756: 23) (1) b.

∞∑

n=2

2

n2 − 1

Quiz 13

Name:


Name:

E-mail (optional):



(Page 756: 29) (1) a.∞∑

n=1

n√2 (Page 756: 31) (1) b.

∞∑

n=1

tan-1 n

Quiz 14

Name:


Name:

E-mail (optional):



(Page 770: 9) (1) a.∞∑

n=1

cos2 n

n2 + 1(Page 770: 11) (1) b.

∞∑

n=2

n2 + 1

n3 − 1

Quiz 15

Name:


Name:

E-mail (optional):



(Page 765: 17) (1) a.∞∑

n=1

n

n2 + 1(Page 760: 19) (1) b.

∞∑

n=1

ne-n2

Quiz 16

Name:


Name:

E-mail (optional):



(Page 765: 21) (1) a.∞∑

n=2

1

n lnn(Page 760: 23) (1) b.

∞∑

n=1

1

n3 + n

Quiz 17

Name:


Name:

E-mail (optional):



(Page 770: 17) (1) a.∞∑

n=1

1√n2 + 1

(Page 770: 31) (1) b.∞∑

n=1

sin(

1n

)

Quiz 18

Name:


Name:

E-mail (optional):



(Page 770: 27) (1) a.∞∑

n=1

(

1 + 1n

)2e-n (Page 770: 29) (1) b.

∞∑

n=1

1

n!

Section 7.2: Exam 1


Name:

E-mail (optional):


(3) 2. If f(x) = 2x3 + x+ 4, find (f-1)′(7).

Solve for x.

(3) a. 3x = 7 (3) b. log4 x = -12

(3) c. logx 9 = 2

(3) d. tanx = 1 (3) e. tan(

sin-1 18

)

122

3. Differentiate.

(3) a. f(x) = etan x (3) b. g(x) = tan-1x2 (3) c. h(x) = ln(x3+x)


(3) a. limx→0

1− cosx

sin x(3) b. lim

x→0x

x

123

5. Integrate.

(3) a.

∫ 1

0

xex2dx

(3) b.

∫

x2 ln xdx

(3) c.

∫

sin-1 xdx

124

(3) d.

∫

1

x2 + 4dx

(3) e.

∫

4√sin x cosxdx

(3) f.

∫

sin2 x sec xdx


(3) 6. If f(x) = 2x3 + x+ 4, find (f-1)′(7).

f ′(x) = 6x2 + 4

f-1(7) = 1

(f-1)′(7) = 1

10

7. Solve for x.

(3) a. 3x = 7

ln 3x = ln 7

x ln 3 = ln 7

x = ln 7ln 3

(3) b. log4 x = -12

x = 4- 12

x = 12

(3) c. logx 9 = 2

x2 = 9

x = 3

(3) d. tanx = 1

x = π4

(3) e. tan(

sin-1 18

)

Let

θ = sin-1 18

y = sin θ = 18

x2 + y2 = 1

x =√638

tan θ = y

x= 1√

63

8. Differentiate.

(3) a. f(x) = etan x

f ′(x) = sec2 xetan x

(3) b. g(x) = tan-1x2

125

g′(x) =2x

1 + x4

(3) c. h(x) = ln(x3 + x)

h′(x) =3x2 + 1

x3 + x

126


(3) a. limx→0

1− cosx

sin x

limx→0

1− cosx

sin x

L’H= lim

x→0

sin x

cosx

= 0

(3) b. limx→0

xx

= limx→0

elnxx

limx→0

ln xx

= limx→0

x ln x

= limx→0

ln x1x

L’H= lim

x→0

1x

- 1x2

= limx→0

-x

= 0

= limx→0

elnxx

= e0

= 1

10. Integrate.

(3) a.

∫ 1

0

xex2dx

= 12ex

2

∣

∣

∣

∣

1

0

= 12(e− 1)

(3) b.

∫

x2 ln xdx

Use integration by parts letting u = ln x,du = 1

xdx, v = 1


∫

x2 lnxdx = 13x3 ln x −

∫

13x3 · 1

xdx =

13x3 ln x−

∫

13x2 dx = 1

3x3 ln x− 1

9x3 +C

(3) c.

∫

sin-1 xdx

Use integration by parts letting u =sin-1 x, du = dx√

1−x2 , v = x, and dv = dx.

∫

sin-1 xdx = x sin-1−∫

x√1− x2

dx =

x sin-1+√1− x2 + C

(3) d.

∫

1

x2 + 4dx

= 12tan-1 x

2+ C

(3) e.

∫

4√sin x cos xdx

= 45(sin x)

54 + C

(3) f.

∫

sin2 x sec xdx

=

∫

1− cos2 x

cosxdx

127

=

∫

(

sec x− cosx)

dx= ln | sec x+ tan x| − sin x+ C

Section 7.3: Exam 2


Name:

E-mail (optional):


(30) 11. Integrate. Choose six and only six. Make your choices clear.

a.

∫

cosxesinx dx b.

∫ π

4

0

tan2 θ dθ c.

∫

1

tan-1 x(x2 + 1)dx

128

d.

∫

sin x sec3 xdx e.

∫

x√x+ 1

dx f.

∫

x2 + x+ 3

(x+ 2)(x2 + 1)dx

129

g.

∫ 1

0

1√x2 + 1

dx h.

∫

2x

x4 + 2x2 + 2dx

130

(10) 12. Determine whether each of the following is convergent or divergent.

a.

∫ ∞

1

ln x

xdx b.

∫ ∞

1

| sinx|x2

dx

(5) 13. Find the arc length of the graph of the function y = ln(cosx) from 0 to π4.

(5) 14. Let R be the region bounded by the curves y =√x, y = 0, and x = 1. Also, let

S be the solid formed by revolving R about the x-axis. Find the surface area of S.



a.

∫

cosxesinx dx = esinx + C

b.

∫ π

4

0

tan2 θ dθ

=

∫ π

4

0

(

sec2 θ − 1)

dθ

=(

tan θ − θ)

∣

∣

∣

∣

π

4

0

= 1− π4

c.

∫

1

tan-1 x(x2 + 1)dx

= ln | tan-1 x|+ C

d.

∫

sin x sec3 xdx

=

∫

tan x sec2 xdx

= 12tan2 x+ C

e.

∫

x√x+ 1

dx

Let u =√x+ 1.

Then x = u2 − 1 and dx = 2u du.

∫

x√x+ 1

dx

=

∫

2u(u2 − 1)

udu

=

∫

(

2u2 − 2)

du

= 23u3 − 2u+ C

= 23

√

(x+ 1)3 − 2(x+ 1) + C

f.

∫

x2 + x+ 3

(x+ 2)(x2 + 1)dx

=

∫(

1

x+ 2+

1

x2 + 1

)

dx

= ln |x+ 2|+ tan-1 x+ C

g.

∫ 1

0

1√x2 + 1

dx

131

1

x

θ

√ x2 + 1

x = tan θ

dx = sec2 θ dθ

∫ 1

0

1√x2 + 1

dx

=

∫ π

4

0

cos θ sec2 θ dθ

=

∫ π

4

0

sec θ dθ

= ln | sec θ + tan θ|∣

∣

∣

∣

π

4

0

= ln(√2 + 1)

h.

∫

2x

x4 + 2x2 + 2dx

=

∫

2x

(x2 + 1)2 + 1dx

= tan-1(x2 + 1) + C


a.

∫ ∞

1

ln x

xdx

Note that lnxx

≥ 1xfor all x ≥ e. Since

∫ ∞

1

1xdx is divergent,

∫ ∞

1

lnxxdx is divergent by the

comparison theorem.

b.

∫ ∞

1

| sin x|x2

dx

Note that | sinx|x2 ≤ 1

x2 for all x. Since

∫ ∞

1

1x2 dx is convergent (2 > 1),

∫ ∞

1

| sinx|x2 dx is

convergent by the comparison theorem.


y′ = - tanx

√

(y′)2 + 1

=√tan2 x+ 1

=√sec2 x

= sec x

∫ π

4

0

sec xdx

= ln | sec x+ tan x|∣

∣

∣

∣

π

4

0

= ln(√2 + 1)

132



y =√x

y′ = 12√x

√

(y′)2 + 1

=√

14x

+ 1

=√

4x+14x

∫ 1

0

2π√x√

4x+14x

dx

=

∫ 1

0

2π√x√4x+1√4x

dx

=

∫ 1

0

π√4x+ 1dx

=

∫ 1

0

π(4x+ 1)12 dx

= 16π(4x+ 1)

32

∣

∣

∣

∣

1

0

= π6(5√5− 1)


Name:

E-mail (optional):



a.

∫

etan-1 x

x2 + 1dx b.

∫ π

4

0

(

tan2 θ + θ)

dθ c.

∫

tan-1 x

xdx

133

d.

∫

tanx

cos2 xdx e.

∫

√x+ 1

xdx f.

∫

x2 + x+ 3

(x+ 2)(x2 + 1)dx

134

g.

∫

√3

2

0

1√1− x2

dx h.

∫

2

x2 + 2x+ 2dx

135


a.

∫ ∞

1

ln x

x2dx b.

∫ ∞

1

| sinx|√x3

dx




Section 7.4: Exam 3


Name:

E-mail (optional):



(5) a.

∞∑

n=1

1

n2(5) b.

∞∑

n=0

(-1)n2n

n!

(5) c.

∞∑

n=2

(-1)n+1

n lnn(5) d.

∞∑

n=0

n!

en2

136

(5) e.∞∑

n=1

sin 1n

1n

(5) f.∞∑

n=1

2n

nn

(5) g.

∞∑

n=0

sinn

en

137


(5) a.∞∑

n=1

(-1)n(x− 2)n

n

(5) b.∞∑

n=0

n!xn

en

138

(Page 795: 5) (5) 25. Find a power series representation for the following function anddetermine the interval of convergence.

f(x) =1

1− x3

(5) 26. Find the Maclaurin series for f(x) = cosx.


139


(5) a.∞∑

n=1

1

n2

This series converges since it is a p-series with p = 2.

(5) b.∞∑

n=0

(-1)n2n

n!

Since limn→∞

∣

∣

∣

∣

(-1)n2n+1

(n+ 1)!· (-1)

nn!

2n

∣

∣

∣

∣

= limn→∞

2

n+ 1= 0,

∞∑

n=0

(-1)n2n

n!converges absolutely by the

Ratio Test.

(Page 784: 11) (5) c.∞∑

n=2

(-1)n+1

n lnn

(Page 784: 25) (5) d.∞∑

n=0

n!

en2

(5) e.∞∑

n=1

sin 1n

1n

Since limn→∞

sin 1n

1n

= 1, the series diverges by the Divergence Test.

(5) f.

∞∑

n=1

2n

nn

Since limn→∞

n

√

2n

nn= lim

n→∞

2

n= 0, the series converges by the Root Test.

(5) g.

∞∑

n=0

sinn

en

Since

∣

∣

∣

∣

sinn

en

∣

∣

∣

∣

≤ 1

enand

∞∑

n=0

1

en=

1

1− 1e

=e

e− 1, the series converges by the Comparison

Test.

140


(5) a.∞∑

n=1

(-1)n(x− 2)n

n

Since∞∑

n=1

(-1)n

nconverges and

∞∑

n=1

1



(5) b.∞∑

n=0

n!xn

en

Since limn→∞

∣

∣

∣

∣

(n+ 1)!xn+1

en+1· en

n!xn

∣

∣

∣

∣

= limn→∞

(n+ 1)|xn|e




f(x) =1

1− x3

(5) 30. Find the Maclaurin series for f(x) = cosx.


Name:

E-mail (optional):



(5) a.

∞∑

n=1

1

nn!(5) b.

∞∑

n=0

(-e)n

(n+ 1)!

(5) c.∞∑

n=2

1

lnnn(5) d.

∞∑

n=0

n!

n2e

141

(5) e.∞∑

n=1

n tan 1n

(5) f.∞∑

n=1

n

πn

(5) g.

∞∑

n=1

n

sinn

142


(5) a.∞∑

n=1

(-1)n(x− 2)n

n!

(5) b.∞∑

n=0

n!xn

en!

143


f(x) = ln |1− x|

(5) 34. Find the Maclaurin series for f(x) = tan-1 x.

Section 7.5: Final

Final Exam Math 1572 Fall 2007

Name:

E-mail (optional):


35. Differentiate.

(3) a. h(x) = ln(sin x) (3) b. g(x) = (ex)3

(3) c. f(x) = tan-1x

(3) 36. Find the half-life of a radioactive that decays from 100 g to 30 g in 3 days.

37. Integrate.

(3) a.

∫

xex dx (3) b.

∫

tan3 θ dθ (3) c.

∫

x2

√x− 1

dx

144

38. Let C be the plane curve with parametric equations x = cos t and y = 2 sin t where0 ≤ t ≤ 2π.

(3) a. Find the equation of the line that is tangent to the curve at the point correspondingto t = π

6.

(3) b. Find the area of the region enclosed by C.

39. (3) a. Sketch the graph of the polar curve r = 2 cos 3θ.

(3) b. Find the area of the region enclosed by the graph of the polar curve r = 2 cos 3θ.

145


(3) a.∞∑

n=0

1

en(3) b.

∞∑

n=0

(-1)n n

n2 + 1

(3) c.

∞∑

n=1

1

nn!(3) d.

∞∑

n=1

sin(

1n

)

146

(3) 41. Find the radius of convergence and the interval of convergence of the power series∞∑

n=1

(x+ 1)n

n.

(3) 42. Express1

1− xas a power series.


∞∑

n=1


∫

[ ∞∑

n=1

nxn−1

]

dx.

Section 7.6: Final

Final Exam Math 1572 Spring 2007

Name:

E-mail (optional):


44. Differentiate.

(2) a. f(x) = tan-1 ex (2) b. f(x) = ln(lnx)

45. Integrate.

(3) a.

∫

ln xdx (3) b.

∫

tan95 θ sec4 θ dθ

147

(3) c.

∫ √1− x2 dx (3) d.

∫

x2

√1− x3

dx

(3) e.

∫

x

x+ 7dx (3) f.

∫

tan2 xdx

148


(2) a.∞∑

n=1

cosnπ

n(2) b.

∞∑

n=1

tan-1 n

n

(2) c.

∞∑

n=2

1

n(lnn2)(2) d.

∞∑

n=1

en

nn

149


(2) a.∞∑

n=0

n!xn

en

(2) b.∞∑

n=1

(x− 3)n

n

48. Let f(x) = ex.

(2) a. Give the Maclaurin series for f and find the radius of convergence.

150

49. Let C be the graph of y = ln x from x = 1 to x = e. Also, let R be the region boundedby the curves x = 1, x = e, C, and the x-axis. Finally, let S be the solid formed by revolvingR about the x-axis. For each of the following, give the integral that represents the givenvalue. Do not evaluate the integral.

(2) a. The arc length of C.

(2) b. The area of R.

(2) c. The surface area of S.

(2) d. The volume of S.

151

50. Consider the polar curve r = 5 cos 2θ.


✲✛

✻

❄

(2) b. Find the area enclosed by the curve.

(2) c. Give the integral that represents the length of the curve. Do not evaluate theintegral.

152

(2) 51. Find the solution of the differential equation that satisfies the given conditions.

dy

dx=

-y2 − 2

2xy; x > 0, y(1) = 2


153

(Page 606: 27) (2) 53. Find the centroid of the region bounded by the curves y =√x

and y = x.

(Page 599: example 2) (2) 54. Find the hydrostatic force on one end of a cylindricaldrum with radius 3 ft if the drum is submerged in water 10 ft deep. Recall that the weightdensity of water is 62.5 lb/ft3.

154

Chapter 8: Math 1572 Spring 2016


Quiz 19(01-15-16)

Name:


For each of the following, determine whether or not the function is 1–1. If so, find theinverse.

(1) 1. f(x) = x3 − x

Since f(0) = f(-1) = f(1) = 0, f is not 1–1.

(1) a. f(x) = 2√x− 1

The function f is 1–1.

Proof : Suppose that x1 , x2 ∈ R such that f(x1) = f(x2). Then

f(x1) = f(x2)

2√x1 − 1 = 2

√x2 − 1

√x

1− 1 =

√x

2− 1

x1 − 1 = x2 − 1

x1= x

2

as desired.

y = 2√x− 1

x = 2√y − 1

x2=

√y − 1

y − 1 = x2

4

y = x2

4+ 1

f-1(x) = x2

4+ 1

Quiz 20(01-20-16)

Name:


(1) b. Give the domain and range of the following function.

g(x) =1

1 + ex

D: (-∞,∞)

R: (0, 1)

(1) c. Solve for x.

log2(3x+ 1) = 4

24 = (3x+ 1)

16 = (3x+ 1)

3x = 15

x = 5

d. Answer the following as true or false (write the entire word) and justify your answer.

(1) i. ln 42

= ln 2

True.

ln 42

= 12ln 4

= ln 412

= ln 2

(1) ii. ln 162

= ln 8

False.

ln 162

= 12ln 16

= ln 1612

= ln 4

6= ln 8

Quiz 21(01-25-16)

Name:


e. Calculate.

(1) i. csc-1 2 = π6

(1) ii. sin (tan-1 3)

y

x= 3

y = 3x

x2 + y2 = 1

x2 + 9x2 = 1

10x2 = 1

x2 = 110

x = 1√10

y = 3√10

sin (tan-1 3)

= 3√10

(1) f. Differentiate.

f(x) = e√

x

f(x) = ex12

f ′(x) = ex12 · 1

2x

- 12

f ′(x) =e√

x

2√x

Quiz 22(01-27-16)

Name:


g. Differentiate.

(1) i. f(x) = ln(x5 + 3x)

f ′(x) =5x4 + 3

x5 + 3x

h. g(x) = ln | sin x|

g′(x) =cos x

sin x= cot x

Differentiate.

(1) i. y = xx2

y = xx2

ln y = ln xx2

ln y = x2 ln x

y′

y= 2x ln x+ x2

x

y′

y= 2x ln x+ x

y′ = y(2x lnx+ x)

y′ = xx2

(2x ln x+ x)

y′ = xx2+1

(2 lnx+ 1)

Quiz 23(02-02-16)

Name:


(1) j. Differentiate the following.

f(x) = tan-1(sin x)

f ′(x) =cos x

1 + sin2 x

k. Calculate the following limits.

(1) i. limx→∞

ln x

xL′H= lim

x→∞

1x

1= 0

(1) ii. limx→∞

x100

ex

L′H= lim

x→∞

100x99

ex

L′H= lim

x→∞

9900x98

ex

...

L′H= lim

x→∞

100!

ex

= 0

Quiz 24(02-10-16)

Name:


l. Integrate.

(1) i.

∫ π

4

0

tan100 θ sec2 θ dθ

= 1101

tan101 θ

∣

∣

∣

∣

π

4

0

= 1101

(1) ii.

∫

sin2 x

cos xdx

=

∫

1− cos2 x

cosxdx

=

∫(

1

cos x− cos2 x

cos x

)

dx

=

∫

(

sec x− cosx)

dx

= ln | sec x+ tan x| − sin x+ C

Quiz 25(02-17-16)

Name:


(2) m. Evaluate the following integral.

∫

1

(x2 + 1)32

dx

1

x

θ

√ x2 + 1

x = tan θ

dx = sec2 θdθ

∫

1

(x2 + 1)32

dx

=

∫

cos3 θ · sec2 θ dθ

=

∫

cos θ dθ

= sin θ + C

=x√

x2 + 1+ C

Quiz 26(02-19-16)

Name:


(2) n. Evaluate the following integral.

∫

x2 + 2x+ 6

(x+ 1)(x2 + 4)dx

A

(x+ 1)+

Bx+ C

x2 + 4=

x2 + 2x+ 6

(x+ 1)(x2 + 4)dx

A(x2 + 4) + (Bx+ C)(x+ 1) = x2 + 2x+ 6

Let x = -1.

5A = 5

A = 1

From the x2 term:

A +B = 1

B = 0

From the x term:

B + C = 2

C = 2

∫

x2 + 2x+ 6

(x+ 1)(x2 + 4)dx =

∫(

1

(x+ 1)+

2

x2 + 4

)

dx = ln |x+ 1|+ tan-1 (x

2

)

+ C

Quiz 27(02-26-16)

Name:


(4) o. Integrate.

i.

∫

sin-1 xdx

Use integration by parts letting u = sin-1 x, du = dx√1−x2 , v = x, and dv = dx.

∫

sin-1 xdx = x sin-1−∫

x√1− x2

dx = x sin-1+√1− x2 + C

ii.

∫

3x2 + x+ 1

(x− 2)(x2 + 1)dx =

∫(

3

x− 2+

1

x2 + 1

)

dx = 3 ln |x− 2|+ tan-1x+ C

iii.

∫

1

x√x2 − 1

dx = sec-1x+ C

iv.

∫

x2√x− 1dx

Let u =√x− 1.

x = u2 + 1

dx = 2udu∫

x2√x− 1dx =

∫

2u2 (u2 + 1)2du =

∫

2u6 + 4u4 + 2u2 du = 27u7 + 4

5u5 + 2

3u3 + C

= 27(x− 1)3

√x− 1 + 4

5(x− 1)2

√x− 1 + 2

3(x− 1)

√x− 1 + C

Quiz 28(03-16-16)

Name:


(1) p. Let R be the region bounded by the curves y =√x, x = 1, and the x-axis. Find the

center of mass of R.

∫ 1

0

√xdx = 2

3x

32

∣

∣

∣

∣

1

0

= 23

x = 32

∫ 1

0

x√xdx = 3

2

(

25

√x5)

∣

∣

∣

∣

1

0

= 35

y = 32

∫ 1

0

12xdx = 3

8x2

∣

∣

∣

∣

1

0

= 38

(

35, 38

)

Quiz 29(03-22-16)

Name:


q. Find the solution of the differential equation that satisfies the given conditions.

(1) i. dy

dx= 6x2 + 1; y(1) = 4

dy

dx= 6x2 + 1

dy = (6x2 + 1) dx

∫

dy =

∫

(

6x2 + 1)

dx

y = 2x3 + x+ C

4 = 2 + 1 + C

C = 1

y = 2x3 + x+ 1

(1) ii.dy

dx=

x+ 2

xy2; y(1) = 3

dy

dx=

x+ 2

xy2

y2 dy =x+ 2

xdx

∫

y2 dy =

∫

x+ 2

xdx

∫

y2 dy =

∫

(

1 + 2x

)

dx

13y3 = x+ 2 ln |x|+ C

9 = 1 + 2 ln 1 + C

C = 8

13y3 = x+ 2 ln |x|+ 8

(1) r. A tank contains 1000 gallons of brine which contains .25 pounds of salt per gallon.Brine which contains .5 pounds of salt per gallon flows into the tank at a rate of 10 gallonsper minute. The solution is kept thoroughly mixed and drains from the tank at a rate of10 gallons per minute. Express the amount of salt in the tank as a function of time.


dQ

dt= 5− Q

100

dQ

dt= 500−Q

100

dQ

500−Q= dt

100

∫

dQ

500−Q=

∫

dt100

- ln |500−Q| = t100

+ C

Since Q(0) = 250, C = - ln 250.

- ln |500−Q| = t100

− ln 250

- ln(500−Q) = t100

− ln 250

ln(500−Q) = - t100

+ ln 250

500−Q = e- t100+ln 250

166

Q = 500− e- t100+ln 250

Q = 500− 250e- t25

Quiz 30(03-25-16)

Name:


s. Determine whether each of the following sequences converges or diverges.

(1) i.

{

lnnn

n

}∞

n=1

limn→∞

lnnn

n

= limn→∞

n lnn

n

= limn→∞

lnn

= ∞

(1) ii.

{

lnn

en

}∞

n=1

limn→∞

lnn

en

= limx→∞

ln x

ex

L’H= lim

x→∞

1x

ex

= limx→∞

1

xex

= 0

Quiz 31(03-30-16)

Name:


t. For each of the following, determine whether the given series converges or diverges.

(1) i.

∞∑

n=1

1

n!

Note that for n ≥ 4, 1n!

≤ 1n2 . Since

∞∑

n=1

1n2 converges,

∞∑

n=1

1n!

converges by the Comparison

Test.

Note that for n ≥ 4, 1n!

≤ 12n. Since

∞∑

n=1

12n

converges,∞∑

n=1

1n!

converges by the Comparison

Test.

Since limn→∞

1(n+1)!

1n!

= limn→∞

n!(n+1)!

= limn→∞

1n+1

= 0,∞∑

n=1

1n!

converges by the Ratio Test.

(1) ii.∞∑

n=1

n!

nn

Note that n!nn = n(n−1)(n−2)···2·1

n·n·n·n···n·n = nn· n−1

n· n−2

n· · · 2

n· 1n≤ 2

n2 . Since∞∑

n=1

2n2 converges,

∞∑

n=1

n!nn

converges by the Comparison Test.

Since limn→∞

(n+1)!

(n+1)(n+1)

n!nn

= limn→∞

(n + 1)!nn

(n+ 1)(n+1)n!= lim

n→∞

(n + 1)n!nn

(n+ 1)(n+1)n!= lim

n→∞

nn

(n+ 1)n= lim

n→∞

( n

n + 1

)n

= 1e,

∞∑

n=1

n!nn converges by the Ratio Test.

Quiz 32(04-01-16)

Name:


u. For each of the following, determine whether the given series converges or diverges.

(1) i.

∞∑

n=0

(

n+ 1

n

)n

Since limn→∞

(

n+1n

)n= e, the series diverges by the divergence test.

(1) ii.∞∑

n=0

( n

n + 1

)n

Since limn→∞

(

nn+1

)n= 1

e, the series diverges by the divergence test.

Quiz 33

Name:


(3) 1. For each of the following, determine whether the series converges absolutely,converges conditionally, or diverges.

a.

∞∑

n=0

en

n!

Since limn→∞

en+1

(n+1)!· n!en

= limn→∞

en+1

= 0, the series∞∑

n=0

en

n!converges by the Ratio test.

b.∞∑

n=1

( n

n + 1

)n2

Since limn→∞

n

√

(

nn+1

)n2

= limn→∞

(

nn+1

)n= 1

e, the series converges by the Root Test.

c.∞∑

n=1

(-1)n

nn!

For n ≥ 2, 1nn! ≤ 1

2n. Since

∞∑

n=1

12n

is a convergent geometric series,∞∑

n=1

1nn! converges by the

Comparison Test. Therefore,∞∑

n=1

(-1)n

nn! is absolutely convergent. Also, since limn→∞

n

√

∣

∣

∣

(-1)n

nn!

∣

∣

∣=

limn→∞

n

√

1nn! = lim

n→∞1

n(n−1)! = 0, the series is is absolutely convergent by the Root Test.

d.

∞∑

n=0

cosnπ

n+ 1= 1

1− 1

2+ 1

3− 1

4+ · · ·

The Alternating Harmonic Series converges by the Alternating Series Test but fails to beabsolutely convergent by the Integral Test.

e.∞∑

n=1

nn

en

This series diverges by the Root Test since limn→∞

n

√

nn

en= lim

n→∞ne= ∞.

171

f.∞∑

n=1

1

(lnn)n

Since lnn ≥ 2 for all n ≥ 8 > e2, 1(lnn)n

≤ 12n

for all n ≥ 8. Hence,∞∑

n=1

1(lnn)n

converges by

the Comparison Test since∞∑

n=1

12n

is a convergent geometric series. Also, since limn→∞

n

√

1(lnn)n

= limn→∞

1lnn

= 0, the series converges by the Root Test.

g.∞∑

n=0

( ∞∑

k=0

1

2n+1 · 2k

)

∞∑

n=0

( ∞∑

k=0

1

2n+1 ·2k

)

=∞∑

n=0

(

1

2n+1

∞∑

k=0

1

2k

)

=∞∑

n=0

2

2n+1 =∞∑

n=0

12n

= 2

h.

∞∑

n=1

(

n2 + 1

n3

)n

Since limn→∞

n

√

(

n2+1n3

)n= lim

n→∞n2+1n3 = 0,

∞∑

n=1

(

n2+1n3

)nconverges by the Root Test.

i.∞∑

n=1

ln

(

n+ 1

n

)

Since limn→∞

ln n+1n

1n

= limx→∞

ln x+1x

1x

L′H= lim

x→∞

xx+1

· x−(x+1)x2

- 1x2

= limx→∞

x

x+ 1= 1 and the series

∞∑

n=1

1n

diverges, the series∞∑

n=1

ln(

n+1n

)

diverges by the Limit Comparison Test.

j.∞∑

n=1

n

√

n + 1

n

For all n ∈ N, n

√

n+1n

≥ 1. So limn→∞

n

√

n+1n

6= 0 which implies that the series diverges by the

Divergence Test.

k.∞∑

n=1

n!

nn

For n ≥ 3, n!nn = 1·2·3···n

n·n·n···n ≤ 2n2 . Since

∞∑

n=1

1n2 is a convergent p-series,

∞∑

n=1

n!nn converges by the

Comparison Test.

172

l.∞∑

n=1

esinn

n2

Since esinn

n2 ≤ 1n2 for all n ∈ N and

∞∑

n=1

1n2 is a convergent p-series,

∞∑

n=1

esinn

n2 converges by the

Comparison Test.

Quiz 34(04-15-16)

Name:



(1) a.∞∑

n=0

n!xn

en

Since limn→∞

∣

∣

∣

∣

(n+ 1)!xn+1

en+1· en

n!xn

∣

∣

∣

∣

= limn→∞

(n+ 1)|xn|e



(1) b.

∞∑

n=1

(-1)n(x− 2)n

n

Since

∞∑

n=1

(-1)n

nconverges and

∞∑

n=1

1



Quiz 35(04-20-16)

Name:


(1) 1. Express f(x) = ln(1− x) as a power series.

f ′(x) = -11−x

= -∞∑

n=0

xn

f(x) = -∞∑

n=0

1n+1

xn+1 = -∞∑

n=1

1nxn


∞∑

n=1


∫

[ ∞∑

n=1

nxn−1

]

dx.

Let f(x) =∞∑

n=1

nxn−1.

∫

f(x)dx =

∫[ ∞∑

n=1

nxn−1

]

dx =∞∑

n=1

xn + C = 11−x

+ C

f(x) = 1(1−x)2

Quiz 36(04-29-16)

Name:


1. Let C be the plane curve defined by the parametric equations x = cos2 θ and y = sin3 θwhere -π

2≤ θ ≤ π

2. Also, let R be the region bounded by C and the y-axis.

a. Sketch C.

✲✛

✻

❄

1-1

1

-1

(1) b. Find the area of R.

x′ = -2 cos θ sin θ∣

∣

∣

∣

∣

∫ π

2

-π2

-2 cos θ sin θ sin3 θ dθ

∣

∣

∣

∣

∣

= 4

∫ π

2

0

cos θ sin4 θ dθ

= 45sin5 θ

∣

∣

∣

∣

π

2

0

= 45

(1) c. Find the length of C.

x′ = -2 cos θ sin θ

y′ = 3 sin2 θ cos θ

∫ π

2

-π2

√4 cos2 θ sin2 θ + 9 sin4 θ cos2 θ dθ

= 2

∫ π

2

0

√4 cos2 θ sin2 θ + 9 sin4 θ cos2 θ dθ

= 2

∫ π

2

0

cos θ sin θ√4 + 9 sin2 θ dθ

= 227

√

(

4 + 9 sin2 θ)3

∣

∣

∣

∣

π

2

0

= 227(13

√13− 8)

176

2. Let R be the region inside the cardioid r = 1 + cos θ and outside the circle r = 1.

a. Sketch the cardioid and the circle.

✲✛

✻

❄

-1-2-3 1 2 3

-1

-2

-3

1

2

3

(1) b. Find the area of R.

12

∫ π

2

-π2

[

(1 + cos θ)2 − 1]

dθ

=

∫ π

2

0

[

(1 + cos θ)2 − 1]

dθ

=

∫ π

2

0

(

cos2 θ + 2 cos θ)

dθ

=

∫ π

2

0

[

12(1 + cos 2θ) + 2 cos θ

]

dθ

=

∫ π

2

0

(

12+ 1

2cos 2θ + 2 cos θ

)

dθ

=(

12θ + 1

4sin 2θ + 2 sin θ

)

∣

∣

∣

∣

π

2

0

= π4+ 2

(1) c. Find the length of the boundary of R.

Note 1: Part of the boundary lies on thecardioid and part of the boundary lies onthe circle.

r = 1 + cos θ

r′ = - sin θ

∫ π

2

-π2

√

(1 + cos θ)2 + sin2 θ dθ + π

= 2

∫ π

2

0

√

(1 + cos θ)2 + sin2 θ dθ + π

= 2

∫ π

2

0

√1 + 2 cos θ + cos2 θ + sin2 θ dθ + π

= 2

∫ π

2

0

√2 + 2 cos θ dθ + π

= 2

∫ π

2

0

√

4 cos2(

θ2

)

dθ + π

= 2

∫ π

2

0

2 cos(

θ2

)

dθ + π

= 8 sin(

θ2

)

∣

∣

∣

∣

π

2

0

+π

= 4√2 + π

Section 8.2: Exam 1


177

(10) 3. For the function below, find the inverse. You may assume without proof that thefunction is 1–1.

f(x) =√x+ 1 + 3

x =√y + 1 + 3

x− 3 =√y + 1

(x− 3)2 = y + 1

y = (x− 3)2 − 1

f-1(x) = (x− 3)2 − 1


(10) a. sin-1

√32

= π3

sin π3=

√32

(10) b. tan(

sin-1 3

7

)

Consider the unit circle.

y = 37

x2 + y2 = 1

x2 + 949

= 1

x2 = 4049

x =√407

tan(

sin-1 3

7

)

=37√407

= 3√40

= 32√10

5. Solve for x.

(10) a. ex+7

= 4

ln ex+7

= ln 4

x+ 7 = ln 4

x = ln 4− 7

(10) b. ln(x+ 1) = 3

x+ 1 = e3

x = e3 − 1

6. Differentiate each of the following.

(10) a. f(x) = sin (ex)


(10) b. g(x) = ln(x4 + x)

g′(x) =4x3 + 1

x4 + x

(10) c. h(x) = sec-1(x3)

h′(x) =3x2

x3√x6 − 1

=3

x√x6 − 1

(10) d. y =√x

x

ln y = ln√x

x

ln y = x ln√x

ln y = 12x ln x

ddx

ln y = ddx

(

12x lnx

)

178

y′

y= 1

2

(

ln x+ x · 1x

)

y′ = 12y (lnx+ 1)

y′ = 12

√x

x(ln x+ 1)

(10) 7. Find the half-life of a radioactive substance that decays from 3 lb to 34lb in 6

years.

Q = Q0

(

12

)

t

λ

34= 3

(

12

)

6λ

14=(

12

)

6λ

(

12

)2=(

12

)

6λ

2 = 6λ

2λ = 6

λ = 3

Alternatively:

14=(

12

)

6λ

ln 14= ln

(

12

)

6λ

ln 14= 6

λln 1

2

λ ln 14= 6 ln 1

2

λ =6 ln 1

2

ln 14

= 3


(10) a. limx→1

ln x

x− 1L′H= lim

x→1

1x

1= 1

(10) b. limx→0

xe1x = lim

x→0

e1x

1x

L′H= lim

x→0

- 1x2 e

1x

- 1x2

= limx→0

e1x = ∞

9. Integrate.

(10) a.

∫ 1

0

xex dx


∫ 1

0

xex dx = xex∣

∣

∣

∣

1

0

−∫ 1

0

ex dx = xex∣

∣

∣

∣

1

0

− ex∣

∣

∣

∣

1

0

= e− 0− (e− 1) = 1

(10) b.

∫

tan-1xdx

Use integration by parts letting u = tan-1 x, du = dxx2+1

, v = x, and dv = dx.

179

∫

tan-1 xdx = x tan-1 x −∫

x

x2 + 1dx = x tan-1 x− 1

2ln |x2 + 1|+ C

Section 8.3: Exam 2


10. Integrate.

(10) a.

∫

tan θ dθ = ln | sec θ|+ C

(10) b.

∫

sec5 θ tan3 θ dθ

=

∫

sec4 θ tan2 θ sec θ tan θ dθ

=

∫

sec4 θ (sec2 θ − 1) sec θ tan θ dθ

=

∫

(sec6 θ − sec4 θ) sec θ tan θ dθ

= 17sec7 θ − 1

5sec5 θ + C

(10) c.

∫ √4− x2 dx

√4− x2

x2

θ

x = 2 sin θ

dx = 2 cos θ dθ

∫ √4− x2 dx

=

∫

(2 cos θ)(2 cos θ)dθ

=

∫

4 cos2 θ dθ

= 2

∫

(

1 + cos 2θ)

dθ

= 2θ + sin 2θ

= 2θ + 2 sin θ cos θ

= 2 sin-1 (x

2

)

+ 2 · x2·√4−x2

2+ C

= 2 sin-1 (x

2

)

+ x√4−x2

2+ C

(10) d.

∫

2x2 − 5x+ 1

(x+ 1)(x− 1)2dx

Ax+1

+ Bx−1

+ C(x−1)2

= 2x2−5x+1(x+1)(x−1)2

A(x−1)2+B(x+1)(x−1)+C(x+1) = 2x2−5x+1

Let x = 1.

2C = -2

C = -1

Let x = -1.

4A = 8

A = 2

Ax2 +Bx2 = 2x2

A+B = 2

B = 0

∫

2x2 − 5x+ 1

(x+ 1)(x− 1)2dx

=

∫

(

2x+1

− 1(x−1)2

)

dx

= 2 ln |x+ 1|+ 1x−1

+ C

180

(10) e.

∫

x√x+ 1

dx

i.

Let u =√x+ 1.

x = u2 − 1

dx = 2udu

∫

x√x+ 1

dx

=

∫

u2 − 1

u· 2udu

=

∫

(

2u2 − 2)

du

= 23u3 − 2u+ C

= 23

√

(x+ 1)3 − 2√x+ 1 + C

ii.

Let u = x+ 1.

x = u− 1

dx = du

∫

x√x+ 1

dx

=

∫

u− 1√u

du

=

∫

(

u12 − u

- 12

)

du

= 23u

32 − 2u

12 + C

= 23

√

(x+ 1)3 − 2√x+ 1 + C

iii.

1

√x

√x+ 1

θ

√x = tan θ

x = tan2 θ

dx = 2 tan θ sec2 θdθ∫

x√x+ 1

dx

=

∫

tan2 θ√tan2 θ + 1

· 2 tan θ sec2 θ dθ

=

∫

tan2 θ√sec2 θ

· 2 tan θ sec2 θ dθ

=

∫

2 tan2 θ tan θ sec θ dθ

= 2

∫

(sec2 θ − 1) tan θ sec θ dθ

= 2

∫

(

sec2 θ tan θ sec θ − tan θ sec θ)

dθ

= 2(

13sec3 θ − sec θ

)

+ C

= 2(

13

√x+ 1

3 −√x+ 1

)

+ C

= 23

√

(x+ 1)3 − 2√x+ 1 + C

iv.

Use integration by parts.

u = x

du = dx

v = 2√x+ 1

181

dv = 1√x+1

dx

∫

x√x+ 1

dx

= 2x√x+ 1−

∫

2√x+ 1dx

= 2x√x+ 1− 4

3(x+ 1)

32+ C

= 23

√

(x+ 1)3 − 2√x+ 1 + C

v.

∫

x√x+ 1

dx=

∫

x+ 1− 1√x+ 1

dx=

∫(

x+ 1√x+ 1

− 1√x+ 1

)

dx=

∫(√

x+ 1− 1√x+ 1

)

dx

=

∫

(

(x+ 1)12 − (x+ 1)

- 12

)

dx = 23(x+ 1)

32 − 2 (x+ 1)

12dx = 2

3

√

(x+ 1)3 − 2√x+ 1 + C

182

11. Determine whether each of the following is convergent or divergent.

(10) a.

∫ ∞

1

x2

ex3 dx

limt→∞

∫ t

1

x2

ex3 dx = limt→∞

-13e-x

3

∣

∣

∣

∣

t

1

= limt→∞

(

-13e-t

3 − - 13e

)

= limt→∞

(

13e

− 1

3et3

)

= 13e

(10) b.

∫ ∞

1

1

x4 + ex + 1dx

Note that1

x4 + ex + 1≤ 1

x4. Since

∫ ∞

1

1

x4dx converges,

∫ ∞

1

1

x4 + ex + 1dx converges by

the Comparison Theorem for Improper Integrals.


y = 23x

32 , 0 ≤ x ≤ 3

y′ = x12

√

(y′)2 + 1 =√x+ 1

∫ 3

0

√x+ 1dx = 2

3(x+ 1)

32

∣

∣

∣

∣

3

0

= 143

(10) 13. For the following, find the surface area of the solid obtained by rotating thegiven curve about the x-axis.

y = 13x3, 0 ≤ x ≤ 1

y′ = x2

2πy√

(y′)2 + 1 = 23πx3

√x4 + 1

∫ 1

0

23πx3

√x4 + 1dx = 1

9π(x4 + 1)

32

∣

∣

∣

∣

1

0

= 19π(

2√2− 1

)

183

14. A barrel in the shape of a right circular cylinder has radius 1 ft and height 4 ft. Thebarrel is full of water. Recall that the density of water is 62.5 lb/ft3.

(10) a. If the barrel is sitting upright on its bottom, what is the is the hydrostatic forceacting on the bottom?

62.5× 4× π lb = 250π lb

(10) b. Suppose the barrel is laying on its side. Find the hydrostatic force acting on theend (formerly the bottom).

Partition the unit circle into n horizontal strips of height 2n.

The depth is represented by 1− x for -1 ≤ x ≤ 1.

The the length of a horizontal strip is represented by 2√1− x2.

The hydrostatic force on the end is given by the following.

limn→∞

n∑

i=1

62.5 ·(

1− 2in

)

· 2n· 2√

1−(

2in

)2

= 125

∫ 1

-1

(1− x)√1− x2 dx

= 125

∫ 1

-1

(√1− x2 − x

√1− x2

)

dx

= 125

[∫ 1

-1

√1− x2 dx−

∫ 1

-1

x√1− x2 dx

]

= 125π2

(See the note below.)

Note 2: The integral

∫ 1

-1

√1− x2 dx can be evaluated using trigonometric substitution and

∫ 1

-1

x√1− x2 dx can be evaluated using substitution. Alternatively, note that

∫ 1

-1

√1− x2 dx

is the area of the upper half of the unit circle which is π2and that f(x) = x

√1− x2dx is an

odd function which means that

∫ 1

-1

x√1− x2 dx = 0.

Section 8.4: Exam 3


184

(10) 15. Find the centroid of the region bounded by the curves y =√x, y = 0, and x =

4.

A =

∫ 4

0

√xdx = 2

3

√x3

∣

∣

∣

∣

4

0

= 163

x = 316

∫ 4

0

x√xdx = 3

16

∫ 4

0

√x3 dx = 3

1625·√x5

∣

∣

∣

∣

4

0

= 316

· 645= 12

5

y = 316

∫ 4

0

12(√x)

2dx = 3

16

∫ 4

0

12xdx = 3

16· 14x2

∣

∣

∣

∣

4

0

= 34

(

125, 34

)


y2 dy

dx= 1

x; y(1) = 3

y2dy = dxx

∫

y2 dy =

∫

1xdx

13y3 = ln |x|+ C

y(1) = 3

C = 9

13y3 = ln |x|+ 9

y = 3√

3 ln |x|+ 27



dQ

dt= -

Q

50

dQ

Q= -

dt

50

∫

dQ

Q=

∫

-dt

50

ln |Q| = - t50

+ C

Since Q(0) = 10, C = ln 10.

ln |Q| = t50

+ ln 10

Q = et50+ln 10

Q = 10e- t50

18. Determine whether each of the following sequences converges or diverges. If it converges,find the limit.

(10) a.{

ln(

n+1n

)}∞

n=1

185

Since the function f(x) = ln x is continuous, limn→∞

ln(

n+1n

)

= ln(

limn→∞

(

n+1n

)

)

= ln 1 = 0.

186

(10) b. {1, 0, 1, 0, 0, 1, 0, 0, 0, 1, . . .}

Since this sequence has a subsequence that converges to 0 and a subsequence that converges1, the sequence does not converge.

(10) 19. Give the sum of the following geometric series.

∞∑

n=0

53n = 5

1− 13

= 152

(10) 20. Is the following series absolutely convergent, conditionally convergent, or diver-gent?

∞∑

n=1

(-1)n1

lnn

Since limn→∞

1lnn

= 0, the series converges by the Alternating Series Test. Also, note that for

n ≥ 3, 1n≤ 1

lnn. Since

∞∑

n=1

1nis a divergent p-series,

∞∑

n=1

1lnn

diverges by the Comparison Test.

So the series is conditionally convergent.


(10) a.

∞∑

n=1

4n

n!

Since limn→∞

(

4n+1

(n+1)!· n!4n

)

= limn→∞

4n+1

= 0,∞∑

n=1

4n

n!converges by the Ratio Test.

(10) b.∞∑

n=1

lnn

n

Since 1n

≤ lnnn

for all n ≥ 3 and∞∑

n=1

1n

is a divergent p-series,∞∑

n=1

lnnn

diverges by the

Comparison Test. The Integral Test can also be used.

(10) c.

∞∑

n=1

en

2n=

∞∑

n=1

(e

2

)n

Since this series is a geometric series with r = e2> 1, the series diverges. The Ratio Test,

Root Test, Divergence Test, and the Integral Test can also be used.

187

(10) d.∞∑

n=1

lnn

en

Since limn→∞

(

ln(n+1)en+1 · en

lnn

)

= limn→∞

(

ln(n+1)e lnn

)

= 1elimn→∞

(

ln(n+1)lnn

)

= 1elimx→∞

(

ln(x+1)lnx

) L’H= 1

elimn→∞

(

xx+1

)

= 1e

< 1,∞∑

n=1

lnnen

converges by the Ratio Test. Also, since lnnen

≤ nen

and∞∑

n=1

nen

converges by the Ratio

Test and the Root Test,∞∑

n=1

lnnen


(10) e.∞∑

n=1

n√2

Note that for all n ∈ N, 1 ≤ n√2. So lim

n→∞n√2 6= 0 (the limit is in fact 1). Hence, the series

diverges by the Divergence Test.

(10) f.

∞∑

n=1

n

n2 + 2n

Note that nn2+2n

≤ n2n

for all n ∈ N. Since limn→∞

(

n+1

2n+1 · 2n

n

)

= 12,

∞∑

n=1

n2n

converges by the

Ratio Test. So∞∑

n=1

nn2+2n


(10) g.∞∑

n=1

(

4

n

)n

Since limn→∞

n

√

(

4n

)n= lim

n→∞4n= 0,

∞∑

n=1

(

4n

)nconverges by the Root Test.

Section 8.5: Final Exam


Name:



(10) a. g(x) = esinx (10) b. g(x) = tan-1(x2)

188

(10) c. y = xx2(10) 23. Calculate the following limit.

limx→0

x

sin-1x

189

24. Integrate.

(10) a.

∫

x ln xdx (10) b.

∫

x7

x8 + 1dx

(10) c.

∫

tan2 θ dθ (10) d.

∫

esinx

sec xdx

190

(10) e.

∫

1

x2 − 1dx (10) f.

∫ 1

0

√1− x2 dx

(10) 25. Find the centroid of the region bounded by the curves y =√x, y = 0, and x =

1.

191


y2 dy

dx= 1

x; y(1) = 1

(10) 27. A tank contains 100 gallons of brine which contains 0.10 pounds of salt pergallon. Pure water flows into the tank at a rate of 4 gallons per minute. The solution iskept thoroughly mixed and drains from the tank at a rate of 4 gallons per minute. Expressthe amount of salt in the tank as a function of time.

192

(10) 28. Find the interval of convergence and radius of convergence of the following powerseries.

∞∑

n=1

(-1)n(x− 4)n

n

(10) 29. Let f(x) = ln x. Find the Taylor series of f centered at 1. What is the radiusof convergence?

193

30. (10) a. Given that ex =∞∑

n=0

xn

n!, prove that d

dxex = ex.

(10) b. Find the sum of the series∞∑

n=0

(√2)

n

n!.

(10) 31. Let C be the plane curve with parametric equations x = cos t and y = sin t.Sketch C and find the equation of the line that is tangent to C at the point where t = π

6.

✲✛

✻

❄

1-1

1

-1

194

(10) 32. Let C be the curve with polar equation r = 3 cos 2θ, 0 ≤ θ ≤ 2π. Sketch C andfind the area of the region it encloses.

✲✛

✻

❄

-1-2-3 1 2 3

-1

-2

-3

1

2

3

(10) 33. Let C be the curve with polar equation r = 2 sin θ, 0 ≤ θ ≤ π. Sketch C and findits length.

✲✛

✻

❄

-1-2-3 1 2 3

-1

-2

-3

1

2

3

195

Chapter 9: Spring 2018


Quiz 3701-12-18

Name:


1. Let f(x) = 4 3√x+ 1.


Proof : Suppose that x1 , x2 ∈ R such that f (x1) = f (x2). Then

f (x1) = f (x

2)

4 3√x1 + 1 = 4 3

√x2 + 1

4 3√x1 = 4 3

√x2

3√x1 = 3

√x2

(

3√x

1

)3=(

3√x

2

)3

x1 = x2 .

Therefore, f is 1–1.

(1) b. Find f-1.

y = 4 3√x+ 1

x = 4 3√y + 1

4 3√y = x− 1

3√y = x−1

4

y =(

x−14

)3

(1) 2. Suppose that g is an invertible differentiable function such that g′(x) = x4 +x2 +1

and g(2) = 10. Find [g-1]′(10).

[g-1]′(10) = 1

g′[g-1 (10)]= 1

g′(2)= 1

21

Quiz 3801-17-18

Name:


1. Calculate and simplify each of the following.

(1) a. tan(

sin-1 1

5

)

Let θ = sin-1 1

5.

✲✛

✻

❄

15

θ

(

x, 15

)

x

y = 15

x2 + y2 = 1

x2 + 125

= 1

x = 2√6

5

tan θ = y

x= 1

2√6

(1) b. sin(

tan-1 1

5

)

Let θ = tan-1 1

5.

✲✛

✻

❄

y

θ

(x, y)

x

tan θ = y

x

15= y

x

x = 5y

x2 + y2 = 1

25y2 + y2 = 1

y = 1√26

sin θ = y = 1√26

(1) 2. Differentiate.

f(x) = tan-1 (√

x+ 1)

f ′(x) = 1(√x+1)2+1

· 12(x+1)

- 12 = 12(x+2)

√x+1

198

(1) 3. Integrate.∫

x√1− x4

dx = 12sin

-1

x2 + C

Quiz 3901-22-18

Name:


1. Differentiate.

(1) a. f(x) = ln√x+ 1

f(x) = 12ln(x+ 1)

f ′(x) = 12(x+1)

(1) b. g(x) = sin ex2

g′(x) = 2xex2cos ex

2

2. Integrate.

(1) a.

∫

x2

x3 + 1dx = 1

3ln |x3 + 1|+ C

(1) b.

∫

e2x + 1

exdx =

∫

ex + e-x dx = ex − e-x + C

Quiz 4001-24-18

Name:


Differentiate.

(1) 1. y = x√

x

ln y = ln x√

x

ln y =√x ln x

y′

y= 1

2x

- 12 ln x+√x · 1

x

y′

y= lnx

2√x+ 1√

x

y′ = y(

lnx+22√x

)

y′ = x√

x

(

lnx+22√x

)

(1) 2. A bacteria culture starts with 100 bacteria and grows at a rate proportional to itssize. After 5 hours the culture contains 400 bacteria. Express the number of bacteria in theculture as a function of time.

P (t) = 100ert

400 = 100e5r

4 = e5r

5r = ln 4

r = 15ln 4

P (t) = 100et ln 45 = 100 · 4

t5

(1) 3. Calculate the following limit.

limx→∞

ex

ln x

L′H= lim

x→∞

ex

1x

= limx→∞

xex

= ∞

Quiz 4102-09-18

Name:



(1) a.

{

n2

en

}∞

n=1

limn→∞

n2

ex

L′H= lim

n→∞

2n

ex

L′H= lim

n→∞

2

ex

= 0

(1) b.

{

n!

nn

}∞

n=1

limn→∞

n!

nn

= limn→∞

n(n− 1)(n− 2) · · ·2 · 1n · n · n · · ·n · n

limn→∞

(

n

n· (n− 1)

n· (n− 2)

n· · · 2

n· 1n

)

≤ limn→∞

1

n

= 0

2. Integrate.

(1) a.

∫

tan3 x sec2 xdx = 14tan4 x+ C

(1) b.

∫

tan2 xdx =

∫

(

sec2 x− 1)

dx = tanx− x+ C

Quiz 4202-14-18

Name:


1. Integrate.

(1) a.

∫

x2 + x+ 2

(x+ 1)(x2 + 1)dx

Solution 1:

Use partial fraction decomposition.

x2 + x+ 2

(x+ 1)(x2 + 1)=

A

x+ 1+

Bx+ C

x2 + 1

x2+x+2 = A(x2+1)+ (Bx+C)(x+1)

x2+x+2 = Ax2+Bx2+Bx+Cx+A+C

Comparing coefficients yields:

A + B = 1B + C = 1A + C = 2

Let x = -1.

Then we have

2A = 2

A = 1

which implies that

B = 0

and

C = 1.

∫

x2 + x+ 2

(x+ 1)(x2 + 1)dx

=

∫(

1

x+ 1+

1

x2 + 1

)

dx

= ln |x+ 1|+ tan-1x+ C

Solution 2:

∫

x2 + x+ 2

(x+ 1)(x2 + 1)dx

=

∫(

x2 + 1

(x+ 1)(x2 + 1)+

x+ 1

(x+ 1)(x2 + 1)

)

dx

=

∫(

1

x+ 1+

1

x2 + 1

)

dx

= ln |x+ 1|+ tan-1x+ C

(1) b.

∫

x3 + 3x2 + 1

x3 + 1dx

=

∫(

x3 + 1

x3 + 1+

3x2

x3 + 1

)

dx

=

∫(

1 +3x2

x3 + 1

)

dx

= x+ ln |x3 + 1|+ C

Quiz 43

Name:


(1) 1. Integrate.

∫ 3

0

x√x+ 1dx

Use substitution.

u =√x+ 1

x+ 1 = u2

x = u2 − 1

dx = 2udu

∫ 3

0

x√x+ 1dx

=

∫ 2

1

2u2(u2 − 1)du

=

∫ 2

1

(

2u4 − 2u2)

du

=(

25u5 − 2

3u3)

∣

∣

∣

∣

2

1

= 645− 16

3−(

25− 2

3

)

= 11615

(1) 2. Determine whether the following improper integral is convergent or divergent.

∫ ∞

1

xex2dx

Divergent.

∫ ∞

1

xex2dx = lim

t→∞

∫ t

1

xex2dx = lim

t→∞12ex

2

∣

∣

∣

∣

t

1

= limt→∞

12

(

et2 − e

)

= ∞

∫ ∞

1

xe-x2dx

Convergent.

∫ ∞

1

xe-x2dx = lim

t→∞

∫ t

1

xe-x2dx = lim

t→∞-12e-x

2

∣

∣

∣

∣

t

1

= limt→∞

-12

(

e-t2 − e-1

)

= limt→∞

-12

(

1

et2 − 1

e

)

= 12e

Quiz 44

Name:


1. For each of the following, determine whether the series is absolutely convergent, condi-tionally convergent, or divergent.

(1) a.

∞∑

n=1

(-1)nlnn

n

Since limn→∞

lnnn

L′H= lim

n→∞1n= 0,

∞∑

n=1

(-1)nlnn

nconverges the Alternating Series Test.

Since

∫ ∞

1

ln x

xdx = lim

t→∞

∫ t

1

ln x

xdx = lim

t→∞12(ln x)2

∣

∣

∣

∣

t

1

= limt→∞

12(ln t)2 = ∞,

∞∑

n=1

lnn

ndiverges

by the Integral Test.

Therefore,

∞∑

n=1

(-1)nlnn

nis conditionally convergent.

(1) b.∞∑

n=1

en

n!

Since limn→∞

en+1

(n+1)!

en

n!

= limn→∞

en+1

(n+1)!· n!en

= limn→∞

en+1

= 0,

∞∑

n=1

en

n!converges absolutely by the Ratio

Test.

(1) c.

∞∑

n=1

( n

n+ 1

)n

Since limn→∞

n

√

( n

n + 1

)n

= limn→∞

n

n+ 1= 1, the Root Test is inconclusive. However, since

limn→∞

( n

n+ 1

)n

= 1e,

∞∑

n=1

( n

n + 1

)n

diverges by the Divergence Test.

d.∞∑

n=1

( n

n + 1

)n2

205

Since limn→∞

n

√

( n

n + 1

)n2

= limn→∞

( n

n+ 1

)n

= 1e,

∞∑

n=1

( n

n+ 1

)n2

converges absolutely by the

Root Test.

206

2. Recall that the weight density of water is 62.5 lb/ft3. An aquarium is in the shape ofa rectangular prism. Its length is 2 ft, its width is 1 ft, and its height is 1 ft. Find thehydrostatic force on each of the following.

a. the bottom

F = δdA = 62.5× 1× 2 = 125

125 lb

b. an end

Partition the end of the tank into n horizontal strips of height 1n.

The depth is represented by x for 0 ≤ x ≤ 1.

The the length of a horizontal strip is 1.


limn→∞

n∑

i=1

62.5 · in· 1n· 1 = 62.5

∫ 1

0

xdx = 62.5(

12x2)

∣

∣

∣

∣

1

0

= 31.25

31.25 lb

c. a side

Partition the side of the tank into n horizontal strips of height 1n.

The depth is represented by x for 0 ≤ x ≤ 1.

The the length of a horizontal strip is 2.


limn→∞

n∑

i=1

62.5 · in· 1n· 2 = 125

∫ 1

0

xdx = 125(

12x2)

∣

∣

∣

∣

1

0

= 62.5

62.5 lb

Also, note that since the area of a side is twice the area of an end, the hydrostatic force ona side is twice the hydrostatic force on an end.

Quiz 45

Name:


(1) 1. Find the radius of convergence and the interval of convergence of the following.

∞∑

n=1

(x− 2)n

n

Since∞∑

n=1

(-1)n

nconverges and

∞∑

n=1

1

ndiverges, the interval of convergence must be [1, 3).

Therefore, the radius of convergence is 1. Alternatively, the ratio test can be used.


dy

dx= 3y, y(0) = 1, y > 0

dy

dx= 3y

1ydy = 3dx

∫

1ydy =

∫

3dx

ln y = 3x+ C

y(0) = 1

ln 1 = 0 + C

C = 0

ln y = 3x

y = e3x

Quiz 46

Name:


(2) 1. Give the interval of convergence and the radius of convergence of

∞∑

n=0

x2n

(2n)!.

Since limn→∞

∣

∣

∣

∣

x2(n+1)

[2(n+ 1)]!· (2n)!x2n

∣

∣

∣

∣

= limn→∞

∣

∣

∣

∣

x2

(2n+ 2)(2n+ 1)

∣

∣

∣

∣

= 0, the interval of convergence is

R and the radius of convergence is ∞.

(1) 2. Let f(x) =∞∑

n=0

x2n

(2n)!. Find f ′(x).

f ′(x) =

∞∑

n=1

x2n−1

(2n− 1)!

3. For each of the following, determine whether the series is absolutely convergent, condi-tionally convergent, or divergent.

a.∞∑

n=0

(-e)n

4n

This is a geometric series with r = -e4. So it converges to

1

1 + e4

= 44+e

. Also, note that

∞∑

n=0

en

4n=

1

1− e4

= 44−e

. So the series is absolutely convergent.

b.∞∑

n=1

( n

n + 1

)n

Since limn→∞

( n

n + 1

)n

= 1e6= 0, the series diverges by the Divergence Test.

c.∞∑

n=1

sin(

nπ2

)

√n

209

Since∞∑

n=1

sin(

nπ2

)

√n

=∞∑

n=0

(-1)n√2n+ 1

and limn→∞

1√2n+ 1

= 0, the series converges by the

Alternating Series Test. Since

∫ ∞

1

1√2x+ 1

dx = limt→∞

∫ t

1

1√2x+ 1

dx = limt→∞

√2x+ 1

∣

∣

∣

∣

t

1

=

limt→∞

(√2t+ 1−

√3)

= ∞, the series is not absolutely convergent. Therefore, the series is

conditionally convergent.

Quiz 47

Name:


(1) 1. Find the radius of convergence and the interval of convergence of the following.

∞∑

n=1

(x+ 3)n

n

Since∞∑

n=1

(-1)n

nconverges and

∞∑

n=1

1

ndiverges, the interval of convergence must be [-4, -2).


2. Sketch the plane curve parameterized by the following.

x = cos t, y = sin2 t, 0 ≤ t ≤ π

x2 + y = 1

y = 1− x2

✲✛

✻

❄

14

12

34

1- 14

- 12

- 34

-1

14

12

34

1

- 14

- 12

- 34

-1

Quiz 48

Name:


(2) 1. Let C be the plane curve defined by the parametric equations x = ln t + t andy = t2 + t. Find the equation of the line that is tangent to C at the point (1, 2).

dxdt

= 1t+ 1

dy

dt= 2t+ 1

dy

dx=

2t+ 11t+ 1

The point (1, 2) corresponds to t = 1.

When t = 1, dy

dx= 3

2.

y − 2 = 32(x− 1)

Quiz 49

Name:


1. Let C be the plane curve defined by the parametric equations x = 23t3 and y = t2 +1 for

0 ≤ t ≤ 2√2.

x = 23t3

dx

dt= 2t2

y = t2 + 1

dy

dt= 2t

dy

dx=

dy

dtdxdt

=2t

2t2=

1

t

(

dx

dt

)2

= 4t4

(

dy

dt

)2

= 4t2

(

dx

dt

)2

+

(

dy

dt

)2

= 4t4 + 4t2

√

(

dx

dt

)2

+

(

dy

dt

)2

=√4t4 + 4t2

√

(

dx

dt

)2

+

(

dy

dt

)2

= 2t√t2 + 1

(1) a. Find the equation of the line that is tangent to C at the point where t =√3.

t =√3

x = 2√3

y = 4

dy

dx= 1√

3

y − 4 = 1√3(x− 2

√3)


∫ 2√2

0

2t√t2 + 1dt = 2

3

√

(t2 + 1)3∣

∣

∣

∣

2√2

0

= 18− 23= 52

3


2π

∫ 2√2

0

(t2 + 1) · 2t√t2 + 1dt

= 2π

∫ 2√2

0

2t√

(t2 + 1)3 dt

= 2π(

25

√

(t2 + 1)5)

∣

∣

∣

∣

2√2

0

= 45π(√

(t2 + 1)5)

∣

∣

∣

∣

2√2

0

213

= 45π(√

(t2 + 1)5)

∣

∣

∣

∣

2√2

0

= 45π(243− 1)

= 968π5

2. Consider the polar curve r = cos θ2+ 1 for 0 ≤ θ ≤ 4π.

a. Sketch the curve.

✲✛

✻

❄

b. Find the area of the region inside the large loop and outside the small loop.

2

[

12

∫ π

0

(

cos θ2+ 1)2

dθ − 12

∫ 2π

π

(

cos θ2+ 1)2

dθ

]

=

∫ π

0

(

cos θ2+ 1)2

dθ −∫ 2π

π

(

cos θ2+ 1)2

dθ

∫

(

cos θ2+ 1)2

dθ

=

∫

(

cos2 θ2+ 2 cos θ

2+ 1)

dθ

=

∫

[

12(1 + cos θ) + 2 cos θ

2+ 1]

dθ

=

∫

[

12cos θ + 2 cos θ

2+ 3

2

]

dθ

= 12sin θ + 4 sin θ

2+ 3

2θ + C

∫ π

0

(

cos θ2+ 1)2

dθ −∫ 2π

π

(

cos θ2+ 1)2

dθ

214

=(

12sin θ + 4 sin θ

2+ 3

2θ)

∣

∣

∣

∣

π

0

−(

12sin θ + 4 sin θ

2+ 3

2θ)

∣

∣

∣

∣

2π

π

=[(

0 + 4 + 3π2

)

− (0 + 0 + 0)]

−[

(0 + 0 + 3π)−(

0 + 4 + 3π2

)]

= 4 + 3π2− (3π

2− 4)

= 8

Quiz 50

Name:


1. Consider the following polar curve for 0 ≤ θ ≤ 2π.

r = cos θ + 1 r = sin θ + 1


r = cos θ + 1

✲✛

✻

❄

r = sin θ + 1

✲✛

✻

❄

(1) b. Find the equation of the line that is tangent to the curve at the point where θ = π4.

r = cos θ + 1

x = (cos θ + 1) cos θ = cos2 θ + cos θ

y = (cos θ + 1) sin θ = cos θ sin θ + sin θ

dy

dx= - sin2 θ+cos2 θ+cos θ

-2 cos θ sin θ−sin θ

θ = π4

x =√2+12

y =√2+12

dy

dx=

√22

-√22− 1

= 1−√2

y −√2+12

= (1−√2)(

x−√2+12

)

r = sin θ + 1

x = (sin θ + 1) cos θ = cos θ sin θ + cos θ

y = (sin θ + 1) sin θ = sin2 θ + sin θ

dy

dx= 2 sin θ cos θ+cos θ

- sin2 θ+cos2 θ−sin θ

θ = π4

216

x =√2+12

y =√2+12

dy

dx=

1 +√22

-√22

= -1−√2

y −√2+12

= (-1−√2)(

x−√2+12

)

(1) c. Give the integral that represents the area of the region enclosed by the curve. Donot evaluate the integral.

r = cos θ + 1

12

∫ 2π

0

(cos θ + 1)2 dθ

r = sin θ + 1

12

∫ 2π

0

(sin θ + 1)2 dθ

(1) d. Give the integral that represents the length of the curve. Do not evaluate theintegral.

r = cos θ + 1

drdθ

= - sin θ

∫ 2π

0

√

(cos θ + 1)2 + (- sin θ)2 dθ

r = sin θ + 1

drdθ

= cos θ

∫ 2π

0

√

(sin θ + 1)2 + (cos θ)2 dθ

Quiz 51

Name:


1. Give the equation of the indicated conic section and sketch its graph.

(1) a. Parabola

Focus: (-4, 3)

Directrix: x = 0

Vertex: (-2, 3)

p = -2

(y − 3)2 = -8(x+ 2)

(1) b. Ellipse

Foci: (-2, 1) and (4, 1)

Vertices: (-4, 1) and (6, 1)

c = 3

a = 5

b = 4

Center: (1, 1)

(x−1)2

25+ (y−1)2

16= 1

(1) c. Hyperbola

Foci: (-2, 0) and (2, 0)

Vertices: (-1, 0) and (1, 0)

a = 1

c = 2

b =√3

x2 − y2

3= 1

218

2. Consider the polar curve r = cos θ2+ 1 for 0 ≤ θ ≤ 4π.

a. Sketch the curve.

✲✛

✻

❄

b. Find the area of the region inside the large loop and outside the small loop.

2

[

12

∫ π

0

(

cos θ2+ 1)2

dθ − 12

∫ 2π

π

(

cos θ2+ 1)2

dθ

]

=

∫ π

0

(

cos θ2+ 1)2

dθ −∫ 2π

π

(

cos θ2+ 1)2

dθ

∫

(

cos θ2+ 1)2

dθ

=

∫

(

cos2 θ2+ 2 cos θ

2+ 1)

dθ

=

∫

[

12(1 + cos θ) + 2 cos θ

2+ 1]

dθ

=

∫

[

12cos θ + 2 cos θ

2+ 3

2

]

dθ

= 12sin θ + 4 sin θ

2+ 3

2θ + C

∫ π

0

(

cos θ2+ 1)2

dθ −∫ 2π

π

(

cos θ2+ 1)2

dθ

=(

12sin θ + 4 sin θ

2+ 3

2θ)

∣

∣

∣

∣

π

0

−(

12sin θ + 4 sin θ

2+ 3

2θ)

∣

∣

∣

∣

2π

π

=[(

0 + 4 + 3π2

)

− (0 + 0 + 0)]

−[

(0 + 0 + 3π)−(

0 + 4 + 3π2

)]

219

= 4 + 3π2− (3π

2− 4)

= 8

Total Points: 42

Section 9.2: Exam 1


(10) 3. Let f(x) = 3√x5 + 5. Find f

-1.

y = 3√x5 + 5

x = 3√

y5 + 5

x3 = y5 + 5

y5 = x3 − 5

y = 5√x3 − 5

f-1(x) = 5

√x3 − 5.

(10) 4. Suppose that f is a differentiable and invertible function such that f(2) = 7 and

f ′(2) = 3. Find(

f-1)′(7).

(

f-1)′

(7) = 1

f ′[f -1 (7)]= 1

f ′(2)= 1

3


(10) a. sin(

sin-1 1

3

)

= 13

(10) b. sin-1 (

sin 5π2

)

= sin-1 (1

2

)

= π2

(10) c. tan(

cos-1 3

5

)

Let θ = cos-1 3

5.

3

45

θ

tan θ = 43

(10) 6. How long will it take 10 g of a substance with a half-life of 30 days to decay to 4g?

Q = 10(

12

)

t30

10(

12

)

t30

= 4

(

12

)

t30

= 25

ln

[

(

12

)

t30

]

= ln 25

t30ln 1

2= ln 2

5

t =30 ln 2

5

ln 12

220

7. Differentiate.

(10) a. f(x) = esinx

f ′(x) = esinx · cosx

(10) b. f(x) = sin-1

ex

f ′(x) = ex√1−e2x


(10) a. limx→∞

ln x

xL′H= lim

x→∞

1x

1= 0

(10) b. limx→0

x ln x

= limx→0

lnx1x

L′H= lim

x→0

1x-1x2

= limx→∞

-x

= 0

9. Give a formula or rule for each of the following sequences.

(10) a. {-1, 4, -9, 16, -25, 36, . . .}

={

(-1)n

n2}∞

n=1

(10) b. {1, 3, 4, 7, 11, 18, 29, . . .}

x1 = 1

x2 = 3

For n ≥ 2,

xn+1 = x

n+ x

n−1 .

10. Integrate.

(10) a.

∫

x3

x4 + 1dx = 1

4ln |x4 + 1|+ C (10) b.

∫

tan-1x

x2 + 1dx = 1

2

(

tan-1x)2

+ C

(10) c.

∫

tan-1xdx

Use integration by parts letting u = tan-1x, du = 1

x2+1dx, v = x, and dv = dx.

∫

tan-1xdx = x tan

-1x−

∫

xx2+1

dx = x tan-1x− 1

2ln |x2 + 1|+ C

221

(10) d.

∫ 2

0

√4− x2 dx

√4− x2

x2

θ

x = 2 sin θ

dx = 2 cos θdθ

∫ 2

0

√4− x2 dx

=

∫ π

2

0

2 cos θ · 2 cos θ dθ

=

∫ π

2

0

4 cos2 θ dθ

=

∫ π

2

0

2(1 + 2 cos 2θ)dθ

= 2(

θ + sin 2θ)

∣

∣

∣

∣

π

2

0

= π

222

Section 9.3: Exam 2

1. (10) a. Find A and B such thatA

x+

B

x+ 2=

2

x(x+ 2).

A

x+

B

x+ 2=

2

x(x+ 2)

A(x+ 2) +Bx = 2

Ax+Bx+ 2A = 2

2A = 2

A = 1

B = -1

(10) b. Use your answer from the first part to evaluate the following integral.

∫

2

x(x+ 2)dx =

∫(

1

x− 1

(x+ 2)

)

dx = ln |x| − ln |x+ 2|+ C

(10) c. Use your answer from the first part to find the sum of the following series.

∞∑

n=1

2

n(n+ 2)=

∞∑

n=1

(

1

n− 1

(n + 2)

)

sn=

n∑

k=1

2

k(k + 2)=

n∑

k=1

(

1

k− 1

(k + 2)

)

= 11−1

3+1

2−1

4+1

3−1

5+1

4−1

6+· · ·+ 1

n−1− 1

n+1+ 1

n− 1

n+2

= 1 + 12− 1

n+1− 1

n+2

limn→∞

sn= lim

n→∞

(

1 + 12− 1

n+1− 1

n+2

)

= 32

2. Integrate.

(10) a.

∫

sec3 x tan3 xdx=

∫

sec2 x tan2 x sec x tanxdx=

∫

sec2 x (sec2 x− 1) sec x tan xdx

=

∫

(sec4 x− sec2 x) sec x tan xdx = 15sec5 x− 1

3sec3 x+ C

(10) b.

∫ 3

0

x√x+ 1

dx

Let u =√x+ 1. Then x = u2 − 1 and dx = 2udu.

∫ 3

0

x√x+ 1

dx =

∫ 2

1

u2 − 1

u· 2udu = 2

∫ 2

1

(

u2 − 1)

du = 2(

13u3 − u

)

∣

∣

∣

∣

2

1

= 2[(

83− 2)

−(

13− 1)]

= 83

(10) c.

∫ ∞

2

1

x ln xdx = lim

t→∞

∫ t

2

1

x ln xdx = lim

t→∞ln (ln x)

∣

∣

∣

∣

t

2

= limt→∞

[ln (ln t)− ln (ln 2)] = ∞

223

3. All parts of this problem refer to the following sequence.

{

(-1)n 1n

}∞n=1

(10) a. Is the sequence bounded?

Yes. All terms of the sequence are in the interval[

-1, 12

]

.

(10) b. Is the sequence monotone?

No. Since the terms are alternating from negative to positive, the sequence is neitherincreasing nor decreasing.

(10) c. Does the sequence converge?

Yes. limn→∞

(-1)n 1n= 0


y = 23x

32 ; 0 ≤ x ≤ 8

y′ = x12

√

(y′)2 + 1 =√x+ 1

∫ 8

0

√x+ 1dx = 2

3

√

(x+ 1)3∣

∣

∣

∣

8

0

= 18− 23= 52

3

(10) 5. Let R be the region bounded by the x-axis, x = 1, and y = 2√x.

a. Find the surface area of the solid obtained by rotating R about the x-axis.

2πy√

(y′)2 + 1 = 4π√x√

1x+ 1 = 4π

√x√

x+1x

= 4π√x+ 1

∫ 1

0

4π√x+ 1dx = 8

3π√

(x+ 1)3∣

∣

∣

∣

1

0

= 83π(

2√2− 1

)

224

b. Find the center of mass (centroid) of R. y′ = 1√x

A =

∫ 1

0

2√xdx = 4

3x

32

∣

∣

∣

∣

1

0

= 43

x = 34

∫ 1

0

2x√xdx = 3

2

∫ 1

0

x32 dx = 3

5x

52

∣

∣

∣

∣

1

0

= 35

y = 34

∫ 1

0

12(2√x)

2dx = 3

2

∫ 1

0

xdx = 34x2

∣

∣

∣

∣

1

0

= 34

Centroid:(

35, 34

)


dy

dx=

x

y√x+ 1

; y(0) = 4

dy

dx=

x

y√x+ 1

y dy =x√x+ 1

dx

∫

y dy =

∫

x√x+ 1

dx

First, evaluate the integral

∫

x√x+ 1

dx.

Let u =√x+ 1. Then x = u2 − 1 and

dx = 2udu.

∫

x√x+ 1

dx

=

∫

u2 − 1

u· 2udu

= 2

∫

(

u2 − 1)

du

= 2(

13u3 − u

)

+ C

= 2(

13

√

(x+ 1)3 −√x+ 1

)

+ C

= 23

√

(x+ 1)3 − 2√x+ 1 + C

∫

y dy =

∫

x√x+ 1

dx

12y2 = 2

3

√

(x+ 1)3 − 2√x+ 1 + C

Use the initial condiction y(0) = 4 tosolve for C.

8 = 23− 2 + C

C = 283

12y2 = 2

3

√

(x+ 1)3 − 2√x+ 1 + 28

3

(10) 7. A tank in the shape of a rectangular prism is filled with water. The base of thetank is 4 ft by 6 ft and the height is 2 ft. Recall that the weight density of water is 62.5lb/ft3. Find the hydrostatic force on one end of tank.

limn→∞

n∑

i=1

62.5 ·(

2in

)

· 2n· 4 = 250

∫ 2

0

xdx = 125x2

∣

∣

∣

∣

2

0

= 500

500 lb

225


(10) a.

∞∑

n=0

(

1

π

)n

Convergent.

This is a geometric series with r = 1π. So the series converges to

1

1− 1π

= ππ−1

.

(10) b.

∞∑

n=1

9

√

1

n10

Convergent. Note that this is a p-series with p = 109

(10) c.

∞∑

n=2

1

n lnn

Divergent.

Use the integral test.

∫ ∞

2

1

x ln xdx = lim

t→∞

∫ t

2

1

x ln xdx = lim

t→∞ln (ln x)

∣

∣

∣

∣

t

2

= limt→∞

[ln (ln t)− ln (ln 2)] = ∞

226

Section 9.4: Exam 3



(10) a. i.

∞∑

n=0

(e

2

)n

This is a geometric series with r = e2> 1. So the series diverges.

ii.∞∑

n=0

(

2

e

)n

This is a geometric series with r = e2. So the series converges to

1

1− e2

= 2e−2

.

(10) b. i.∞∑

n=1

n− 1

n + 1

Since limn→∞

n− 1

n+ 1= 1 6= 0, the series diverges by the Divergence Test.

ii.∞∑

n=2

n+ 1

n− 1

Since limn→∞

n+ 1

n− 1= 1 6= 0, the series diverges by the Divergence Test.

(10) c. i.∞∑

n=1

1

nπ

This is a p-series with p = π > 1. So the series converges.

ii.

∞∑

n=1

1

ne

This is a p-series with p = e > 1. So the series converges.

227

(10) d. i.∞∑

n=1

( n

n+ 1

)n2

Since limn→∞

n

√

( n

n+ 1

)n2

= limn→∞

( n

n+ 1

)n

= 1e< 1, the series converges by the Root Test.

ii.

∞∑

n=1

( n

n + 1

)n

Since limn→∞

( n

n + 1

)n

= 1e6= 0, the series diverges by the Divergence Test.

(10) e. i.

∞∑

n=1

πn

(n + 1)!

Since limn→∞

[

πn+1

(n+ 2)!· (n+ 1)!

πn

]

= limn→∞

π

n+ 2= 0, the series converges by the Ratio Test.

ii.

∞∑

n=1

πn+1

n!

Since limn→∞

[

πn+2

(n+ 1)!· n!

πn+1

]

= limn→∞

π

n + 1= 0, the series converges by the Ratio Test.

(10) 2. Determine whether the given series is absolutely convergent, conditionally con-vergent, or divergent.

iii.

∞∑

n=2

(-1)n

n2 lnn

Since limn→∞

1

n2 lnn= 0, the series converges by the Alternating Series Test. Also, note that

1

n2 lnn≤ 1

n2for all n. Since

∞∑

n=1

1

n2p-series with p = 2, it converges. Hence,

∞∑

n=1

1

n2 lnn

converges by the Comparison Test. Therefore,∞∑

n=2

(-1)n

n2 lnnis absolutely convergent.

iv.∞∑

n=2

(-1)n

n lnn

Since limn→∞

1

n lnn= 0, the series converges by the Alternating Series Test.

228

Since

∫ ∞

2

1

x lnxdx = lim

t→∞

∫ t

2

1

x ln xdx = lim

t→∞ln (ln x)

∣

∣

∣

∣

t

2

= limt→∞

[ln (ln t)− ln (ln 2)] = ∞,

∞∑

n=2

1

n lnndiverges by the Integral Test. Therefore,

∞∑

n=2

(-1)n

n lnnis conditionally convergent.

(10) 3. Let f(x) = x3 + x2 − x+ 1 [x3 − x2 + x+ 1]. Find the Taylor series of f centeredat 1. What is the radius of convergence?

a. f(x) = x3 + x2 − x+ 1

f ′(x) = 3x2 + 2x− 1

f ′′(x) = 6x+ 2

f ′′′(x) = 6

f (4)(x) = 0

∞∑

n=0

f(n)(1)n!

(x− 1)n

= 2 + 4(x− 1) + 82!(x− 1)2 + 6

3!(x− 1)3

= 2 + 4(x− 1) + 4(x− 1)2 + (x− 1)3

RoC: R

IoC: ∞

b. f(x) = x3 − x2 + x+ 1

f ′(x) = 3x2 − 2x+ 1

f ′′(x) = 6x− 2

f ′′′(x) = 6

f (4)(x) = 0

∞∑

n=0

f(n)(1)n!

(x− 1)n

= 2 + 2(x− 1) + 42!(x− 1)2 + 6

3!(x− 1)3

= 2 + 2(x− 1) + 2(x− 1)2 + (x− 1)3

RoC: R

IoC: ∞

4. Let g(x) = ex.

(10) a. Give the Maclaurin series for g.

g(n)(x) = ex

∞∑

n=0

g(n)(0)xn

n!=

∞∑

n=0

xn

n!

(10) b. Give the 4th-degree Taylor polynomial of g at 0.

T4(x) =4∑

n=0

xn

n!= 1 + x+ x2

2+ x3

3+ x4

24

229

(10) c. Use your answer from the previous part to approximate 1e[√e]. Do not simplify

your answer.

i. 1e

T4(-1) = 1− x+ 12− 1

3+ 1

24

ii.√e

T4

(

12

)

= 1 + 12+ 1

8+ 1

24+ 1

384

(10) 5. Find the interval of convergence and radius of convergence of the following powerseries.

a.

∞∑

n=1

(x− 5)n

n!

Since limn→∞

[

(x− 5)n+1

(n+ 1)!· n!

(x− 5)n

]

= limn→∞

x− 5

n + 1= 0, the interval of convergence is R and

the radius of convergence is ∞.

b.

∞∑

n=1

(x− 5)n

n

Since

∞∑

n=1

(-1)n

nconverges and

∞∑

n=1

1

ndiverges, the interval of convergence must be [4, 6).


(10) 6. Evaluate the following indefinite integral as an infinite series.

a.

∫

1

1− xdx =

∫

( ∞∑

n=0

xn

)

dx =∞∑

n=0

xn+1

n+1+ C

b.

∫

1

1− x2dx =

∫

[ ∞∑

n=0

(x2)n

]

dx =

∫

( ∞∑

n=0

x2n

)

dx =

∞∑

n=0

x2n+1

2n+1+ C

Section 9.5: Final Exam


230

Name:

Directions: Show all of your work and justify all of your answers. An answer withoutjustification will receive a zero. The use of calculators, phones, electronic devices, or outsidesources will result in a score of 0 on the exam. Sign the attendance sheet.

(10) 7. Find the inverse of the following function.

f(x) = (x+ 7)3

(10) 8. Suppose that f is an invertible and differentiable function such that f(2) = 3 and f ′(2) = 15.

Find(

f-1)′(3).

231


(10) a. y = tan (ex) (10) b. y =√ln x


limx→∞

ln x

x+ tan-1x


∫ ∞

1

1

ln x+ xdx

232

12. Integrate.

(10) a.

∫

x

x2 + 8x+ 15dx

(10) b.

∫

x+ 4√x+ 5

dx

(10) c.

∫

xex dx

233

13. let R be the region bounded by the curves y = 6√x, y = 0, and x = 1.


(10) b. Find the centroid of R.

(10) c. Find the surface area of the solid formed by revolving R about the x-axis.

234


(10) a. the bottom (10) b. a side

(10) 15. Find the limit of the sequence defined as follows. Hint: List several terms of thesequence.

x1= 9

For n > 1, xn+1 =

√x

n.

235

16. Consider the polar curve r = 2 sin θ.


✲✛

✻

❄

(10) b. Find the length of the curve.

236

17. Determine whether each of the following series converges or diverges.

(10) a.

∞∑

n=1

3

√

1

n2

(10) b.∞∑

n=1

cos n+ 1

en

(10) 18. For the following power series, find the interval of convergence and radius ofconvergence.

∞∑

n=1

(x− 2)n

n

237

(10) 19. Give the equation of the indicated conic section and sketch its graph.

Ellipse

Center: (4, 3)

Vertex: (4, 8)

Focus: (4, 6)✲✛

✻

❄

(10) 20. Identify the following conic. Find all vertices, foci, asymptotes, and directrices.

y2 − 12x− 6y = 39

(10) 21. Find the inverse of the following function.

f(x) = 3√x+ 7

(10) 22. Suppose that f is an invertible and differentiable function such that f(5) = 2 and f ′(5) = 3.

Find(

f-1)′(2).


(10) a. y = sin-1

(ex) (10) b. y = xsinx

238

(10) 24. A bacterial culture grows exponentially from 100 to 5000 in 2 hours. Find theexponential growth rate.


limx→∞

ln x

ln x+ x


∫ ∞

1

1

ex2 dx

239

27. Integrate.

(10) a.

∫

1

x2 − 3x+ 2dx

(10) b.

∫

x+ 1√x+ 3

dx

(10) c.

∫

ln xdx

240

28. let R be the region bounded by the curves y = 14√x, y = 0, and x = 1.



(10) c. Find the surface area of the solid formed by revolving R about the x-axis.

241


(10) a. the bottom (10) b. an end

(10) 30. Find the limit of the sequence defined as follows. Hint: List several terms of thesequence.

x1 =√5

For n > 1, xn+1 =

√5x

n.

242

31. Consider the polar curve r = cos 2θ.


✲✛

✻

❄

(10) b. Find the area of the region enclosed by the curve.

243


(10) a.

∞∑

n=1

1

n2

(10) b.

∞∑

n=1

n!

2n

(10) c.∞∑

n=1

(

1

n+ 1

)n

244

(10) 33. Let f(x) = ln x. Find the Taylor series of f centered at 1. What is the radiusof convergence?

245

Chapter 10: Spring 2019

Section 10.1: Exam 1

34. Define f : (-∞, -1) ∪ (-1,∞) → R by f(x) =1

x3 + 1.


Proof : Suppose that u, v ∈ (-∞, -1) ∪(-1,∞) such that f(u) = f(v). Then

f(u) = f(v)

1

u3 + 1=

1

v3 + 1

u3 + 1 = v3 + 1

u3 = v3

u = v.

(10) b. Find f-1.

y =1

x3 + 1

x3 + 1 = 1y

x3 = 1y− 1

x = 3

√

1y− 1

f-1(y) = 3

√

1y− 1

35. Define f : (-∞, -1) ∪ (-1,∞) → R by f(x) =1

3√x+ 1

.


Suppose that u, v ∈ (-∞, -1) ∪ (-1,∞)such that f(u) = f(v). Then

f(u) = f(v)

13√u+ 1

=1

3√v + 1

3√u+ 1 = 3

√v + 1

u+ 1 = v + 1

u = v.

(10) b. Find f-1.

y = 13√x+1

3√x+ 1 = 1

y

x+ 1 = 1y3

x = 1y3

− 1

f-1(y) = 1

y3− 1

246

36. Define f : R → R by f(x) = x5 + x + 1. Given that f is invertible, find each of thefollowing.

(10) a. f-1(-1) = -1 (10) b. f

-1(1) = 0

(10) c. [f-1]′(-1)

f ′(x) = 5x4 + 1

[f-1]′(-1) = 1

f ′[f -1 (-1)]= 1

f ′(-1)= 1

6

(10) d. [f-1]′(1)

f ′(x) = 5x4 + 1

[f-1]′(1) = 1

f ′[f -1 (1)]= 1

f ′(0)= 1


(10) a. tan-1√3 = π

3

(10) b. sec-1√2 = π

4

(10) c. sec(

tan-1 43

)

Let θ = tan-1 43.

Then tan θ = 43.

3

45

θ

sec θ = 53

sec(

tan-1 43

)

= 53

38. Solve each of the following.

(10) a. 3x

= 25

ln 3x

= ln 25

x ln 3 = ln 25

x = ln 25ln 3

(Page 409: 35) (10) b. e2x − ex − 6 = 0

(ex − 3)(ex + 2) = 0

ex − 3 = 0

ex = 3

x = ln 3

39. Differentiate.

(10) a. f(x) = ex3+x

f ′(x) = ex3+x(3x2 + 1)

(10) b. g(x) = esin-1 x

g′(x) =esin

-1 x

√1− x2

(10) c. g(x) = etan-1 x

g′(x) =etan

-1 x

1 + x2

(10) d. h(x) = tan-1 (ln x)

h′(x) =1x

1 + (ln x)2=

1

x [1 + (ln x)2]

(10) e. h(x) = sin-1 (ln x)

h′(x) =1x

√

1− (ln x)2=

1

x√

1− (ln x)2

247

(10) f. y = x(x2)

ln y = ln x(x2)

ln y = x2 ln x

ddx

(

ln y)

= ddx

(

x2 ln x)

y′

y= 2x ln x+ x

y′ = y(2x lnx+ x)

y′ = x(x2)

(2x lnx+ x)

y′ = x(x2+1)

(2 lnx+ 1)


(10) a. limx→∞

ln x

x

L’H

= limx→∞

1x

1= 0

(10) b. limx→0

sin-1 x

ex − 1

L’H

= limx→0

1√1−x2

ex= 1

(10) c. limx→0

tan-1 x

ex − 1

L’H

= limx→0

11+x2

ex= 1

41. Integrate.

(10) a.

∫

1√4− x2

dx = sin-1 x2+ C

(10) b.

∫

ex

1 + e2xdx = tan-1 ex + C

(10) c.

∫

1 + e2x

exdx

=

∫

(

e-x + ex)

dx

= -e-x + ex + C

= ex − 1ex

+ C

(10) d.

∫

ln x

xdx = 1

2(ln x)2 + C

248



42. A bacterial culture grows exponentially from 200 to 300 bacteria in 4 hours.

(10) a. Find the growth function.

Q(t) = Q0ekt

Q(t) = 200ekt

Q(4) = 300

200e4k = 300

e4k = 32

4k = ln 32

k = 14ln 3

2

Q(t) = 200e( 14 ln 3

2)t= 200

(

32

)t

4

(10) b. When will the number of bacte-ria reach 500?

Q(t) = 500

200(

32

)t

4 = 500

(

32

)t

4 = 52

ln[

(

32

)t

4

]

= ln 52

t4ln(

32

)

= ln 52

t =4 ln 5

2

ln 32

43. Integrate.

(10) a.

i.

∫

x cosxdx

Integrate by parts.

u = xdu = dxv = sin xdv = cosx dx∫

x cosxdx

= x sin x−∫

sin xdx

= x sin x+ cosx+ C

ii.

∫

x sin xdx

Integrate by parts.

u = xdu = dxv = - cos xdv = sin x dx∫

x sin xdx

= -x cosx−∫

- cos xdx

= -x cosx+ sin x+ C

249

(10) b.

i.

∫

tan2 xdx

=

∫

(

sec2 x− 1)

dx

= tanx− x+ C

ii.

∫

sec3 x tan xdx

=

∫

sec2 x sec x tan xdx

= 13sec3 x+ C

(10) c.

i.

∫

cos(tan-1 x)

1 + x2dx = sin(tan-1 x) + C ii.

∫

sin(tan-1 x)

1 + x2dx = - cos(tan-1 x) + C

(10) d.

i.

∫

3x+ 3

x2 + x− 2dx

3x+ 3

x2 + x− 2=

3x+ 3

(x+ 2)(x− 1)

A

x+ 2+

B

x− 1=

3x+ 3

(x+ 2)(x− 1)

A(x− 1) +B(x+ 2) = 3x+ 3

Let x = 1.

B = 2

Let x = -2.

A = 1

∫

3x+ 3

x2 + x− 2dx

=

∫(

1

x+ 2+

2

x− 1

)

dx

= ln |x+ 2|+ 2 ln |x− 1|+ C

ii.

∫

3x

x2 + x− 2dx

3x

x2 + x− 2=

3x

(x+ 2)(x− 1)

A

x+ 2+

B

x− 1=

3x

(x+ 2)(x− 1)

A(x− 1) +B(x+ 2) = 3x

Let x = 1.

B = 1

Let x = -2.

A = 2

∫

3x

x2 + x− 2dx

=

∫(

2

x+ 2+

1

x− 1

)

dx

= 2 ln |x+ 2|+ ln |x− 1|+ C

250

(10) e.

i.

∫ 4

0

√16− x2 dx

Use trigonometric substitution.

√16− x2

x4

θ

x = 4 sin θ

dx = 4 cos θ dθ

θ = sin-1 x4

∫ 4

0

√16− x2 dx

=

∫ π

2

0


=

∫ π

2

0

16 cos2 θ dθ

=

∫ π

2

0

8(1− cos 2θ)dθ

=(

8θ − 4 sin 2θ)

∣

∣

∣

∣

π

2

0

= 4π

Alternatively, note that the integral rep-resents the area of one-fourth of the circlecentered at (0, 0) with radius 4.

ii.

∫ 3

0

√9− x2 dx

Use trigonometric substitution.

√9− x2

x3

θ

x = 3 sin θ

dx = 3 cos θ dθ

θ = sin-1 x3

∫ 4

0

√9− x2 dx

=

∫ π

2

0


=

∫ π

2

0

9 cos2 θ dθ

=

∫ π

2

0

92(1− cos 2θ)dθ

=(

92θ − 9

4sin 2θ

)

∣

∣

∣

∣

π

2

0

= 9π4

Alternatively, note that the integral rep-resents the area of one-fourth of the circlecentered at (0, 0) with radius 3.

251

(10) f.

i.

∫

tan-1 xdx

Integrate by parts.

u = tan-1 xdu = 1

x2+1dx

v = xdv = dx

∫

tan-1 xdx

= x tan-1 x−∫

x

x2 + 1dx

= x tan-1 x− 12ln(x2 + 1) + C

ii.

∫

sin-1 xdx

Integrate by parts.

u = sin-1 xdu = 1√

1−x2 dxv = xdv = dx

∫

sin-1 xdx

= x sin-1 x−∫

x√1− x2

dx

= x sin-1 x+√1− x2 + C

(10) g.

i.

∫

x+ 4√x+ 3

dx

Use substitution.

u = x+ 3

du = dx∫

x+ 4√x+ 3

dx

=

∫

u+ 1√u

du

=

∫

(√u+ 1√

u

)

du

= 23u

32 + 2u

12 + C

= 23

√

(x+ 3)3 + 2√x+ 3 + C

Alternatively:

∫

x+ 4√x+ 3

dx

=

∫(

x+ 3√x+ 3

+1√x+ 3

)

dx

=

∫(√

x+ 3 +1√x+ 3

)

dx

= 23

√

(x+ 3)3 + 2√x+ 3 + C

252

ii.

∫

x+ 3√x+ 2

dx

Use substitution.

u = x+ 2

du = dx∫

x+ 3√x+ 2

dx

=

∫

(√u+ 1√

u

)

du

= 23u

32 + 2u

12 + C

= 23

√

(x+ 2)3 + 2√x+ 2 + C

Alternatively:

∫

x+ 3√x+ 2

dx

=

∫(

x+ 2√x+ 2

+1√x+ 2

)

dx

=

∫(√

x+ 2 +1√x+ 2

)

dx

= 23

√

(x+ 2)3 + 2√x+ 2 + C

(10) h.

i.

∫ π

4

0

esinx

sec xdx

∫ π

4

0

esinx

sec xdx

=

∫ π

4

0

esinx cosxdx

= esinx

∣

∣

∣

∣

π

4

0

= e

√2

2 − 1

ii.

∫ π

4

0

etan x

cos2 xdx

=

∫ π

4

0

etan x sec2 xdx

= etan x

∣

∣

∣

∣

π

4

0

= e− 1

253

44. Determine whether each of the following is convergent or divergent.

(10) a.

∫ ∞

3

lnx

xdx

∫ ∞

3

ln x

xdx = lim

t→∞

∫ t

3

ln x

xdx = lim

t→∞12(ln x)2

∣

∣

∣

∣

t

3

= 12limt→∞

[

(ln t)2 − (ln 3)2]

= ∞

Alternatively, note that for x ≥ 3, lnxx

> 1x. Since

∫ ∞

3

1

xdx diverges,

∫ ∞

3

ln x

xdx diverges

by the Comparison Theorem for Improper Integrals.

(10) b.

i.

∫ ∞

1

1

x3 +√x+ 1

dx

Note that for x ≥ 1, 1x3+

√x+1

< 1x3 . Since

∫ ∞

1

1

x3dx converges,

∫ ∞

1

1

x3 +√x+ 1

dx con-

verges by the Comparison Theorem for Improper Integrals.

ii.

∫ ∞

1

1

x2 + 3√x+ 1

dx

Note that for x ≥ 1, 1x2+ 3

√x+1

< 1x2 . Since

∫ ∞

1

1

x2dx converges,

∫ ∞

1

1

x2 + 3√x+ 1

dx con-

verges by the Comparison Theorem for Improper Integrals.

254



45. Let R be the region in the plane bounded by the curves x = 1, y = 0, and y = 2√x3.


∫ 1

0

2√x3 dx = 4

5

√x5

∣

∣

∣

∣

1

0

= 45


x = 54

∫ 1

0

2x√x3 dx = 5

2

∫ 1

0

√x5 dx

= 52

(

27

√x7)

∣

∣

∣

∣

1

0

= 57

y = 54

∫ 1

0

12

(

2√x3)2

dx = 52

∫ 1

0

x3 dx

= 58x4

∣

∣

∣

∣

1

0

= 58

Centroid:(

57, 58

)

(10) c. Find the perimeter of R.

y = 2√x3 = 2x

32

y′ = 3x12 = 3

√x

√

(y′)2 + 1 =√9x+ 1

∫ 1

0

√9x+ 1dx = 2

27(9x+ 1)

32

∣

∣

∣

∣

1

0

= 227(10

√10− 1)

Perimeter: 227(10

√10− 1) + 2

(10) 46. Let R be the region in the plane bounded by the curves x = 2, y = 0, and y =√x. Find the surface area of the solid formed by revolving R about the x-axis.

y =√x

y′ = 12√x

2πx√

(y′)2 + 1 = 2πx√

14x

+ 1

= 2πx√

4x+14x

= π√4x+ 1

∫ 2

0

π√4x+ 1dx = π

6(4x+ 1)

32

∣

∣

∣

∣

2

0

= 13π3


dy

dx= 2xy, y(0) = e3, y > 0

dy

dx= 2xy

dy

y= 2x dx

∫

1ydy =

∫

2xdx

ln y = x2 + C

y(0) = e3

C = 3

ln y = x2 + 3

y = ex2+3

255

(10) 48. Recall that the weight density of water is 62.5 lb/ft3. An aquarium is in theshape of a rectangular prism. Its length is 5 ft, its width is 3 ft, and its height is 2 ft. Findthe hydrostatic force on

a. an end

limn→∞

n∑

i=1

62.5 · 2in· 2n· 3 = 187.5

∫ 2

0

xdx = 187.5(

12x2)

∣

∣

∣

∣

2

0

= 375

375 lb

b. a side

limn→∞

n∑

i=1

62.5 · 2in· 2n· 5 = 312.5

∫ 2

0

xdx = 312.5(

12x2)

∣

∣

∣

∣

2

0

= 625

156.25 lb

(10) 49. A tank contains 500 gallons of brine which contains .20 pounds of salt pergallon. Pure water flows into the tank at a rate of 10 gallons per minute. The solution iskept thoroughly mixed and drains from the tank at a rate of 10 gallons per minute. Expressthe amount of salt in the tank as a function of time.


dQ

dt= -

Q

500· 10 = -

Q

50

dQ

Q= -

dt

50

∫

dQ

Q=

∫

-dt

50

ln |Q| = - t50

+ C

lnQ = - t50

+ C

Since Q(0) = 100, C = ln 100.

lnQ = - t50

+ ln 100

Q = 100e- t50

50. Give a formula or rule for each of the following sequences.

(10) a. {14, 29, 316, 425, 536, 649, 764, . . .}

{

n

(n+ 1)2

}∞

n=1

(10) b. {2, 3, 6, 18, 108, 1944, . . .}

x1 = 2

x2 = 3

xn+2

= xn· x

n+1

256

51. For each of the following, determine whether the sequence converges or diverges.

(10) a.

{

n!

(n+ 1)!

}∞

n=1

limn→∞

n!

(n+ 1)!= lim

n→∞

1

n+ 1= 0

Converges

(10) b.

{

n+ 1√n

}∞

n=1

limn→∞

n+ 1√n

= limn→∞

(√n + 1√

n

)

= ∞

Diverges

(10) 52. Find the sum of the following series.

∞∑

n=1

2

n2 + 2n=

∞∑

n=1

2

n(n+ 2)=

∞∑

n=1

(

1

n− 1

n + 2

)

For each n ∈ N, Sn= 1− 1

3+ 1

2− 1

4+ 1

3− 1

5+ 1

4− 1

6+ · · ·+ 1

n−2− 1

n+ 1

n−1− 1

n+1+ 1

n− 1

n+2

= 1 + 12− 1

n+1− 1

n+2. So lim

n→∞S

n= 3

2.

∞∑

n=1

2

n2 + 2n= 3

2

53. Consider the following sequence.Advice: Study it for a minute.

{

56, 56+ 1

12, 56+ 1

12+ 1

24, 56+ 1

12+ 1

24+ 1

48, 56+ 1

12+ 1

24+ 1

48+ 1

96, 56+ 1

12+ 1

24+ 1

48+ 1

96+ 1

192, . . .

}

(10) a. Explain why the sequence is increasing.

Note that for each n ∈ N, xn+1

= xn+ 1

6·2n .

(10) b. Explain why the sequence is bounded above by 1.

The first term is x1= 5

6. If 1

6is added to x

1, the sum is 1. However, x

2= x

1+ 1

12. If 1

12is

added to x2 , the sum is 1. However, x3 = x2 +124. For each n ∈ N, x

n+1 = xn+ 1

2(1 − x

n)

= 1− 16·2n−1 < 1.

(10) c. Use the previous two parts to conclude that the sequence converges. What is thelimit of the sequence?

By the previous two parts, the sequence is bounded and increasing. Recall that boundedmonotonic sequences are convergent. Also, by the argument above, lim

n→∞x

n= lim

n→∞

(

1− 16·2n−1

)

= 1− 0 = 1.

Contentsjktartir.people.ysu.edu/1572/exam.pdf · 2019-04-19 · 1 Chapter 1: Math 1572 Summer 2002 Section 1.1: Review Solutions to Review Problems 1. For each of the following, ﬁnd

Documents