C H A P T E R 5Integration
Section 5.1 Antiderivatives and Indefinite Integration . . . . . . . . . 395
Section 5.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407
Section 5.3 Reimann Sums and Definite Integrals . . . . . . . . . . . 420
Section 5.4 The Fundamental Theorem of Calculus . . . . . . . . . . 429
Section 5.5 Integration by Substitution . . . . . . . . . . . . . . . . . 441
Section 5.6 Numerical Integration . . . . . . . . . . . . . . . . . . . 460
Section 5.7 The Natural Logarithmic Function: Integration . . . . . . 470
Section 5.8 Inverse Trigonometric Functions: Integration . . . . . . . 480
Section 5.9 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 489
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510
C H A P T E R 5Integration
Section 5.1 Antiderivatives and Indefinite Integration
395
1.ddx�
3x3 � C� �
ddx
�3x�3 � C� � �9x�4 ��9x 4
3.ddx�
13
x3 � 4x � C� � x2 � 4 � �x � 2��x � 2�
7.
Check:ddx�
25
x5�2 � C� � x3�2
y �25
x5�2 � C
dydx
� x3�2
2.ddx�x4 �
1x
� C� � 4x3 �1x2
4.
� x1�2 � x�3�2 �x2 � 1
x3�2
ddx�
2�x2 � 3�3�x
� C� �ddx�
23
x3�2 � 2x�1�2 � C�
5.
Check:ddt
t3 � C � 3t2
y � t3 � C
dydt
� 3t2 6.
Check:dd�
�� � C � �
r � �� � C
drd�
� �
8.
Check:ddx�
�1x2 � C� � 2x�3
y �2x�2
�2� C �
�1x2 � C
dydx
� 2x�3
Given Rewrite Integrate Simplify
9.34
x 4�3 � Cx 4�3
4�3� C�x1�3 dx�3�x dx
11. �2
�x� C
x�1�2
�1�2� C�x�3�2 dx� 1
x�x dx
13. �1
4x2 � C12�
x�2
�2� � C12�x�3 dx� 1
2x3 dx
10. �1x
� Cx�1
�1� C�x�2 dx� 1
x2 dx
12.14
x 4 �32
x2 � Cx 4
4� 3�x2
2 � � C��x3 � 3x� dx�x�x2 � 3� dx
14.�19x
� C19�
x�1
�1� � C19�x�2 dx� 1
�3x2� dx
396 Chapter 5 Integration
15.
Check:ddx�
x2
2� 3x � C� � x � 3
��x � 3� dx �x2
2� 3x � C
17.
Check:ddx�
14
x 4 � 5x � C� � x3 � 5
��x3 � 5� dx �14
x 4 � 5x � C
19.
Check:ddx�
25
x5�2 � x2 � x � C� � x3�2 � 2x � 1
��x3�2 � 2x � 1� dx �25
x5�2 � x2 � x � C
16.
Check:ddx�5x �
x2
2� C� � 5 � x
��5 � x� dx � 5x �x2
2� C
18.
Check:ddx
x 4 � 2x 3 � x � C � 4x 3 � 6x 2 � 1
��4x3 � 6x2 � 1� dx � x 4 � 2x 3 � x � C
20.
Check:ddx�
47
x7�4 � x � C� � x3�4 � 1 � 4�x3 � 1
��4�x3 � 1� dx � ��x3�4 � 1� dx �47
x7�4 � x � C
21.
Check:ddx��
12x2 � C� �
1x3
� 1x3 dx � �x�3 dx �
x�2
�2� C � �
12x2 � C 22.
Check:ddx��
13x3 � C� �
1x4
� 1x4 dx � �x�4 dx �
x�3
�3� C � �
13x3 � C
25.
Check:
� �x � 1��3x � 2�
ddx�x3 �
x2
2� 2x � C� � 3x2 � x � 2
� x3 �x2
2� 2x � C
��x � 1��3x � 2� dx � ��3x2 � x � 2� dx
23.
Check: �x2 � x � 1
�x
ddx�
25
x5�2 �23
x3�2 � 2x1�2 � C� � x3�2 � x1�2 � x�1�2
�2
15x1�2�3x2 � 5x � 15� � C�
25
x5�2 �23
x3�2 � 2x1�2 � C�x2 � x � 1
�x dx � ��x3�2 � x1�2 � x�1�2� dx
24.
Check: �x2 � 2x � 3
x4 ddx��
1x
�1x2 �
1x3 � C� � x�2 � 2x�3 � 3x�4
��1x
�1x2 �
1x3 � C�
x�1
�1�
2x�2
�2�
3x�3
�3� C �x2 � 2x � 3
x4 dx � ��x�2 � 2x�3 � 3x�4� dx
26.
Check:
� �2t2 � 1�2
ddt�
45
t5 �43
t3 � t � C� � 4t4 � 4t2 � 1
�45
t5 �43
t3 � t � C
��2t2 � 1�2 dt � ��4t4 � 4t2 � 1� dt
27.
Check:ddy�
27
y7�2 � C� � y5�2 � y2�y
�y2�y dy � �y5�2 dy �27
y7�2 � C 28.
Check:ddt�
13
t3 �34
t4 � C� � t2 � 3t3 � �1 � 3t�t2
��1 � 3t�t2 dt � ��t2 � 3t3� dt �13
t3 �34
t4 � C
29.
Check:ddx
�x � C� � 1
�dx � �1 dx � x � C 30.
Check:ddt
�3t � C� � 3
�3 dt � 3t � C
Section 5.1 Antiderivatives and Indefinite Integration 397
31.
Check:ddx
��2 cos x � 3 sin x � C� � 2 sin x � 3 cos x
��2 sin x � 3 cos x� dx � �2 cos x � 3 sin x � C 32.
Check:ddt�
13
t3 � cos t � C� � t2 � sin t
��t2 � sin t� dt �13
t3 � cos t � C
33.
Check:ddt
�t � csc t � C� � 1 � csc t cot t
��1 � csc t cot t� dt � t � csc t � C 34.
Check:d
d��13
� 3 � tan � � C� � � 2 � sec2 �
��� 2 � sec2 �� d� �13
� 3 � tan � � C
35.
Check:ddx
��2 cos x � 5e x � C� � 2 sin x � 5e x
� �2 sin x � 5e x� dx � �2 cos x � 5e x � C
37.
Check:dd�
�tan � � cos � � C� � sec2 � � sin �
��sec2 � � sin �� d� � tan � � cos � � C
39.
Check:ddy
�tan y � C� � sec2 y � tan2 y � 1
��tan2 y � 1� dy � �sec2 y dy � tan y � C
41.
Check:ddx�x2 �
4x
ln 4� C� � 2x � 4x
� �2x � 4x� dx � x2 �4x
ln 4� C
36.
Check:ddx
�x3 � 2e x � C� � 3x2 � 2e x
� �3x2 � 2e x� dx � x3 � 2e x � C
38.
Check:
� sec y�tan y � sec y�
ddy
�sec y � tan y � C� � sec y tan y � sec2 y
� sec y � tan y � C
�sec y�tan y � sec y� dy � ��sec y tan y � sec2 y� dy
40.
�cos x
1 � cos2 x
Check: ddx
�csc x � C � csc x cot x �1
sin x�
cos xsin x
� �csc x cot x dx � �csc x � C
� cos x1 � cos2 x
dx � �cos xsin2 x
dx � �� 1sin x��
cos xsin x� dx
42.
Check:ddx
�sin x �3x
ln 3� C� � cos x � 3x
� �cos x � 3x� dx � sin x �3x
ln 3� C
43.
Check:ddx �
x2
2� 5 ln�x� � C� � x �
5x
� �x �5x� dx �
x2
2� 5 ln�x� � C 44.
Check:ddx
�4 ln�x� � tan x � C� �4x
� sec2 x
� �4x
� sec2 x� dx � 4 ln�x� � tan x � C
45.
2
2
3
3
2 2
x
3C
2C
0C
y
f �x� � cos x 46.
x4 6 8−2
−2
−4
2
4
6
C = 0
C = 3
C = 2−
y
f �x� � �x 47.
x
6
4
2
−2
−4
6
C = 3
C = 0
C = 2−8
y
f �x� � ln x
398 Chapter 5 Integration
48.
y
6
4
2
−4
x2−2−4
C = 3
C = 0
C = 2−
4
f �x� �12
e x 49.
Answers will vary.
5
4
3
32123
y
x
2x)x)
22x) )xf
ff ′
f �x� � 2x � C
f� �x� � 2 50.
x
y
4
6
8
−2−4 2 4
f x( ) 2= +x2
2
f x( ) = x2
2
f ′
f �x� �x2
2� C
f��x� � x
51.
Answers will vary.
3
4
3
2
231
2
x
y3
3
3x
xx)f
xx3
x)f )3
)
f
f �x� � x �x3
3� C
f��x� � 1 � x2 52.
x−2−4
−2
−4
4
f x( ) = −
f x( ) 1= − +
1
1
x
x
f ′
y
f �x� � �1x
� C
f��x� �1x2 53.
y � x2 � x � 1
1 � �1�2 � �1� � C ⇒ C � 1
y � ��2x � 1� dx � x2 � x � C
dydx
� 2x � 1, �1, 1�
54.
y � x2 � 2x � 1
2 � �3�2 � 2�3� � C ⇒ C � �1
y � �2�x � 1� dx � x2 � 2x � C
dydx
� 2�x � 1� � 2x � 2, �3, 2� 55.
y � sin x � 4
4 � sin 0 � C ⇒ C � 4
y � �cos x dx � sin x � C
dydx
� cos x, �0, 4� 56.
y � 3 ln x
3 � 3 ln e � C ⇒ C � 0
y � �3x dx � 3 ln x � C
dydx
�3x, x > 0, �e, 3�
57. (a) Answers will vary.
x
y
−3 5
−3
5
(b)
y �x2
4� x � 2
2 � C
2 �42
4� 4 � C
y �x2
4� x � C
−4 8
−2
6dydx
�12
x � 1, �4, 2�
Section 5.1 Antiderivatives and Indefinite Integration 399
58. (a)
x− 4
−5
5
4
y (b)
y �x3
3� x �
73
C �73
3 � �13
� 1 � C
3 ���1�3
3� ��1� � C
y �x3
3� x � C
−4 4
−5
( 1, 3)−
5dydx
� x2 � 1, ��1, 3�
59. (a)
3
−2
6
−4
y
x
(b)
y � sin x � 4
4 � sin�0� � C ⇒ C � 4
y � �cos x dx � sin x � C
7
6
−1
−6
dydx
� cos x, �0, 4�
60. (a)
1 7x
y
(1, 3)
(b)
y �1x
� 2
3 �11
� C ⇒ C � 2
y � ��1x 2 dx � ��x�2 dx �
�x�1
�1� C �
1x
� C
−1
5
8−1
dydx
��1x2 , x > 0, �1, 3�
61. (a) 9
3
−9
−3
(b)
y � x2 � 6
�2 � ��2�2 � C � 4 � C ⇒ C � �6
y � �2x dx � x2 � C
dydx
� 2x, ��2, �2�
62. (a) 20
60
0
(b)
y �43
x3�2 �43
12 �43
�4�3�2 � C �43
�8� � C �323
� C ⇒ C �43
y � �2x1�2 dx �43
x 3� 2�C
dydx
� 2�x, �4, 12� (c) 20
60
0
(c) 12
15
−8
−15
400 Chapter 5 Integration
63.
f �x� � 2x2 � 6
f�0� � 6 � 2�0�2 � C ⇒ C � 6
f �x� � �4x dx � 2x2 � C
f��x� � 4x, f �0� � 6
65.
h�t� � 2t4 � 5t � 11
h�1� � �4 � 2 � 5 � C ⇒ C � �11
h�t� � ��8t3 � 5� dt � 2t 4 � 5t � C
h��t� � 8t3 � 5, h�1� � �4
67.
f �x� � x2 � x � 4
f �2� � 6 � C2 � 10 ⇒ C2 � 4
f �x� � ��2x � 1� dx � x2 � x � C2
f��x� � 2x � 1
f��2� � 4 � C1 � 5 ⇒ C1 � 1
f��x� � �2 dx � 2x � C1
f �2� � 10
f��2� � 5
f � �x� � 2
69.
f �x� � �4x1�2 � 3x � �4�x � 3x
f �0� � 0 � 0 � C2 � 0 ⇒ C2 � 0
f �x� � ���2x�1�2 � 3� dx � �4x1�2 � 3x � C2
f��x� � �2
�x� 3
f��4� � �22
� C1 � 2 ⇒ C1 � 3
f��x� � �x�3�2 dx � �2x�1�2 � C1 � �2
�x� C1
f �0� � 0
f��4� � 2
f � �x� � x�3�2
64.
g�x� � 2x3 � 1
g�0� � �1 � 2�0�3 � C ⇒ C � �1
g�x� � �6x2 dx � 2x3 � C
g��x� � 6x2, g�0� � �1
66.
f �s� � 3s 2 � 2s 4 � 23
f �2� � 3 � 3�2�2 � 2�2�4 � C � 12 � 32 � C ⇒ C � 23
f �s� � ��6s � 8s 3� ds � 3s 2 � 2s 4 � C
f��s� � 6s � 8s 3, f �2� � 3
68.
f �x� �1
12x4 � 6x � 3
f �0� � 0 � 0 � C2 � 3 ⇒ C2 � 3
f �x� � ��13
x3 � 6� dx �1
12x4 � 6x � C2
f��x� �13
x3 � 6
f��0� � 0 � C1 � 6 ⇒ C1 � 6
f��x� � �x2 dx �13
x3 � C1
f �0� � 3
f��0� � 6
f ��x� � x2
70.
f �x� � �sin x � 2x � 6
f �0� � 0 � 0 � C2 � 6 ⇒ C2 � 6
f �x� � ���cos x � 2� dx � �sin x � 2x � C2
f��x� � �cos x � 2
f��0� � �1 � C1 � 1 ⇒ C1 � 2
f��x� � �sin x dx � �cos x � C1
f �0� � 6
f��0� � 1
f ��x� � sin x
Section 5.1 Antiderivatives and Indefinite Integration 401
71.
f �x� � e x � x � 4
f �0� � 5 � e0 � 0 � C2 ⇒ C2 � 4
f �x� � � �e x � 1� dx � e x � x � C2
f��0� � 2 � e0 � C1 ⇒ C1 � 1
f��x� � � e x dx � e x � C1
f � �x� � e x 72.
f �x� � �2 ln�x� � 6x � 3
f �1� � 3 � 6 � C2 ⇒ C2 � �3
f �x� � � ��2x
� 6� dx � �2 ln�x� � 6x � C2
f��1� � 4 � �2 � C1 ⇒ C1 � 6
f��x� � � 2x2 dx � � 2x�2 dx �
�2x
� C1
f � �x� �2x2
73. (a)
(b) h�6� � 0.75�6�2 � 5�6� � 12 � 69 cm
h�t� � 0.75t2 � 5t � 12
h�0� � 0 � 0 � C � 12 ⇒ C � 12
h�t� � ��1.5t � 5� dt � 0.75t2 � 5t � C 74.
P�7� � 100�7�3�2 � 500 2352 bacteria
P�t� �23
�150�t3�2 � 500 � 100t3�2 � 500
P�1� �23
k � 500 � 600 ⇒ k � 150
P�0� � 0 � C � 500 ⇒ C � 500
P�t� � �kt1�2 dt �23
kt 3�2 � C
dPdt
� k�t, 0 ≤ t ≤ 10
75. Graph of is given.
(a)
(b) No. The slopes of the tangent lines are greater than 2on Therefore, f must increase more than 4 unitson
(c) No, because f is decreasing on
(d) f is an maximum at because andthe first derivative test.
(e) f is concave upward when is increasing on and f is concave downward on Pointsof inflection at x � 1, 5.
�1, 5�.�5, �.��, 1�f�
f��3.5� 0x � 3.5
4, 5.f �5� < f �4�
0, 4.0, 2.
f��4� �1.0
f�f �0� � �4. (f) is a minimum at
(g)
x
2
2
4
4
6
6 8−2
−6
y
x � 3.f �
76. Since is negative on is decreasing onSince is positive on is increasing
on has a relative minimum at Since ispositive on f is increasing on
x1 2 3−2−3
1
2
3
−2
−3f
f ′
f ′′
y
��, �.��, �,f��0, 0�.f��0, �.
f��0, �,f ���, 0�.f���, 0�,f � 77.
The ball reaches its maximum height when
s�158 � � �16�15
8 �2� 60�15
8 � � 6 � 62.25 feet
t �158 seconds
32t � 60
v�t� � �32t � 60 � 0
s�t� � �16t2 � 60t � 6, Position function
s�0� � 6 � C2
s�t� � ���32t � 60� dt � �16t 2 � 60t � C2
v�0� � 60 � C1
v�t� � ��32 dt � �32t � C1
a�t� � �32 ft�sec2
402 Chapter 5 Integration
78.
f �t� � �16t2 � v0t � s0
f �0� � 0 � 0 � C2 � s0 ⇒ C2 � s0
f �t� � s�t� � ���32t � v0� dt � �16t2 � v0t � C2
f��t� � �32t � v0
f��0� � 0 � C1 � v0 ⇒ C1 � v0
f��t� � v�t� � ��32 dt � �32t � C1
f �0� � s0
f��0� � v0
f� �t� � a�t� � �32 ft�sec2 79. From Exercise 78, we have:
when time to reachmaximum height.
v0 187.617 ft�sec
v02 � 35,200
�v0
2
64�
v02
32� 550
s� v0
32� � �16� v0
32�2
� v0� v0
32� � 550
t �v0
32�s��t� � �32t � v0 � 0
s�t� � �16t2 � v0t
80.
(a)
Choosing the positive value,
(b)
� �8�65 �64.498 ft�sec
v�1 � �654 � � �32�1 � �65
4 � � 8
v�t� � s��t� � �32t � 8
t �1 � �65
4 2.266 seconds.
t �1 ± �65
4
�8�2t2 � t � 8� � 0
s�t� � �16t2 � 8t � 64 � 0
s0 � 64 ft
v0 � 8 ft�sec
82. From Exercise 81, (Using the canyon floor as position 0.)
t2 �16004.9
⇒ t �326.53 18.1 sec
4.9t2 � 1600
f �t� � 0 � �4.9t2 � 1600
f �t� � �4.9t2 � 1600.
81.
f �0� � s0 � C2 ⇒ f �t� � �4.9t2 � v0t � s0
f �t� � ���9.8t � v0� dt � �4.9t2 � v0t � C2
v�0� � v0 � C1 ⇒ v�t� � �9.8t � v0
v�t� � ��9.8 dt � �9.8t � C1
a�t� � �9.8
83. From Exercise 81,
(Maximum height when )
f � 109.8� 7.1 m
t �109.8
9.8t � 10
v � 0.v �t� � �9.8t � 10 � 0
f �t� � �4.9t2 � 10t � 2.
Section 5.1 Antiderivatives and Indefinite Integration 403
84. From Exercise 81, If
then
for this t-value. Hence, and we solve
v02 � 3880.8 ⇒ v0 62.3 m�sec.
4.9 v02 � �9.8�2 198
�4.9 v02 � 9.8 v0
2 � �9.8�2 198
�4.9 v0
2
�9.8�2 �v0
2
9.8� 198
�4.9� v0
9.8�2
� v0� v0
9.8� � 2 � 200
t � v0�9.8
v�t� � �9.8t � v0 � 0
f �t� � 200 � �4.9t2 � v0t � 2,
f �t� � �4.9t2 � v0t � 2.
85.
since the stone was dropped,
Thus, the height of the cliff is 320 meters.
v�20� � �32 m�sec
v�t� � �1.6t
s0 � 320
s�20� � 0 ⇒ �0.8�20�2 � s0 � 0
s�t� � ���1.6t� dt � �0.8t2 � s0
v0 � 0. v�t� � ��1.6 dt � �1.6t � v0 � �1.6t,
a � �1.6
86.
When
v2 � v02 � 2GM�1
y�
1R�
v2 �2GM
y� v0
2 �2GM
R
12
v2 �GM
y�
12
v02 �
GMR
C �12
v02 �
GMR
12
v02 �
GMR
� C
y � R, v � v0.
12
v2 �GM
y� C
�v dv � �GM� 1y2 dy 87.
(a)
(b) when or
(c) when
v�2� � 3�1���1� � �3
t � 2.a�t� � 6�t � 2� � 0
3 < t < 5.0 < t < 1v�t� > 0
a�t� � v��t� � 6t � 12 � 6�t � 2�
� 3�t2 � 4t � 3� � 3�t � 1��t � 3�
v�t� � x��t� � 3t2 � 12t � 9
0 ≤ t ≤ 5x�t� � t3 � 6t2 � 9t � 2,
404 Chapter 5 Integration
88.
(a)
(b) when and
(c) when
v�73� � �3�7
3� � 5��73
� 3� � 2��23� � �
43
t �73
.a�t� � 6t � 14 � 0
3 < t < 5.0 < t < 53
v�t� > 0
a�t� � v��t� � 6t � 14
v�t� � x��t� � 3t2 � 14t � 15 � �3t � 5��t � 3� � t3 � 7t2 � 15t � 9
0 ≤ t ≤ 5x�t� � �t � 1��t � 3�2, 89.
a�t� � v��t� � �12
t�3�2 ��1
2t 3�2, acceleration
x�t� � 2t1�2 � 2, position function
x�1� � 4 � 2�1� � C ⇒ C � 2
x�t� � �v�t� dt � 2t1�2 � C
t > 0v�t� �1
�t� t�1�2,
90. (a)
(b) for k � 0, 1, 2, . . .t � k�,v�t� � 0 � sin t
f �t� � �cos t � 4
f �0� � 3 � �cos�0� � C2 � �1 � C2 ⇒ C2 � 4
f �t� � �v�t� dt � �sin t dt � �cos t � C2
v�t� � �a�t� dt � �cos t dt � sin t � C1 � sin t �since v0 � 0�
a�t� � cos t
91. (a)
(b)
s�13� �275234
�13�2
2�
25036
�13� 189.58 m
s�t� � at 2
2�
25036
t �s�0� � 0�
a �550468
�275234
1.175 m�sec2
55036
� 13a
v�13� �80036
� 13a �25036
v�0� �25036
⇒ v�t� � at �25036
v�t� � at � C
a�t� � a �constant acceleration�
v�13� � 80 km�hr � 80 �10003600
�80036
m�sec
v�0� � 25 km�hr � 25 �10003600
�25036
m�sec
Section 5.1 Antiderivatives and Indefinite Integration 405
92.
s�t� � �8.25t 2 � 66t
v�t� � �16.5t � 66
a�t� � �16.5
� 132 when a �332
� 16.5.
s�66a � � �
a2�
66a �
2
� 66�66a �
�at � 66 � 0 when t �66a
.
v�t� � 0 after car moves 132 ft.
s�t� � �a2
t2 � 66t �Let s�0� � 0.�
v�t� � �at � 66
a�t� � �a
15 mph � 22 ft�sec
30 mph � 44 ft�sec
v�0� � 45 mph � 66 ft�sec (a)
(b)
(c)
It takes 1.333 seconds to reduce the speed from 45 mphto 30 mph, 1.333 seconds to reduce the speed from 30 mph to 15 mph, and 1.333 seconds to reduce the speedfrom 15 mph to 0 mph. Each time, less distance is neededto reach the next speed reduction.
0
73.33 117.33
132
feet feet
45 m
ph =
66
ft/se
c
30 m
ph =
44
ft/se
c15
mph
= 2
2 ft/
sec
0 m
ph
s� 4416.5� 117.33 ft
t �44
16.5 2.667
�16.5t � 66 � 22
s� 2216.5� 73.33 ft
t �22
16.5 1.333
�16.5t � 66 � 44
93. Truck:
Automobile:
At the point where the automobile overtakes the truck:
when sec.t � 10 0 � 3t�t � 10�
0 � 3t2 � 30t
30t � 3t2
s�t� � 3t2 �Let s�0� � 0.�
v�t� � 6t �Let v�0� � 0.�
a�t� � 6
s�t� � 30t �Let s�0� � 0.�
v�t� � 30 (a)
(b) v�10� � 6�10� � 60 ft�sec 41 mph
s�10� � 3�10�2 � 300 ft
94.
(a) (b)
(c)
The second car was going faster than the first until the end.
s2�30� 1970.3 feet
s1�30� 953.5 feet
In both cases, the constant of integration is 0 because s1�0� � s2�0� � 0.
s2�t� � �v2�t� dt � �0.1208t3
3�
6.7991t2
2� 0.0707t
s1�t� � �v1�t� dt �0.1068
3 t 3 �
0.04162
t2 � 0.3679t
v2�t� � �0.1208t2 � 6.7991t � 0.0707
v1�t� � 0.1068t2 � 0.0416t � 0.3679
�1 mi�hr��5280 ft�mi��3600 sec�hr� �
2215
ft�sec
t 0 5 10 15 20 25 30
0 3.67 10.27 23.47 42.53 66 95.33
0 30.8 55.73 74.8 88 93.87 95.33v2�ft�sec�
v1�ft�sec�
406 Chapter 5 Integration
95. True 96. True 97. True 98. True
99. False. For example, because x3
3� C �x2
2� C1��x2
2� C2�.�x � x dx �x dx � �x dx
100. False. has an infinite number of antiderivatives, eachdiffering by a constant.
f 101.
Answer: f �x� �x3
3 � 4x �
163
f �2� � 0 ⇒ 83
� 8 � C1 � 0 ⇒ C1 �163
f �x� �x3
3 � 4x � C1
f��2� � 0 ⇒ 4 � C � 0 ⇒ C � �4
f��x� � x2 � C
f ��x� � 2x
102.
continuous at
continuous at
f �x� � ��x � 12x � 51
,,
,
0 ≤ x < 22 ≤ x < 33 ≤ x ≤ 4
x � 3 ⇒ 6 � 5 � C3 � 1f
x � 2 ⇒ �2 � 1 � 4 � C2 ⇒ C2 � �5f
f �0� � 1 ⇒ C1 � 1
f �x� � ��x � C1
2x � C2
C3
,,
,
0 ≤ x < 22 < x < 33 < x ≤ 4
1 2
1
2
3 4x
y
f��x� � ��120
,,,
0 ≤ x < 22 < x < 33 < x ≤ 4
103.
is continuous: Values must agree at
The left and right hand derivatives at do not agree. Hence is not differentiable at x � 2.f
x � 2
f �x� � �x � 2,3x2
� 2,2
0 ≤ x < 2 2 ≤ x ≤ 5
4 � 6 � C2 ⇒ C2 � �2
x � 2:f
f �1� � 3 ⇒ 1 � C1 � 3 ⇒ C1 � 2
f �x� � �x � C1,3x2
� C22 ,
0 ≤ x < 2 2 ≤ x ≤ 5
f��x� � �1,3x,
0 ≤ x < 2 2 ≤ x ≤ 5
104.
Thus, for some constant k. Since,and
Therefore,
[Note that and satisfy theseproperties.]
c�x� � cos xs�x� � sin x
s�x�2 � c�x�2 � 1.
k � 1.c�0� � 1,s�0� � 0s�x�2 � c�x�2 � k
� 0
� 2s�x�c�x� � 2c�x�s�x�
ddx
s�x�2 � c�x�2 � 2s�x�s��x� � 2c�x�c��x�
Section 5.2 Area 407
105.ddx
�ln�Cx�� �ddx
�ln�C� � ln�x�� � 0 �1x
�1x
106.ddx
�ln�x� � C� �1x
� 0 �1x
107.
Note: and satisfy these conditions.
Differentiate with respect to
Differentiate with respect to
Letting
Hence,
Adding,
Integrating,
Clearly for if then which contradicts that are nonconstant.
Now,
Thus, and we have f �x�2 � g�x�2 � 1.C � 1
� C 2.
� � f �x�2 � g�x�2�� f � y�2 � g� y�2�
� f �x�2f � y�2 � g �x�2g� y�2 � f �x�2g� y�2 � g�x�2f � y�2
C � f �x � y�2 � g�x � y�2 � � f �x� f �y� � g�x�g�y��2 � � f �x�g� y� � g�x�f �y��2
f, gf �x�2 � �g�x�2 ⇒ f �x� � g�x� � 0,C � 0,C � 0,
f �x�2 � g�x�2 � C.
2 f �x�f��x� � 2g �x�g��x� � 0.
2g�x�g��x� � 2g�x�f �x�g��0�.
2 f �x�f��x� � �2 f �x�g�x�g��0�
g� �x� � f �x�g��0� � g�x�f��0� � f �x�g��0�.
f� �x� � f �x�f��0� � g�x�g��0� � �g�x�g��0�y � 0,
y.��g� �x � y� � f �x�g�� y� � g�x�f�� y�
y.��f� �x � y� � f �x�f�� y� � g�x�g�� y�
�g�x� � sin xf �x� � cos x�
f� �0� � 0
g�x � y� � f �x�g� y� � g�x�f � y�
f �x � y� � f �x�f � y� � g�x�g� y�
Section 5.2 Area
1. �5
i�1�2i � 1� � 2�
5
i�1i � �
5
i�11 � 2�1 � 2 � 3 � 4 � 5� � 5 � 35
2. �6
k�3k�k � 2� � 3�1� � 4�2� � 5�3� � 6�4� � 50 3. �
4
k�0
1k2 � 1
� 1 �12
�15
�1
10�
117
�15885
4. �5
j�3 1j
�13
�14
�15
�4760
5. �4
k�1c � c � c � c � c � 4c
6. �4
i�1��i � 1�2 � �i � 1�3� � �0 � 8� � �1 � 27� � �4 � 64� � �9 � 125� � 238
7. �9
i�1 13i
8. �15
i�1
51 � i
9. �8
j�1�5� j
8 � 3 10. �4
j�1�1 � � j
42
11.2n
�n
i�1��2i
n 3
� �2in 12.
2n
�n
i�1�1 � �2i
n� 1
2
13.3n
�n
i�1�2�1 �
3in
2
14.1n
�n�1
i�0�1 � � i
n2
15. � 2�20�21�2 � 420�
20
i�12i � 2�
20
i�1i 16. � 2�15�16�
2 � 45 � 195�15
i�1�2i � 3� � 2�
15
i�1i � 3�15�
408 Chapter 5 Integration
17.
� �19�20��39�6 � 2470
�20
i�1�i � 1�2 � �
19
i�0i2 18.
� �10�11��21�6 � 10 � 375
�10
i�1�i2 � 1� � �
10
i�1i2 � �
10
i�11
19.
� 12,040
� 14,400 � 2480 � 120
�152�16�2
4� 2
15�16��31�6
�15�16�
2
�15
i�1 i�i � 1�2 � �
15
i�1i 3 � 2�
15
i�1i 2 � �
15
i�1i 20.
�102�11�2
4�
10�11�2
� 3080
�10
i�1i�i2 � 1� � �
10
i�1i3 � �
10
i�1i
21. sum seq (TI-82)
��20��21��41�
6� 60 � 2930
�20
i�1�i2 � 3� �
20�20 � 1��2�20� � 1�6
� 3�20�
2 � 3, x, 1, 20, 1� � 2930�x 22. sum seq (TI-82)
��15�2�16�2
4� 15�16� � 14,160
�15
i�1�i3 � 2i� �
�15� 2�15 � 1� 2
4� 2
15�15 � 1�2
3 � 2x, x, 1, 15, 1� � 14,160�x
23.
s � �1 � 3 � 4 �92��1� �
252 � 12.5
S � �3 � 4 �92 � 5��1� �
332 � 16.5 24.
s � �4 � 4 � 2 � 0��1� � 10
S � �5 � 5 � 4 � 2��1� � 16
25.
s � �2 � 2 � 3��1� � 7
S � �3 � 3 � 5��1� � 11 26.
s � �2 �43 � 1 �
45 �
23� �
295 � 5.8
S � �4 � 2 �43 � 1 �
45� �
13715 � 9.13
27.
s�4� � 0�14 ��1
4 �14 ��1
2 �14 ��3
4�14 �
1 � �2 � �38
� 0.518
S�4� ��14 �
14 ��1
2 �14 ��3
4 �14 � �1�1
4 �1 � �2 � �3 � 2
8� 0.768
28.
s�4� � 4�e�0.5 � e�1 � e�1.5 � e�2� 12 � 2.666
S�4� � 4�e�0 � e�0.5 � e�1 � e�1.5� 12 � 4.395
29.
s�5� �1
6 5�15 �
17 5�
15 �
18 5�
15 �
19 5�
15 �
12�
15 �
16
�17
�18
�19
�1
10� 0.646
S�5� � 1�15 �
16 5�
15 �
17 5�
15 �
18 5�
15 �
19 5�
15 �
15
�16
�17
�18
�19
� 0.746
30.
s�5� ��1 � �15
2
�15 ��1 � �2
52
�15 ��1 � �3
52
�15 ��1 � �4
52
�15 � 0 � 0.659
�15�1 �
�245
��21
5�
�165
��95 � 0.859
S�5� � 1�15 ��1 � �1
52
�15 ��1 � �2
52
�15 ��1 � �3
52
�15 ��1 � �4
52
�15
>>
Section 5.2 Area 409
31. limn→�
��81n4n2�n � 1�2
4 �814
limn→�
�n4 � 2n3 � n2
n4 �814
�1� �814
32. limn→�
��64n3n�n � 1��2n � 1�
6 �646
limn→�
�2n3 � 3n2 � nn3 �
646
�2� �643
33. limn→�
��18n2n�n � 1�
2 �182
limn→�
�n2 � nn2 �
182
�1� � 9 34. limn→�
�� 1n2n�n � 1�
2 �12
limn→�
�n2 � nn2 �
12
�1� �12
35.
S�10,000� � 1.0002
S�1000� � 1.002
S�100� � 1.02
S�10� �1210
� 1.2
�n
i�1
2i � 1n2 �
1n2 �
n
i�1�2i � 1� �
1n2�2
n�n � 1�2
� n �n � 2
n� S�n�
36.
S�10,000� � 2.0003
S�1000� � 2.003
S�100� � 2.03
S�10� �2310
� 2.3
�n
j�1
4j � 1n2 �
1n2 �
n
j�1�4j � 1� �
1n2�4n�n � 1�
2� n �
2n � 3n
� S�n�
37.
S�10,000� � 1.99999998
S�1000� � 1.999998
S�100� � 1.9998
S�10� � 1.98
�6n2�2n2 � 3n � 1 � 3n � 3
6 �1n2�2n2 � 2� � S�n�
�n
k�1
6k�k � 1�n3 �
6n3 �
n
k�1�k2 � k� �
6n3�n�n � 1��2n � 1�
6�
n�n � 1�2
38.
S�10,000� � 1.000066657
S�1000� � 1.00066567
S�100� � 1.006566
S�10� � 1.056
�1
3n3�3n3 � 2n2 � 3n � 2� � S�n�
�1
3n3�3n3 � 6n2 � 3n � 4n2 � 6n � 2�
�4n3�n3 � 2n2 � n
4�
2n2 � 3n � 16
�n
i�1
4i2�i � 1�n4 �
4n4 �
n
i�1�i3 � i2� �
4n4�n2�n � 1�2
4�
n�n � 1��2n � 1�6
410 Chapter 5 Integration
39. � 8 limn→�
�1 �1n � 8� lim
n→� �8�n2 � n
n2 � limn→�
16n2�n�n � 1�
2 limn→�
�n
i�1�16i
n2 � limn→�
16n2 �
n
i�1i
40. � limn→�
42�1 �
1n � 2� lim
n→� 4n2�n�n � 1�
2 limn→�
�n
i�1�2i
n �2n � lim
n→� 4n2 �
n
i�1i
41.
� limn→�
� 16�
2 � �3 n� � �1 n2�1 �
13
� limn→�
16�
2n3 � 3n2 � nn3
� limn→�
1n3��n � 1��n��2n � 1�
6 limn→�
�n
i�1 1n3�i � 1�2 � lim
n→� 1n3 �
n�1
i�1i 2
42.
� 2�1 � 2 �43 �
263
� 2 limn→�
�1 � 2 �2n
�43
�2n
�2
3n2
� limn→�
2n3�n3 � �4n��n�n � 1�
2 �4�n��n � 1��2n � 1�
6
� limn→�
2n3��
n
i�1n2 � 4n �
n
i�1i � 4 �
n
i�1i 2
limn→�
�n
i�1�1 �
2in
2
�2n � lim
n→� 2n3 �
n
i�1�n � 2i�2
43. � 2 limn→�
�1 �n2 � n
2n2 � 2�1 �12 � 3� 2 lim
n→� 1n�n �
1n�
n�n � 1�2 lim
n→� �
n
i�1�1 �
in�
2n � 2 lim
n→� 1n��
n
i�11 �
1n
�n
i�1i
44.
� 2 limn→�
�10 �13n
�4n2 � 20
� 2 limn→�
�1 � 3 �3n
� 4 �6n
�2n2 � 2 �
4n
�2n2
� 2 limn→�
1n4 �n4 � 6n2�n�n � 1�
2 � 12n�n�n � 1��2n � 1�6 � 8�n2�n � 1�2
4
� 2 limn→�
1n4 �
n
i�1�n3 � 6n2i � 12ni 2 � 8i 3�
limn→�
�n
i�1�1 �
2in
3
�2n � 2 lim
n→� 1n4 �
n
i�1�n � 2i�3
45. (a)
(c) Since is increasing, on
—CONTINUED—
� �n
i�1 f �2i � 2
n �2n � �
n
i�1��i � 1��2
n�2n
s�n� � �n
i�1 f �xi�1� �x
�xi�1, xi�.f �mi� � f �xi�1�y � x
x31
3
2
1
y (b)
Endpoints:
(d) on
S�n� � �n
i�1 f �xi� �x � �
n
i�1 f �2i
n 2n
� �n
i�1�i�2
n�2n
�xi�1, xi�f �Mi� � f �xi�
0 < 1�2n < 2�2
n < . . . < �n � 1��2n < n�2
n � 2
�x �2 � 0
n�
2n
Section 5.2 Area 411
45. —CONTINUED—
(e) (f)
� limn→�
2�n � 1�
n� 2
� limn→�
� 4n2n�n � 1�
2
limn→�
�n
i�1�i�2
n�2n � lim
n→� 4n2 �
n
i�1i
� limn→�
�2�n � 1�n
�4n � 2
� limn→�
4n2�n�n � 1�
2� n
limn→�
�n
i�1��i � 1��2
n�2n � lim
n→� 4n2 �
n
i�1�i � 1�x 5 10 50 100
1.6 1.8 1.96 1.98
2.4 2.2 2.04 2.02S�n�
s�n�
46. (a)
(c) Since is increasing, on
(d) on
(e)
S�n� � �n
i�1 f �xi � �x � �
n
i�1 f �1 � i�2
n�2n � �
n
i�1�1 � i�2
n�2n
[xi�1, xi�f �Mi� � f �xi�
� � n
i�1f �1 � �i � 1��2
n�2n � �
n
i�1�1 � �i � 1��2
n�2n
s�n� � �n
i�1 f �xi�1� �x
�xi�1, xi�.f �mi � � f �xi�1�y � x
x2 4
1
2
3
4
y
x 5 10 50 100
3.6 3.8 3.96 3.98
4.4 4.2 4.04 4.02S�n�
s�n�
(b)
Endpoints:
1 < 1 � 1�2n < 1 � 2�2
n < . . . < 1 � �n � 1��2n < 1 � n�2
n
1 < 1 �2n
< 1 �4n
< . . . < 1 �2nn
� 3
�x �3 � 1
n�
2n
(f)
� limn→�
�2 �2�n � 1�
n � limn→�
�4 �2n � 4
limn→�
�n
i�1�1 � i�2
n�2n � lim
n→� 2n�n � �2
nn�n � 1�
2
� limn→�
�2 �2n � 2
n�
4n � lim
n→� �4 �
2n � 4
limn→�
�n
i�1�1 � �i � 1��2
n�2n � lim
n→� �2
n�n �2n�
n�n � 1�2
� n
412 Chapter 5 Integration
47.
Area � limn→�
s�n� � 2
� 3 �2n2 �
n
i�1i � 3 �
2�n � 1�n2n2 � 2 �
1n
s�n� � �n
i�1 f � i
n�1n � �
n
i�1��2� i
n � 3�1n
321x
2
1
y�Note: �x �1 � 0
n�
1n�0, 1�y � �2x � 3,
48.
Area � limn→�
S�n� � 6 �272
�392
� 6 �27n2��n � 1�n
2 � 6 �272 �1 �
1n
� �n
i�1�3�2 �
3in � 4�3
n � 18 � 3�3n
2
�n
i�1i � 12
S�n� � � n
i�1f �2 �
3in �
3n
x4 6 8 10 12
2
4
6
8
10
12
y�Note: �x �5 � 2
n�
3n�2, 5�y � 3x � 4,
49.
Area � limn→�
S�n� �73
� � 1n3 �
n
i�1i 2 � 2 �
n�n � 1��2n � 1�6n3 � 2 �
16�2 �
3n
�1n2 � 2
S�n� � � n
i�1f � i
n�1n � �
n
i�1�� i
n2
� 2�1n
x2 3
3
1
1
y
�Note: �x �1n�0, 1�y � x2 � 2,
50.
Area � limn →�
S�n� �92
�2� � 3 � 12
�27n3
n�n � 1��2n � 1�6
�3n
�n� �92
2n2 � 3n � 1
n2 � 3
�27n 3 �
n
i�1i 2 �
3n
�n
i�11
S�n� � �n
i�1 f �3i
n �3n � �
n
i�1��3i
n 2
� 1�3n
y
x1 2
2
4
6
8
10
12
3
�0, 3� �Note: �x �3 � 0
n�
3ny � x2 � 1,
Section 5.2 Area 413
51.
Area � limn →�
s�n� � 30 �83
� 4 �703
� 2313
� 30 �8
6n2 �n � 1��2n � 1� �4n
�n � 1�
�2n
�15n �4
n2 n�n � 1��2n � 1�
6�
4n
n�n � 1�
2
�2n
�n
i�1 �15 �
4i2
n2 �4in
s�n� � �n
i�1 f �1 �
2in �
2n � �
n
i�1 �16 � �1 �
2in
2
�2n
x
y
1−1
2468
101214
18
2 3
�1, 3� �Note: �x �2ny � 16 � x2,
52. Find area of region over the interval
Area �43
12
Area � limn→�
s�n� � 1 �13
�23
� 1 �1n3 �
n
i�1i 2 � 1 �
n�n � 1��2n � 1�6n3 � 1 �
16�2 �
3n
�1n2
s�n� � �n
i�1 f � i
n�1n � �
n
i�1�1 � � i
n2
�1n
x
2
3
−2 2
−1
y
�Note: �x �1n�0, 1�.��1, 1�;y � 1 � x2,
53.
Area � limn →�
s�n� � 189 �814
� 27 �272
�5134
� 128.25
� 189 �814n2�n � 1�2 �
816n2�n � 1��2n � 1� �
272
n � 1
n
�3n�63n �
27n3
n2�n � 1�2
4�
27n2
n�n � 1��2n � 1�6
�9n
n�n � 1�
2
�3n
�n
i�1 �63 �
27i3
n3 �27i2
n2 �9in
s�n� � �n
i�1 f�1 �
3in �
3n � �
n
i�1�64 � �1 �
3in
3
�3n
x5 6
30
y
40
50
60
70
20
10
1 2−1 3
�1, 4� �Note: �x �4 � 1
n�
3ny � 64 � x3,
54.
Since y both increases and decreases on is neither an upper nor lower sum.
Area � limn→�
T�n� � 1 �14
�34
� 1 �1n
�14
�2
4n�
14n2
�2n2 �
n
i�1i �
1n4 �
n
i�1i3 �
n�n � 1�n2 �
1n4�n2�n � 1�2
4
T�n� � �n
i�1 f � i
n�1n � �
n
i�1�2� i
n � � in
3
�1n
T�n��0, 1�,
x0.5 1.0 2.0
0.5
1.0
1.5
2.0
y
�Note: �x �1 � 0
n�
1n�0, 1�y � 2x � x3,
414 Chapter 5 Integration
55.
Again, is neither an upper nor a lower sum.
Area � limn→�
T�n� � 4 � 10 �323
� 4 �23
� 4 � 10�1 �1n �
163 �2 �
3n
�1n2 � 4�1 �
2n
�1n2
�4n
�n� �20n2 �
n�n � 1�2
�32n3 �
n�n � 1��2n � 1�6
�16n4 �
n2�n � 1�2
4
� �n
i�1�2 �
10in
�16i 2
n2 �8i3
n3 �2n �
4n�
n
i�11 �
20n2 �
n
i�1i �
32n3 �
n
i�1i 2 �
16n4 �
n
i�1i 3
� �n
i�1��1 �
4in
�4i2
n2 � ��1 �6in
�12i 2
n2 �8i3
n3 �2n
T�n� � �n
i�1 f ��1 �
2in �
2n � �
n
i�1���1 �
2in
2
� ��1 �2in
3
�2n
T�n�
x1
2
1
1
y
�Note: �x �1 � ��1�
n�
2n��1, 1�y � x2 � x3,
57.
Area � limn→�
S�n� � limn→�
�6 �6n � 6
�6�n � 1�
n� 6 �
6n
�12n2 �
n
i�1i � �12
n2 �n�n � 1�
2
� �n
i�1 f �2i
n �2n � �
n
i�13�2i
n �2nS�n� � �
n
i�1 f �mi� �y
64x
4
2
2
2
y
�Note: �y �2 � 0
n�
2n0 ≤ y ≤ 2f �y� � 3y,
56.
Area � limn→�
s�n� � 2 �52
�43
�14
�7
12
� 2 �52
�5
2n�
43
�2n
�2
3n2 �14
�1
2n�
14n2
� �n
i�1��2 �
5in
�4i 2
n2 �i 3
n3�1n � 2 �
5n2 �
n
i�1i �
4n3 �
n
i�1i 2 �
1n4 �
n
i�1i 3
s�n� � �n
i�1 f ��1 �
in�
1n � �
n
i�1���1 �
in
2
� ��1 �in
3
�1n
x1 2−1−2
−1
1
2
3
y
�Note: �x �0 � ��1�
n�
1n��1, 0�y � x2 � x3,
58.
Area � limn →�
S�n� � 2 � 1 � 3
�2n�n �
1n
n�n � 1�
2 � 2 �n � 1
n
� �n
i�1 12�2 �
2in �
2n �
2n
�n
i�1�1 �
in
S�n� � �n
i�1 g�2 �
2in �
2n
y
x1
1
2
3
4
5
2 3 4 5
g�y� �12
y, 2 ≤ y ≤ 4 �Note: �y �4 � 2
n�
2n
Section 5.2 Area 415
59.
Area � limn→�
S�n� � limn→�
�9 �272n
�9
2n2 � 9
�9n2�2n2 � 3n � 1
2 � 9 �272n
�9
2n2 �27n3 �
n�n � 1��2n � 1�6
� �n
i�1 �3i
n 2
�3n �
27n3 �
n
i�1i 2S�n� � �
n
i�1 f �3i
n �3n
x2 4 6 8 10
2
4
6
−2
−4
y�Note: �y �3 � 0
n�
3n0 ≤ y ≤ 3f �y� � y2,
60.
Area � limn →�
S�n� � 3 � 1 �13
�113
� 3 �n � 1
n�
�n � 1��2n � 1�6n2
�1n
�3n �2n
n�n � 1�
2�
1n2
n�n � 1��2n � 1�
6
�1n
�n
i�1 �3 �
2in
�i2
n2
�1n
�n
i�1 �4 �
4in
� 1 �2in
�i2
n2
�1n
�n
i�1 �4�1 �
in � �1 �
in
2
S�n� � �n
i�1 f�1 �
in�
1n
1
1
2
3
5
2 3 4 5
y
x
f �y� � 4y � y2, 1 ≤ y ≤ 2 �Note: �y �2 � 1
n�
1n
61.
Area � limn →�
S�n� � 6 � 10 �83
� 4 �443
�2n
�n
i�1�3 �
10in
�4i2
n2 �8i3
n3 �2n�3n �
10n
n�n � 1�
2�
4n2
n�n � 1��2n � 1�6
�8n3
n2�n � 1�2
4
�2n
�n
i�14�1 �
4in
�4i2
n2 � �1 �6in
�12i2
n2 �8i3
n3
� �n
i�1�4�1 �
2in
2
� �1 �2in
3
2n
S�n� � �n
i�1 g�1 �
2in �
2n
x
6
y
8
10
2
−2
−4
−2−4
g�y� � 4y2 � y3, 1 ≤ y ≤ 3 �Note: �y �3 � 1
n�
2n
416 Chapter 5 Integration
62.
Area � limn →�
S�n� � 2 �14
� 1 �32
�194
� 2 ��n � 1�2
n24�
12
�n � 1��2n � 1�
n2 �3�n � 1�
2n
�1n�2n �
1n3
n2�n � 1�2
4�
3n2
n�n � 1��2n � 1�6
�3n
n�n � 1�
2
�1n
�n
i�1�2 �
i3
n3 �3i2
n2 �3in
� �n
i�1��1 �
in
3
� 11n
S�n� � �n
i�1h�1 �
in�
1n
2
1
2
3
4
5
4 6 8 10
y
x
h�y� � y3 � 1, 1 ≤ y ≤ 2 �Note: �y �1n
63.
Let
�698
�12��
116
� 3 � � 916
� 3 � �2516
� 3 � �4916
� 3
Area � � n
i�1f �ci� �x � �
4
i�1�ci
2 � 3��12
c4 �74
c3 �54
,c2 �34
,c1 �14
,�x �12
,
ci �xi � xi�1
2.
n � 40 ≤ x ≤ 2,f �x� � x2 � 3, 64.
Let
� 53
� ��14
� 2 � �94
� 6 � �254
� 10 � �494
� 14
Area � �n
i�1 f �ci� �x � �
4
i�1�ci
2 � 4ci��1�
c4 �72
c3 �52
,c2 �32
,c1 �12
,�x � 1,
ci �xi � xi�1
2.
n � 40 ≤ x ≤ 4,f �x� � x2 � 4x,
65.
Let
�
16�tan
32� tan
3
32� tan
5
32� tan
7
32 � 0.345
Area � � n
i�1f �ci� �x � �
4
i�1�tan ci��
16
c4 �7
32c3 �
5
32,c2 �
3
32,c1 �
32,�x �
16,
ci �xi � xi�1
2.
n � 40 ≤ x ≤
4,f �x� � tan x, 66.
Let
�
8�sin
16� sin
3
16� sin
5
16� sin
7
16 � 1.006
Area � �n
i�1 f �ci� �x � �
4
i�1�sin ci��
8
c4 �7
16c3 �
5
16,c2 �
3
16,c1 �
16,�x �
8,
ci �xi � xi�1
2.
n � 40 ≤ x ≤
2,f �x� � sin x,
67.
�Exact value is 16 3.�
�0, 4�f �x� � �x,
n 4 8 12 16 20
5.3838 5.3523 5.3439 5.3403 5.3384Approximate area
Section 5.2 Area 417
68. �2, 6�f �x� �8
x2 � 1,
n 4 8 12 16 20
2.3397 2.3755 2.3824 2.3848 2.3860Approximate area
70. �0, 2�f �x� � cos�x,
n 4 8 12 16 20
1.1041 1.1053 1.1055 1.1056 1.1056Approximate area
69. �1, 3�f �x� � tan�x8 ,
n 4 8 12 16 20
2.2223 2.2387 2.2418 2.2430 2.2435Approximate area
71. �1, 5�f �x� � ln x,
n 4 8 12 16 20
4.0786 4.0554 4.0509 4.0493 4.0485Approximate area
72. �0, 2�f �x� � xe x,
n 4 8 12 16 20
8.1711 8.3341 8.3646 8.3753 8.3802Approximate area
74.
(a) square unitsA � 3
x1 2 3 4
1
2
3
4
y73.
(b) square unitsA � 6
x1 2 3 4
1
2
3
4
y
75. We can use the line bounded by and The sum of the areas of these inscribed rectangles is thelower sum.
x
y
a b
x � b.x � ay � x The sum of the areas of these circumscribed rectangles is theupper sum.
We can see that the rectangles do not contain all of the area inthe first graph and the rectangles in the second graph covermore than the area of the region. The exact value of the arealies between these two sums.
x
y
a b
418 Chapter 5 Integration
76. See the definition of area, page 297.
80. True. (Theorem 5.3)79. True. (Theorem 5.2 (2))
78.
Let area bounded by the x-axis, and Let area of the rectangle bounded by and Thus,In this program, the computer is generating pairs of random points in the rectangle whose area is represented by It is keeping track of how many of these points, lie in the regionwhose area is represented by Since the points are randomly generated, we assume that
The larger is the better the approximation to A1.N2
A1
A2�
N1
N2 ⇒ A1 �
N1
N2A2.
A1.N1,A2.
N2
A2 � � 2��1� � 1.570796.x � 2.x � 0,y � 0,y � 1,A2 �x � 2.x � 0f �x� � sin x,A1 �
x
0.25
0.5
0.75
1.0
π4
π2
π2( ), 1f x x( ) sin( )=
y�0,
2f �x� � sin x,
81. Suppose there are n rows and columns in the figure. The stars on the left total as do the stars on the right. There are stars in total, hence
1 � 2 � . . . � n �12�n��n � 1�.
2�1 � 2 � . . . � n� � n�n � 1�
n�n � 1�1 � 2 � . . . � n,n � 1
77. (a)
Lower sum:
(c)
Midpoint Rule:
(e)
(f) increases because the lower sum approaches the exact value as n increases. decreases because the upper sumapproaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than theexact value, whereas the upper sum is always larger.
S�n�s�n�
M�4� � 223 � 44
5 � 557 � 62
9 �6112315 � 19.403
x1
2
2 3
4
4
6
8
y
s�4� � 0 � 4 � 513 � 6 � 151
3 �463 � 15.333
x1
2
2 3
4
4
6
8
y (b)
Upper sum:
(d) In each case, The lower sum uses left end-points, The upper sum uses right endpoints,
The Midpoint Rule uses midpoints, �i �12��4 n�.�i��4 n�.
�i � 1��4 n�.�x � 4 n.
S�4� � 4 � 513 � 6 � 62
5 � 211115 �
32615 � 21.733
x1
2
2 3
4
4
6
8
y
n 4 8 20 100 200
15.333 17.368 18.459 18.995 19.06
21.733 20.568 19.739 19.251 19.188
19.403 19.201 19.137 19.125 19.125M�n�
S�n�
s�n�
Section 5.2 Area 419
83. (a)
(c) Using the integration capability of a graphing utility,you obtain
A � 76,897.5 ft2.
y � ��4.09 10�5�x3 � 0.016x2 � 2.67x � 452.9 (b)
00 350
500
82. (a)
(b)
(c)
Let As
limn→�
An � limx→0
r2�sin xx � r2�1� � r 2
x → 0.n →�,x � 2 n.
An � n�12
r2 sin 2
n �r2n2
sin 2
n� r2�sin�2 n�
2 n
A �12
bh �12
r�r sin �� �12
r2 sin �
h � r sin �
sin � �hr θ
r
r
h
� �2
n
84. For n odd,
1 row, 1 block
2 rows, 4 blocks
3 rows, 9 blocks
�n � 12
2 blocks
n � 12
rows,n,
n � 5,
n � 3,
n � 1,
For n even,
1 row, 2 blocks
2 rows, 6 blocks
3 rows, 12 blocks
n2 � 2n4
blocksn2
rows,n,
n � 6,
n � 4,
n � 2,
85. (a)
The formula is true for
Assume that the formula is true for
Then we have
which shows that the formula is true for n � k � 1.
� �k � 1��k � 2�
� k�k � 1� � 2�k � 1�
�k�1
i�12i � �
k
i�12i � 2�k � 1�
�k
i�12i � k�k � 1�.
n � k:
n � 1: 2 � 1�1 � 1� � 2.
�n
i�12i � n�n � 1� (b)
The formula is true for because
Assume that the formula is true for
Then we have
which shows that the formula is true for n � k � 1.
��k � 1�2
4�k � 2�2
��k � 1�2
4�k2 � 4�k � 1��
�k2�k � 1�2
4� �k � 1�3
�k�1
i�1i3 � �
k
i�1i3 � �k � 1�3
�k
i�1 i3 �
k2�k � 1�2
4.
n � k:
13 �12�1 � 1�2
4�
44
� 1.
n � 1
�n
i�1i3 �
n2�n � 1�2
4
420 Chapter 5 Integration
86. Assume that the dartboard has corners at
A point in the square is closer to the center than the top edge if
By symmetry, a point in the square is closer to the center than the rightedge if
In the first quadrant, the parabolas and intersect at There are 8 equal regions that make up the total region, as indicated in the figure.
Area of shaded region
Probability �8S
Area square� 2�2�2
3�
56� �
4�23
�53
S � ��2�1
0 �1
2�1 � x2� � x� dx �
2�23
�56
−1 1
−1
1
x
y
2 − 1, 2 − 1( (
(1, 1)
��2 � 1, �2 � 1�.x �12�1 � y2�y �
12�1 � x2�
x ≤12
�1 � y2�.
�x, y�
y ≤12
�1 � x2�.
x2 � y2 ≤ 1 � 2y � y2
�x2 � y2 ≤ 1 � y
�x, y�
(0, 0)
(x, 1)
(x, y)
−1 1
−1
1
x
y�±1, ±1�.
Section 5.3 Reimann Sums and Definite Integrals
1.
� 3�3�23
� 0� � 2�3 � 3.464
� limn →�
3�3��n � 1��2n � 1�3n2 �
n � 12n2 �
� limn →�
3�3
n3 �2n�n � 1��2n � 1�
6�
n�n � 1�2 �
� limn →�
3�3
n3 �n
i�1�2i2 � i�
limn →�
�n
i�1f �ci��xi � lim
n →� �
n
i�1�3i2
n2 3n2�2i � 1�
�xi �3i2
n2 �3�i � 1�2
n2 �3n2�2i � 1�
3n2
3(2)2
n23(n − 1)2
n2. . .
. . .
3
1
2
3
y
x
y = x
f �x� � �x, y � 0, x � 0, x � 3, ci �3i2
n2
Section 5.3 Reimann Sums and Definite Integrals 421
2.
� 2 limn→�
1n4�3n4
4�
n3
2�
n2
4 � � 2 limn→�
�34
�1
2n�
14n2� �
32
� 2 limn→�
1n4�3n4 � 6n3 � 3n2
4�
2n3 � 3n2 � n2
�n2 � n
2 �
� 2 limn→�
1n4�3n2�n � 1�2
4 � 3n�n � 1��2n � 1�6 �
n�n � 1�2 �
� 2 limn→�
1n4 �
n
i�1�3i 3 � 3i 2 � i�
limn→�
�n
i�1 f �ci� �xi � 2 lim
n→� �
n
i�1
3�i 3
n3 �3i 2 � 3i � 1n3 �
�xi �i 3
n3 ��i � 1�3
n3 �3i 2 � 3i � 1
n3
ci �i 3
n3x � 1,x � 0,y � 0,f �x� � 2 3�x,
3. on
�10
4 6 dx � lim
n→� 36 � 36
� �n
i�166
n � �n
i�1
36n
� 36� n
i�1f �ci� �xi � �
n
i�1 f 4 �
6in
6n
Note: �x �10 � 4
n�
6n
, ��� → 0 as n →��4, 10 .y � 6
4. on
�3
�2x dx � lim
n→� 5
2�
252n �
52
� �10 � 25n2n�n � 1�
2� �10 �
252 1 �
1n �
52
�252n
�n
i�1 f �ci � �xi � �
n
i�1 f �2 �
5in
5n � �
n
i�1�2 �
5in
5n � �10 �
25n2 �
n
i�1i
Note: �x �3 � ��2�
n�
5n
, ��� → 0 as n →���2, 3 .y � x
5. on
�1
�1x3 dx � lim
n→� 2n
� 0
� �2 � 61 �1n � 42 �
3n
�1n2 � 41 �
2n
�1n2 �
2n
� �2 �12n2 �
n
i�1i �
24n3 �
n
i�1i 2 �
16n4 �
n
i�1i 3
�n
i�1 f �ci� �xi � �
n
i�1 f �1 �
2in
2n � �
n
i�1�1 �
2in
3
2n � �
n
i�1��1 �
6in
�12i 2
n2 �8i 3
n3 �2n
Note: �x �1 � ��1�
n�
2n
, ��� → 0 as n →���1, 1 .y � x3
422 Chapter 5 Integration
6. on
� 6 � 12 � 8 � 26
�3
13x2 dx � lim
n →� �6 �
12�n � 1�n
�4�n � 1��2n � 1�
n2 �
� 6 � 12 n � 1
n� 4
�n � 1��2n � 1�n2
�6n�n �
4n
n�n � 1�
2�
4n2
n�n � 1��2n � 1�6 �
�6n�
n
i�11 �
4in
�4i2
n2
� �n
i�131 �
2in
2
2n
�n
i�1 f �ci� �xi � �
n
i�1 f 1 �
2in
2n
�1, 3 . Note: �x �3 � 1
n�
2n
, ��� → 0 as n →�y � 3x2
7. on
�2
1�x2 � 1� dx � lim
n→� 10
3�
32n
�1
6n2 �103
� 2 �2n2 �
n
i�1i �
1n3 �
n
i�1i2 � 2 � 1 �
1n �
162 �
3n
�1n2 �
103
�3
2n�
16n2
�n
i�1 f �ci� �xi � �
n
i�1 f 1 �
in
1n � �
n
i�1�1 �
in
2
� 1�1n � �
n
i�1�1 �
2in
�i2
n2 � 1�1n
Note: �x �2 � 1
n�
1n
, ��� → 0 as n →��1, 2 .y � x2 � 1
8. on
� 15 � 27 � 27 � 15
�2
�1 �3x2 � 2� dx � lim
n →� �15 � 27
�n � 1�n
�272
�n � 1��2n � 1�
n2 �
� 15 �27�n � 1�
n�
272
�n � 1��2n � 1�
n2
�3n�3n �
18n
n�n � 1�
2�
27n2
n�n � 1��2n � 1�6
� 2n�
�3n
�n
i�1�31 �
6in
�9i2
n2 � 2�
�3n�
n
i�1�3�1 �
3in
2
� 2
�n
i�1 f �ci� �xi � �
n
i�1 f �1 �
3in
3n
��1, 2 . Note: �x �2 � ��1�
n�
3n
; ��� → 0 as n →�y � 3x2 � 2
9.
on the interval ��1, 5 .
lim���→0
�n
i�1�3ci � 10� �xi � �5
�1�3x � 10� dx 10.
on the interval �0, 4 .
lim���→0
�n
i�16ci�4 � ci�2 �xi � �4
06x�4 � x�2 dx
Section 5.3 Reimann Sums and Definite Integrals 423
11.
on the interval �0, 3 .
lim���→0
�n
i�1
�ci2 � 4 �xi � �3
0
�x2 � 4 dx 12.
on the interval �1, 3 .
lim���→0
�n
i�1 3
ci2 �xi � �3
1
3x2 dx
13. �5
1 1 �
3x dx 14. ��
0 2�x sin x dx 15. �5
03 dx 16. �2
0x2 dx
17. �4
1 2x dx 18. �2
02e�x dx 19. ��
0sin x dx 20. ���4
0tan x dx
21. �2
0y3 dy 22. �2
0�y � 2�2 dy
23. Rectangle
x5
5
3
2
42 31
1
Rectangle
y
A � �3
04 dx � 12
A � bh � 3�4�
24. Rectangle
x
1
2
3
5
−a a
Rectangle
y
A � �a
�a
4 dx � 8a
A � bh � 2�4��a�
26. Triangle
x1 2 3 4
−1
1
2
3
Triangle
y
A � �4
0
x2
dx � 4
A �12
bh �12
�4��2�
25. Triangle
x
Triangle
4
2
42
y
A � �4
0x dx � 8
A �12
bh �12
�4��4�
27. Trapezoid
x
9
6
321
3
Trapezoid
y
A � �2
0�2x � 5� dx � 14
A �b1 � b2
2h � 5 � 9
2 2
29. Triangle
A � �1
�1�1 � �x�� dx � 1
A �12
bh �12
�2��1�1 Triangle
11x
y
28. Triangle
2
2
4
6
8
10
4 6 8 10
y
x
A � �8
0�8 � x�dx � 32
A �12
bh �12
�8��8� � 32
30. Triangle
A � �a
�a
�a � �x�� dx � a2
A �12
bh �12
�2a�a
x
Triangle
−a a
a
y
424 Chapter 5 Integration
32. Semicircle
A � �r
�r
�r2 � x2 dx
A �12
�r2
x−r
−r
r
r Semicircle
y31. Semicircle
A � �3
�3
�9 � x2 dx �9�
2
A �12
�r 2 �12
��3�2
x
4 Semicircle
4224
y
33. �2
4x dx � ��4
2x dx � �6
35. �4
24x dx � 4�4
2x dx � 4�6� � 24
In Exercises 33–40, �4
2x3 dx � 60, �4
2x dx � 6, �4
2dx � 2
37. �4
2�x � 8� dx � �4
2x dx � 8�4
2dx � 6 � 8�2� � �10
39.
�12
�60� � 3�6� � 2�2� � 16
�4
21
2x3 � 3x � 2 dx �
12�
4
2x3 dx � 3�4
2x dx � 2�4
2dx
34. �2
2x3 dx � 0
36. �4
215 dx � 15�4
2dx � 15�2� � 30
38. �4
2�x3 � 4� dx � �4
2x3 dx � 4�4
2dx � 60 � 4�2� � 68
40.
� 6�2� � 2�6� � 60 � �60
�4
2�6 � 2x � x3� dx � 6�4
2dx � 2�4
2x dx � �4
2x3 dx
42. (a)
(b)
(c)
(d) �6
3�5f �x� dx � �5�6
3f �x� dx � �5��1� � 5
�3
3f �x� dx � 0
�3
6f �x� dx � ��6
3f �x� dx � ���1� � 1
�6
0f �x� dx � �3
0f �x� dx � �6
3f �x� dx � 4 � ��1� � 3
44. (a)
(b)
(c)
(d) �1
03f �x� dx � 3�1
0f �x� dx � 3�5� � 15
�1
�13f �x� dx � 3�1
�1f �x� dx � 3�0� � 0
�1
0f �x� dx � �0
�1f �x� dx � 5 � ��5� � 10
�0
�1f �x� dx � �1
�1f �x� dx � �1
0f �x� dx � 0 � 5 � �5
41. (a)
(b)
(c)
(d) �5
03f �x� dx � 3�5
0f �x� dx � 3�10� � 30
�5
5f �x� dx � 0
�0
5f �x� dx � ��5
0f �x� dx � �10
�7
0f �x� dx � �5
0f �x� dx � �7
5f �x� dx � 10 � 3 � 13
43. (a)
(b)
(c)
(d) �6
23f �x� dx � 3�6
2f �x� dx � 3�10� � 30
�6
22g�x� dx � 2�6
2g�x� dx � 2��2� � �4
� �2 � 10 � �12
�6
2�g �x� � f �x� dx � �6
2g�x� dx � �6
2f �x� dx
� 10 � ��2� � 8
�6
2� f �x� � g�x� dx � �6
2f �x� dx � �6
2g�x� dx
45. Lower estimate:
Upper estimate: �32 � 24 � 12 � 4 � 20 �2� � 88
�24 � 12 � 4 � 20 � 36 �2� � �48 46. (a) left endpoint estimate
(b) right endpoint estimate
(c) midpoint estimate
If f is increasing, then (a) is below the actual value and (b) is above.
�0 � 18 � 50 �2� � 136,
�8 � 30 � 80 �2� � 236,
��6 � 8 � 30 (2) � 64,
Section 5.3 Reimann Sums and Definite Integrals 425
49. (a)
(b)
(c)
(d) � f odd��5
�5f �x� dx � 0
� f even��5
�5f �x� dx � 2�5
0f �x� dx � 2�4� � 8
�Let u � x � 2.��3
�2f �x � 2� dx � �5
0f �x� dx � 4
�5
0� f �x� � 2 dx � �5
0f �x� dx � �5
02 dx � 4 � 10 � 14
47. (a) Quarter circle below x-axis:
(b) Triangle:
(c) Triangle Semicircle below x-axis:
(d) Sum of parts (b) and (c):
(e) Sum of absolute values of (b) and (c):
(f) Answer to (d) plus �3 � 2�� � 20 � 23 � 2�2�10� � 20:
4 � �1 � 2�� � 5 � 2�
4 � �1 � 2�� � 3 � 2�
�12 �2��1� �
12 ��2�2 � ��1 � 2���
12 bh �
12 �4��2� � 4
�14 �r 2 � �
14 ��2�2 � ��
50.
4 8
4
8(8, 8)
x
y
�8
0 f �x� dx � 4�4� � 4�4� �
12
�4��4� � 40
f �x� � �4,x,
x < 4 x ≥ 4
48. (a)
(b)
(c)
(d)
(e)
(f) �10
4f (x) dx � 2 � 4 � �2
�11
0f (x) dx � �
12
� 2 � 2 � 2 � 4 �12
� 2
�11
5f (x) dx �
12
�12
�4��2� �12
� �3
�7
0f (x) dx � �
12
�12
�2��2� � 2 �12
�2��2� � 12
� 5
�4
33f (x) dx � 3�2� � 6
x
y
(4, 2)
(11, 1)
(3, 2)
(0, �1)
(8, �2)
−2 2 4 8 12
−1
−2
1
2
�1
0�f �x� dx � ��1
0f �x� dx �
12
51. The left endpoint approximation will be greater than the actual area: >
52. The right endpoint approximationwill be less than the actual area: <
53.
is not integrable on the intervalbecause f has a discontinuity
at x � 4.�3, 5
f �x� �1
x � 4
54. is integrable on but is not continuous on There is discontinuity at To see that
is integrable, sketch a graph of the region bounded by and the x-axis forYou see that the integral equals 0.�1 ≤ x ≤ 1.
f �x� � �x��x
�1
�1
�x�x
dx
x � 0.
x1 2−2
−2
1
2
y��1, 1 .��1, 1 ,f �x� � �x��x
426 Chapter 5 Integration
56.
(b) square unitsA � 43
x1 1314 42
1
2
3
4
y55.
(a) square unitsA � 5
x1 2 3 4
1
2
3
4
y
57.
�2
02e�x 2
dx �12
�2��2� � 2
1
3
1 2 3
y
x
58.
Area �13
1
1 2
−1
y
x
�2
1ln x dx
59. �3
0x�3 � x dx
4 8 12 16 20
3.6830 3.9956 4.0707 4.1016 4.1177
4.3082 4.2076 4.1838 4.1740 4.1690
3.6830 3.9956 4.0707 4.1016 4.1177R�n�
M�n�
L�n�
n
60. �3
0
5x2 � 1
dx
4 8 12 16 20
7.9224 7.0855 6.8062 6.6662 6.5822
6.2485 6.2470 7.2460 6.2457 6.2455
4.5474 5.3980 5.6812 5.8225 5.9072R�n�
M�n�
L�n�
n
61. �3
1 1x dx
4 8 12 16 20
1.2833 1.1865 1.1562 1.1414 1.1327
1.0898 1.0963 1.0976 1.0980 1.0982
0.9500 1.0199 1.0451 1.0581 1.0660R�n�
M�n�
L�n�
n
Section 5.3 Reimann Sums and Definite Integrals 427
64. �3
0x sin x dx
4 8 12 16 20
2.8186 2.9985 3.0434 3.0631 3.0740
3.1784 3.1277 3.1185 3.1152 3.1138
3.1361 3.1573 3.1493 3.1425 3.1375R�n�
M�n�
L�n�
n
62. �4
0 e x dx
4 8 12 16 20
31.1929 41.3106 45.1605 47.1772 48.4169
51.4284 53.0439 53.3508 53.4588 53.5089
84.7910 68.1097 63.0265 60.5768 59.1365R�n�
M�n�
L�n�
n
63. ���2
0sin2 x dx
4 8 12 16 20
0.5890 0.6872 0.7199 0.7363 0.7461
0.7854 0.7854 0.7854 0.7854 0.7854
0.9817 0.8836 0.8508 0.8345 0.8247R�n�
M�n�
L�n�
n
65. True 66. False
�1
0x�x dx � �1
0x dx�1
0
�x dx67. True
68. True 69. False
�2
0��x� dx � �2
70. True. The limits of integration arethe same.
71.
� �4��1� � �10��2� � �40��4� � �88��1� � 272
�4
i�1 f �ci� �x � f �1� �x1 � f �2� �x2 � f �5� �x3 � f �8� �x4
c4 � 8c3 � 5,c2 � 2,c1 � 1,
�x4 � 1�x3 � 4,�x2 � 2,�x1 � 1,
x4 � 8x3 � 7,x2 � 3,x1 � 1,x0 � 0,
�0, 8 f �x� � x2 � 3x,
428 Chapter 5 Integration
72.
� 12
�
4 � �32 �
12 � �32 2�
3 � ��1���� � �0.708
�4
i�1 f �ci� �xi � f �
6 �x1 � f �
3 �x2 � f 2�
3 �x3 � f 3�
2 �x4
c4 �3�
2c3 �
2�
3,c2 �
�
3,c1 �
�
6,
�x4 � ��x3 �2�
3,�x2 �
�
12,�x1 �
�
4,
x4 � 2�x3 � �,x2 ��
3,x1 �
�
4,x0 � 0,
�0, 2� f �x� � sin x,
73.
��b � a��a � b�
2�
b2 � a2
2
� �b � a��a �b � a
2 �
� a�b � a� ��b � a�2
2
� limn→�
�a�b � a� ��b � a�2
n n � 1
2 �
� limn→�
�b � an
�an� � b � an 2
n�n � 1�
2 �
� limn→�
�b � an �
n
i�1a � b � a
n 2 �
n
i�1i�
� limn→�
�n
i�1�a � ib � a
n �b � an
�b
a
x dx � lim���→0
�n
i�1 f �ci� �x
�x �b � a
n, ci � a � i��x� � a � ib � a
n
74.
�13
�b3 � a3�
� a2�b � a� � a�b � a�2 �13
�b � a�3
� limn→�
�a2�b � a� �a�b � a�2�n � 1�
n�
�b � a�3
6 �n � 1��2n � 1�
n2 �
� limn→�
b � an �na2 �
2a�b � a�n
n�n � 1�
2� b � a
n 2 n�n � 1��2n � 1�
6 �
� limn→�
�b � an �
n
i�1a2 �
2ai�b � a�n
� i2b � an 2�
� limn→�
�n
i�1�a � ib � a
n �2b � an
�b
a
x 2 dx � lim���→0
�n
i�1 f �ci� �x
�x �b � a
n, ci � a � i��x� � a � ib � a
n
Section 5.4 The Fundamental Theorem of Calculus 429
75.
is not integrable on the interval As or in each subinterval since there
are an infinite number of both rational and irrational numbers in any interval, no matter how small.
f �ci� � 0f �ci� � 1��� → 0,�0, 1�.
f �x� � �1,0,
x is rationalx is irrational
76.
The limit
does not exist. This does notcontradict Theorem 5.4 becausef is not continuous on �0, 1�.
lim���→0 �
n
i�1 f �ci� �xi
1 2
1
2
x
f(x) = 1x
y
f �x� � �0,1x
,
x � 0 0 < x ≤ 1
77. The function f is nonnegative between and
Hence,
is a maximum for and b � 1.a � �1
�b
a
�1 � x2� dx
−2 1 2
−1
−2
2
x
y
f(x) = 1 − x2
x � 1.x � �1 78. To find use a geometric approach.
Thus,
�2
0x dx � 1�2 � 1� � 1.
x1 2 3−1
−1
1
2
3
y
�20 x dx,
79. Let and The appropriate Riemann Sum is
� limn→�
2n2 � 3n � 1
6n2 � limn→�
�13
�1
2n�
16n2 �
13
limn→�
1n3�12 � 22 � 32 � . . . � n2� � lim
n→� 1n3 �
n�2n � 1��n � 1�6
�n
i�1f �ci � �xi � �
n
i�1� i
n 2 1
n�
1n3 �
n
i�1i 2.
�xi � 1�n.f �x� � x2, 0 ≤ x ≤ 1,
Section 5.4 The Fundamental Theorem of Calculus
1.
is positive.��
0
4x2 � 1
dx
−2
−5 5
5f �x� �
4x2 � 1
3.
�2
�2x�x2 � 1 dx � 0
−5
−5 5
5f �x� � x�x2 � 1
5. �1
02x dx � �x2�
1
0� 1 � 0 � 1
2.
��
0cos x dx � 0 0
−2
�
2f �x� � cos x
4.
is negative.�2
�2x�2 � x dx −2 2
−5
5f �x� � x�2 � x
6. �7
23 dv � �3v�
7
2� 3�7� � 3�2� � 15
430 Chapter 5 Integration
17. �1
�1�3�t � 2� dt � �3
4t 4�3 � 2t�
1
�1� �3
4� 2 � �3
4� 2 � �4
18. �8
1�2
x dx � �2�8
1x�1�2 dx � ��2�2�x1�2�
8
1� �2�2x�
8
1� 8 � 2�2
19. �1
0
x � �x3
dx �13�
1
0�x � x1�2� dx �
13�
x2
2�
23
x3�2�1
0�
13�
12
�23 � �
118
20. �2
0�2 � t��t dt � �2
0�2t1�2 � t3�2� dt � �4
3t3�2 �
25
t5�2�2
0� �t�t
15�20 � 6t��
2
0�
2�215
�20 � 12� �16�2
15
21. �0
�1�t1�3 � t2�3� dt � �3
4t 4�3 �
35
t5�3�0
�1� 0 � �3
4�
35 � �
2720
22.
�12�
35
x5�3 �38
x8�3��1
�8� �x5�3
80�24 � 15x��
�1
�8� �
180
�39� �3280
�144� �456980
��1
�8
x � x2
2 3�x dx �
12�
�1
�8�x2�3 � x5�3� dx
7. �0
�1�x � 2� dx � �x2
2� 2x�
0
�1� 0 � �1
2� 2 � �
52
9. �1
�1�t2 � 2� dt � �t3
3� 2t�
1
�1� �1
3� 2 � ��
13
� 2 � �103
8. �5
2��3v � 4� dv � ��
32
v2 � 4v�5
2� ��
752
� 20 � ��6 � 8� � �392
10. �3
1
�3x2 � 5x � 4� dx � �x3 �5x2
2� 4x�
3
1� �27 �
45
2� 12 � �1 �
5
2� 4 � 38
11. �1
0�2t � 1�2 dt � �1
0�4t2 � 4t � 1� dt � �4
3t3 � 2t2 � t�
1
0�
43
� 2 � 1 �13
13. �2
1� 3
x2 � 1 dx � ��3x
� x�2
1� ��
32
� 2 � ��3 � 1� �12
15. �4
1
u � 2
�u du � �4
1�u1�2 � 2u�1�2� du � �2
3u3�2 � 4u1�2�
4
1� �2
3��4�3
� 4�4� � �23
� 4� �23
12. �1
�1�t3 � 9t� dt � �1
4t4 �
92
t2�1
�1� �1
4�
92 � �1
4�
92 � 0
14. ��1
�2�u �
1u2 du � �u2
2�
1u�
�1
�2� �1
2� 1 � �2 �
12 � �2
16. �3
�3v1�3 dv � �3
4v 4�3�
3
�3�
34
��3�3 �4 � �3��3 �4� � 0
Section 5.4 The Fundamental Theorem of Calculus 431
23. split up the integral at the zero
� �3x � x2�3�2
0� �x2 � 3x�
3
3�2� �9
2�
94 � 0 � �9 � 9� � �9
4�
92 � 2�9
2�
94 �
92
x �32 ��3
0�2x � 3� dx � �3�2
0�3 � 2x� dx � �3
3�2�2x � 3� dx
24.
� �4 � 0� � ��35 �252 � �28 � 8�� �
132
� �x2
2� x�
4
2� �7x �
x2
2 �5
4
� �4
2�x � 1� dx � �5
4�7 � x� dx
�5
2�3 � �x � 4�� dx � �4
2�3 � �x � 4�� dx � �5
4�3 � �x � 4�� dx
25.
� �8 �83 � �9 � 12� � �8
3� 8 �
233
� �4x �x3
3 �2
0� �x3
3� 4x�
3
2
�3
0�x2 � 4� dx � �2
0�4 � x2� dx � �3
2�x2 � 4� dx
26.
�43
� 0 �43
�43
� 0 � 4
� �13
� 2 � 3 � �9 � 18 � 9� � �13
� 2 � 3 � �643
� 32 � 12 � �9 � 18 � 9�
� �x3
3� 2x2 � 3x�
1
0� �x3
3� 2x2 � 3x�
3
1� �x3
3� 2x2 � 3x�
4
3
�4
0�x2 � 4x � 3� dx � �1
0�x2 � 4x � 3� dx � �3
1�x2 � 4x � 3� dx � �4
3�x2 � 4x � 3� dx split up the integral at
the zeros x � 1, 3��
27. ��
0�1 � sin x� dx � �x � cos x�
�
0� �� � 1� � �0 � 1� � 2 � �
29. ���6
���6sec2 x dx � �tan x�
��6
���6�
�33
� ���33 �
2�33
31. �e
1 �2x �
1x dx � �x2 � ln x�
e
1� �e2 � 1� � �1 � 0� � e2 � 2
28. ���4
0
1 � sin2 �cos2 �
d� � ���4
0d� � ���
��4
0�
�
4
30. ���2
��4�2 � csc2 x� dx � �2x � cot x�
��2
��4� �� � 0� � ��
2� 1 �
�
2� 1 �
� � 22
32. �5
1 x � 1
x dx � �5
1 �1 �
1x dx � �x � ln x�
5
1� �5 � ln 5� � �1 � ln 1� � 4 � ln 5
33. ���3
���34 sec � tan � d� � �4 sec ��
��3
���3� 4�2� � 4�2� � 0 34.
� �� 2
4� 1 � �� 2
4� 1 � 2
���2
���2�2t � cos t� dt � �t 2 � sin t�
��2
���2
432 Chapter 5 Integration
47. Since on
A � �2
0�x3 � x� dx � �x4
4�
x2
2 �2
0� 4 � 2 � 6.
�0, 2�,y ≥ 0 48. Since on
A � �3
0�3x � x2� dx � �3
2x2 �
x3
3 �3
0�
92
.
�0, 3�,y ≥ 0
49. �e
1 4x dx � 4 ln x�
e
1� 4 ln e � 4 ln 1 � 4 50. �2
0 e x dx � e x�
2
0� e2 � e0 � e2 � 1
51.
c � 0.4380 or c � 1.7908
c � �1 ± �6 � 4�23 �2
�c � 1 � ±�6 � 4�23
��c � 1�2�
6 � 4�23
c � 2�c � 1 �3 � 4�2
3� 1
c � 2�c �3 � 4�2
3
f �c��2 � 0� �6 � 8�2
3
�2
0�x � 2�x � dx � �x2
2�
4x3�2
3 �2
0� 2 �
8�23
52.
c � 3�92
� 1.6510
c3 �92
9c3 � 2
f �c��3 � 1� � 4
�3
1
9x3 dx � ��
92x2�
3
1� �
12
�92
� 4
432 Chapter 5 Integration
35. �3
ln 2� 12�2
0 �2x � 6� dx � � 2x
ln 2� 6x�
2
0� � 4
ln 2� 12 � � 1
ln 2� 0
37. �1
�1 �e� � sin �� d� � �e� � cos ��
1
�1� �e � cos 1� � �e�1 � cos��1�� � e �
1e
36. �92
�124ln 5
� �72.546 �3
0 �t � 5t� dt � �t2
2�
5t
ln 5�3
0� �9
2�
125ln 5 � �0 �
1ln 5
38. �2e
e
�cos x �1x dx � �sin x � ln x�
2e
e� �sin�2e� � ln�2e�� � �sin�e� � ln�e�� � sin�2e� � sin�e� � ln 2
39. A � �1
0�x � x2� dx � �x2
2�
x3
3 �1
0�
16
40. A � �1
�1�1 � x4� dx � �x �
15
x5�1
�1�
85
41. A � �3
0�3 � x��x dx � �3
0�3x1�2 � x3�2� dx � �2x3�2 �
25
x5�2�3
0� �x�x
5�10 � 2x��
3
0�
12�35
42. A � �2
1
1x2 dx � ��
1x�
2
1� �
12
� 1 �12
44. A � ��
0�x � sin x� dx � �x2
2� cos x�
�
0�
� 2
2� 2 �
� 2 � 42
43. A � ���2
0cos x dx � �sin x�
��2
0� 1
45. Since on
A � �2
0�3x2 � 1� dx � �x3 � x�
2
0� 8 � 2 � 10.
�0, 2�,y ≥ 0 46. Since on
Area � �8
0�1 � x1�3� dx � �x �
34
x 4�3�8
0� 8 �
34
�16� � 20.
�0, 8�,y ≥ 0
Section 5.4 The Fundamental Theorem of Calculus 433
53.
� ±arccos ��
2� ±0.4817
c � ±arcsec� 2
��
sec c � ±2
��
sec2 c �4�
2 sec2 c �8�
f �c���
4� ��
�
4 � � 4
���4
���42 sec2 x dx � �2 tan x�
��4
���4� 2�1� � 2��1� � 4 54.
c � ±0.5971
cos c �3�32�
f �c���
3� ��
�
3 � � �3
���3
���3cos x dx � �sin x�
��3
���3� �3
55.
c �3
ln 4� 2.1640
3c
� ln 4
15 �3c
� 15 � ln 4
�5 �1c �3� � 15 � ln 4
f �c��4 � 1� � 15 � ln 4
� �20 � ln 4� � �5 � 0� � 15 � ln 4
�4
1
�5 �1x dx � �5x � ln x�
4
1
�1, 4�f �x� � 5 �1x, 56.
c � log2� 73 ln 2 � 1.7512
2c �7
3 ln 2
3�2c� �7
ln 2
�10 � 2c��3� � 30 �7
ln 2
f �c��3 � 0� � 30 �7
ln 2
� �30 �8
ln 2 � �0 �1
ln 2 � 30 �7
ln 2
�3
0 �10 � 2x� dx � �10x �
2x
ln 2�3
0
�0, 3�f �x� � 10 � 2x,
57.
Average value
when or x � ±2�3
3� ±1.155.x2 � 4 �
83
4 � x2 �83
�83
5
−1
−3 3
2 33 3
8,−( ( 2 33 3
8,( (12 � ��2��
2
�2�4 � x2� dx �
14�4x �
13
x3�2
�2�
14��8 �
83 � ��8 �
83 � �
83
58.
� 2�3 �13 �
163
1
3 � 1�3
1
4�x2 � 1�x2 dx � 2�3
1�1 � x�2� dx � 2�x �
1x�
3
1
434 Chapter 5 Integration
60.
0
1
40
x �3
ln 4� 2.1640
2x �6
ln 4
12x
�16
ln 4
�16
ln x�4
1�
16
ln 4 � 0.2310
Average value �1
4 � 1 �4
1 12x
dx
�1, 4�f �x� �12x
,
61.
x � 0.690, 2.451
sin x �2�
−1
−
2
20.690,
2.451,
(((
(π
π
2
�23�
2
Average value �2�
1� � 0�
�
0sin x dx � ��
1�
cos x��
0�
2�
62.
x � 0.881
cos x �2�
−0.5
0 2.71
0.881, 2π ((
1.5
Average value �2�
1���2� � 0
���2
0cos x dx � � 2
� sin x�
��2
0�
2�
63. The distance traveled is The area under the curve from is approximately�18 squares��30� � 540 ft.
0 ≤ t ≤ 8�8
0 v�t� dt. 64. The distance traveled is The area under the curve from is approximately�29 squares��5� � 145 ft.
0 ≤ t ≤ 5�5
0 v�t� dt.
65. If is continuous on and then �b
a
f �x� dx � F�b� � F�a�.F�x� � f �x� on �a, b�,�a, b�f
66. (a)
x1 2 3 4 5 6 7
1
2
3
4
A
A A A
1
2 3 4
y
� 8
�12
�3 � 1� �12
�1 � 2� �12
�2 � 1� � �3��1�
� A1 � A2 � A3 � A4
�7
1f �x� dx � Sum of the areas (b)
(c)
Average value
x1 2 3 4 5 6 7
1
2
3
4
5
6
7
y
�206
�103
A � 8 � �6��2� � 20
Average value ��7
1 f �x� dx
7 � 1�
86
�43
434 Chapter 5 Integration
59.
−1
5
1
−1
(0.16, 2.35)
x � ln�e � e�1
2 � 0.1614
e x �12
�e � e�1�
2e x � e � e�1
� e x�1
�1� e � e�1 � 2.3504
Average value �1
1 � ��1� �1
�1 2e x dx � �1
�1 e x dx
��1, 1�f �x� � 2e x,
Section 5.4 The Fundamental Theorem of Calculus 435
67. �2
0f �x� dx � ��area of region A� � �1.5 68.
� 3.5 � ��1.5� � 5.0
�6
2f �x� dx � �area of region B� � �6
0f �x� dx � �2
0f �x� dx
69. �6
0� f �x�� dx � ��2
0f �x� dx ��6
2f �x� dx � 1.5 � 5.0 � 6.5
71.
� 12 � 3.5 � 15.5
�6
0�2 � f �x�� dx � �6
02 dx � �6
0f �x� dx
73. (a)
F �x� � 500 sec2 x
F�0� � k � 500
F�x� � k sec2 x (b)
� 826.99 newtons � 827 newtons
�1500
���3 � 0�
1��3 � 0�
��3
0500 sec2 x dx �
1500� �tan x�
��3
0
70. �2
0�2f �x� dx � �2�2
0f �x� dx � �2��1.5� � 3.0
72. Average value �16�
6
0f �x� dx �
16
�3.5� � 0.5833
74.1
R � 0�R
0k�R2
� r2� dr �kR�R2r �
r3
3�R
0�
2kR2
3
75.1
5 � 0�5
0�0.1729t � 0.1522t2 � 0.0374t3� dt �
15�0.08645t2 � 0.05073t3 � 0.00935t4�
5
0� 0.5318 liter
76. (a)
The area above the -axis equals the area below the-axis. Thus, the average value is zero.x
x
24
−1
0
1 (b)
The average value of appears to be g.S
240
0
10
77. (a)
(b) (c) meters�60
0v�t� dt � ��8.61 10�4t4
4�
0.0782t3
3�
0.208t2
2� 0.0952t�
60
0� 2476
70
−10
−10
90
v � �8.61 10�4t3 � 0.0782t2 � 0.208t � 0.0952
78. (a) histogram
—CONTINUED—
t2
21
4
43
6
8
10
12
14
16
18
65 8 97
N (b)
� 4980 customers
�6 � 7 � 9 � 12 � 15 � 14 � 11 � 7 � 2�60 � �83�60
436 Chapter 5 Integration
78. —CONTINUED—
(c) Using a graphing utility, you obtain
(d)
(e)
The estimated number of customers is
(f) Between 3 P.M. and 7 P.M., the number of customers is approximately
Hence, per minute.3017�240 � 12.6
��7
3N�t� dt �60� � �50.28��60� � 3017.
�85.162��60� � 5110.
�9
0N�t� dt � 85.162
−2
−2 10
16
N�t� � �0.084175t3 � 0.63492t2 � 0.79052 � 4.10317.
79.
F�8� �642
� 5�8� � �8
F�5� �252
� 5�5� � �252
F�2� �42
� 5�2� � �8
F�x� � �x
0�t � 5� dt � �t2
2� 5t�
x
0�
x2
2� 5x
80.
F�8� �84
4� 64 � 16 � 4 � 1068
F�5� �6254
� 25 � 10 � 4 � 167.25
�Note: F�2� � �2
2�t3 � 2t � 2� dt � 0�F�2� � 4 � 4 � 4 � 4 � 0
�x4
4� x2 � 2x � 4
� �x4
4� x2 � 2x � �4 � 4 � 4�
F�x� � �x
2�t3 � 2t � 2� dt � �t4
4� t2 � 2t�
x
2
81.
F�8� � 10�78 �
354
F�5� � 10�45 � 8
F�2� � 10�12 � 5
� �10x
� 10 � 10�1 �1x
F�x� � �x
1
10v2 dv � �x
110v�2 dv �
�10v �
x
182.
F�8� �1
64�
14
� �1564
F�5� �1
25�
14
� �21100
� �0.21
F�2� �14
�14
� 0
F�x� � �x
2
�2t3
dt � ��x
22t�3 dt �
1t2�
x
2�
1x2 �
14
436 Chapter 5 Integration
Section 5.4 The Fundamental Theorem of Calculus 437
83.
F�8� � sin 8 � sin 1 � 0.1479
F�5� � sin 5 � sin 1 � �1.8004
F�2� � sin 2 � sin 1 � 0.0678
F�x� � �x
1cos � d� � sin ��
x
1� sin x � sin 1
84.
F�8� � 1 � cos 8 � 1.1455
F�5� � 1 � cos 5 � 0.7163
F�2� � 1 � cos 2 � 1.4161
F�x� � �x
0sin � d� � �cos ��
x
0� �cos x � cos 0 � 1 � cos x
85.
(a)
(b) g increasing on and decreasing on
(c) g is a maximum of 9 at
(d)
2 4 6 8
2
4
6
8
10
x
y
x � 4.
�4, 8��0, 4�
g�8� � �8
0f �t� dt � 8 � 3 � 5
g�6� � �6
0f �t� dt � 9 � ��1� � 8
g�4� � �4
0f �t� dt � 7 � 2 � 9
g�2� � �2
0f �t� dt � 4 � 2 � 1 � 7
g�0� � �0
0f �t� dt � 0
g�x� � �x
0f �t� dt 86.
(a)
(b) g decreasing on and increasing on
(c) g is a minimum of at
(d)
x
y
−2 2 4 8 10−2
−4
−6
−8
2
4
x � 4.�8
�4, 8��0, 4�
g�8� � �8
0f �t� dt � �2 � 6 � 4
g�6� � �6
0f �t� dt � �8 � 2 � 4 � �2
g�4� � �4
0f �t� dt � �
12�4��4� � �8
g�2� � �2
0f �t� dt � �
12�2��4� � �4
g�0� � �0
0f �t� dt � 0
g�x� � �x
0f �t�dt
87. (a)
(b)ddx�
12
x2 � 2x� � x � 2
�x
0�t � 2� dt � �t2
2� 2t�
x
0�
12
x2 � 2x
88. (a)
(b)ddx�
14
x4 �12
x2� � x3 � x � x�x2 � 1�
�x
0t�t2 � 1� dt � �x
0�t3 � t� dt � �1
4t4 �
12
t2�x
0�
14
x4 �12
x2 �x2
4�x2 � 2�
89. (a)
(b)ddx�
34
x 4�3 � 12� � x1�3 � 3�x
�x
8
3�t dt � �34
t 4�3�x
8�
34
�x4�3 � 16� �34
x4�3 � 12 90. (a)
(b)ddx�
23
x3�2 �163 � � x1�2 � �x
�x
4
�t dt � �23
t3�2�x
4�
23
x3�2 �163
�23
�x3�2 � 8�
438 Chapter 5 Integration
91. (a)
(b)ddx
�tan x � 1� � sec2 x
�x
��4sec2 t dt � �tan t�
x
��4� tan x � 1 92. (a)
(b)ddx
�sec x � 2� � sec x tan x
�x
��3sec t tan t dt � �sec t�
x
��3� sec x � 2
93. (a)
(b)ddx
�e x � e�1� � e x
F �x� � �x
�1 et dt � et�
x
�1� e x � e�1 94. (a)
(b)ddx
�ln x� �1x
F �x� � �x
1 1t dt � ln t�
x
1� ln x
95.
F�x� � x2 � 2x
F�x� � �x
�2�t 2 � 2t� dt 96.
F�x� �x2
x2 � 1
F�x� � �x
1
t2
t2 � 1 dt 97.
F�x� � �x4 � 1
F�x� � �x
�1
�t4 � 1 dt
98.
F�x� � 4�x
F�x� � �x
1
4�t dt 99.
F�x� � x cos x
F�x� � �x
0t cos t dt 100.
F�x� � sec3 x
F�x� � �x
0sec3 t dt
101.
Alternate Solution:
F�x� � ��4x � 1� � 4�x � 2� � 1 � 8
� ��x
0�4t � 1� dt � �x�2
0�4t � 1� dt
� �0
x
�4t � 1� dt � �x�2
0�4t � 1� dt
F�x� � �x�2
x
�4t � 1� dt
F�x� � 8
� 8x � 10
� �2�x � 2�2 � �x � 2�� � �2x2 � x�
� �2t2 � t�x�2
x
F�x� � �x�2
x
�4t � 1� dt 102.
Alternate Solution:
F�x� � ���x�3��1� � �x3� � 0
� ���x
0t3 dt � �x
0t3 dt
� �0
�x
t3 dt � �x
0t3 dt
F�x� � �x
�x
t3 dt
F�x� � 0
F�x� � �x
�x
t3 dt � �t4
4�x
�x� 0
103.
Alternate Solution:
F�x� � �sin x ddx
�sin x� � �sin x�cos x�
F�x� � �sin x
0
�t dt
F�x� � �sin x�1�2 cos x � cos x�sin x
F�x� � �sin x
0
�t dt � �23
t3�2�sin x
0�
23
�sin x�3�2
438 Chapter 5 Integration
Section 5.4 The Fundamental Theorem of Calculus 439
105.
F�x� � sin�x3�2 � 3x2 � 3x2 sin x 6
F�x� � �x3
0sin t 2 dt 106.
F�x� � sin�x2�2�2x� � 2x sin x 4
F�x� � �x2
0sin � 2 d�
107.
has a relative maximum at x � 2.g
g�0� � 0, g�1� �12
, g�2� � 1, g�3� �12
, g�4� � 0x
y
1
1 2 3 4
2
−2
−1
f g
g�x� � �x
0f �t� dt
108. (a)
Horizontal asymptote:
(b)
The graph of does not have a horizontal asymptote.A�x�
limx→�
A�x� � limx→�
�4x �4x
� 8 � � � 0 � 8 � �
�4x2 � 8x � 4
x�
4�x � 1�2
x
� �4t �4t�
x
1� 4x �
4x
� 8
A�x� � �x
1�4 �
4t2 dt
y � 4
limt→�
g�t� � 4
g�t� � 4 �4t2
109.
� 28 units
� 4 � 4 � 20
� 3�1
0�t 2 � 4t � 3� dt � 3�3
1�t 2 � 4t � 3� dt � 3�5
3�t 2 � 4t � 3� dt
� �5
03��t � 3��t � 1�� dtTotal distance � �5
0�x�t�� dt
� 3�t � 3��t � 1�
� 3�t 2 � 4t � 3�
x�t� � 3t 2 � 12t � 9
x�t� � t 3 � 6t 2 � 9t � 2
104.
Alternate Solution: F�x� � �x2��3�2x� � 2x�5
F�x� � �x2
2t�3 dt � � t�2
�2�x2
2� ��
12t2�
x2
2�
�12x4 �
18
⇒ F�x� � 2x�5
440 Chapter 5 Integration
112. P �2��
��2
0sin � d� � ��
2�
cos ����2
0� �
2�
�0 � 1� �2�
� 63.7%
113. True 114. True
115. The function is not continuous on
Each of these integrals is infinite. has a nonremovable discontinuity at x � 0.
f �x� � x�2
�1
�1x�2 dx ��0
�1x�2 dx ��1
0x�2 dx
��1, 1�.f �x� � x�2 116. Let be an antiderivative of Then,
. � f �v�x��v�x� � f �u�x��u�x�
� F�v�x��v�x� � F�u�x��u�x�
ddx ��
v�x�
u�x�f �t� dt� �
ddx
�F�v�x�� � F�u�x��
�v�x�
u�x�f �t� dt � �F�t��
v�x�
u�x�� F�v�x�� � F�u�x��
f �t�.F�t�
117.
By the Second Fundamental Theorem of Calculus, we have
Since must be constant.f �x�f�x� � 0,
� �1
1 � x2 �1
x2 � 1� 0.
f�x� �1
�1�x�2 � 1��1x2 �
1x2 � 1
f �x� � �1�x
0
1t2 � 1
dt � �x
0
1t2 � 1
dt
118.
(a)
(c)
(d) G��0� � 0 � f �0� � �0
0f �t� dt � 0
G��x� � x � f �x� � �x
0f �t� dt
G�0� � �0
0�s�s
0f �t� dt� ds � 0
G�x� � �x
0�s�s
0f �t� dt� ds
(b) Let
G�0� � 0�0
0f �t� dt � 0
G�x� � F�x� � x�x
0f �t� dt
G�x� � �x
0F�s� ds
F�s� � s�s
0f �t� dt.
440 Chapter 5 Integration
110.
Using a graphing utility,
Total distance � �5
0�x�t�� dt � 27.37 units.
x�t� � 3t2 � 14t � 15
x�t� � �t � 1��t � 3�2 � t 3 � 7t 2 � 15t � 9 111.
� 2�2 � 1� � 2 units
� 2t1�2�4
1
� �4
1
1
�t dt
� �4
1�v�t�� dt
Total distance � �4
1�x�t�� dt
Section 5.5 Integration by Substitution
Section 5.5 Integration by Substitution 441
1. 10x dx
2.
3. 2x dx
4. 2x 2 dx
5. tan x
6. sin x cos x dx�cos xsin3 x
dx
sec2 x dx�tan2 x sec2 x dx
�sec 2x tan 2x dx
x2 � 1� x
�x2 � 1 dx
3x2 dxx3 � 1�x2�x3 � 1 dx
5x2 � 1��5x2 � 1�2�10x� dx
du � g��x� dxu � g�x��f �g�x��g��x� dx
7.
Check:ddx�
�1 � 2x�5
5� C� � 2�1 � 2x�4
��1 � 2x�4�2� dx ��1 � 2x�5
5� C 8.
Check:
� �x2 � 9�3�2x�
ddx�
�x2 � 9�4
4� C� �
4�x2 � 9�3
4�2x�
��x2 � 9�3�2x� dx ��x2 � 9�4
4� C
9.
Check:ddx�
23
�9 � x2�3�2 � C� �23
�32
�9 � x2�1�2��2x� � �9 � x2��2x�
��9 � x2�1�2��2x� dx ��9 � x2�3�2
3�2� C �
23
�9 � x2�3�2 � C
10.
Check:ddx�
34
�1 � 2x2�4�3 � C� �34
�43
�1 � 2x2�1�3��4x� � �1 � 2x2�1�3��4x�
��1 � 2x2�1�3��4x� dx �3
4�1 � 2x2�4�3 � C
11.
Check:ddx�
�x4 � 3�3
12� C� �
3�x4 � 3�2
12�4x3� � �x4 � 3�2�x3�
�x3�x4 � 3�2 dx �1
4��x4 � 3�2�4x3� dx �1
4 �x4 � 3�3
3� C �
�x4 � 3�3
12� C
12.
Check:ddx�
�x3 � 5�5
15� C� �
5�x3 � 5�4�3x2�15
� �x3 � 5�4x2
�x2�x3 � 5�4 dx �13��x3 � 5�4�3x2� dx �
13
�x3 � 5�5
5� C �
�x3 � 5�5
15� C
13.
Check:ddx�
�x3 � 1�5
15� C� �
5�x3 � 1�4�3x2�15
� x2�x3 � 1�4
�x2�x3 � 1�4 dx �13��x3 � 1�4�3x2� dx �
13�
�x3 � 1�5
5 � � C ��x3 � 1�5
15� C
442 Chapter 5 Integration
14.
Check:ddx�
�4x2 � 3�3
24� C� �
3�4x2 � 3�2�8x�24
� x�4x2 � 3�2
�x�4x2 � 3�2 dx �18��4x2 � 3�2�8x� dx �
18�
�4x2 � 3�3
3 � � C ��4x2 � 3�3
24� C
15.
Check:ddt�
�t2 � 2�3�2
3� C� �
3�2�t2 � 2�1�2�2t�3
� �t2 � 2�1�2t
�t�t2 � 2 dt �12��t2 � 2�1�2�2t� dt �
12
�t2 � 2�3�2
3�2� C �
�t2 � 2�3�2
3� C
16.
Check:ddt�
16
�t4 � 5�3�2 � C� �16
�32
�t4 � 5�1�2�4t3� � �t4 � 5�1�2�t3�
�t 3�t4 � 5 dt �14��t 4 � 5�1�2�4t3� dt �
14
�t 4 � 5 �3�2
3�2� C �
16
�t 4 � 5�3�2 � C
17.
Check:ddx��
158
�1 � x2�4�3 � C� � �158
�43
�1 � x2�1�3��2x� � 5x�1 � x2�1�3 � 5x 3�1 � x2
�5x�1 � x2�1�3 dx � �52��1 � x2�1�3��2x� dx � �
52
��1 � x2�4�3
4�3� C � �
158
�1 � x2�4�3 � C
18.
Check:ddu�
2�u3 � 5�3�2
9� C� �
29
� 32
�u3 � 5�1�2�3u2� � �u3 � 5�1�2�u2�
�u2�u3 � 5 du �13��u3 � 5�1�2�3u2� du �
13
�u3 � 5�3�2
3�2� C �
2�u3 � 5�3�2
9� C
19.
Check:ddx�
14�1 � x2�2 � C� �
14
��2��1 � x2��3��2x� �x
�1 � x2�3
� x�1 � x2�3 dx � �
12��1 � x2��3��2x� dx � �
12
�1 � x2��2
�2� C �
14�1 � x2�2 � C
20.
Check:ddx�
�14�1 � x 4� � C� �
14
�1 � x 4��2�4x3� �x3
�1 � x 4�2
� x3
�1 � x4�2 dx �14��1 � x4��2�4x3� dx � �
14
�1 � x4��1 � C ��1
4�1 � x4� � C
21.
Check:ddx��
13�1 � x3� � C� � �
13
��1��1 � x3��2�3x2� �x2
�1 � x3�2
� x2
�1 � x3�2 dx �13��1 � x3��2�3x2� dx �
13�
�1 � x3��1
�1 � � C � �1
3�1 � x3� � C
22.
Check:ddx�
13�9 � x3� � C� �
13
��1��9 � x3��2��3x2� �x2
�9 � x3�2
� x2
�9 � x3�2 dx � �13��9 � x3��2��3x2� dx � �
13�
�9 � x3��1
�1 � � C �1
3�9 � x3� � C
Section 5.5 Integration by Substitution 443
23.
Check:d
dx���1 � x2�1�2 � C � �
1
2�1 � x2��1�2��2x� �
x�1 � x2
� x�1 � x2
dx � �1
2��1 � x2��1�2��2x� dx � �
1
2 �1 � x2�1�2
1�2� C � ��1 � x2 � C
24.
Check:d
dx��1 � x4
2� C� �
1
2�
1
2�1 � x4��1�2�4x3� �
x3
�1 � x4
� x3
�1 � x4 dx �
1
4��1 � x4��1�2�4x3� dx �
1
4 �1 � x4�1�2
1�2� C �
�1 � x4
2� C
25.
Check:ddt��
�1 � �1�t�4
4� C� � �
14
�4�1 �1t �
3
�1t 2� �
1t 21 �
1t �
3
�1 �1t �
3
1t 2� dt � ��1 �
1t �
3
�1t 2� dt � �
�1 � �1�t�4
4� C
26.
Check:ddx�
13
x3 �19
x�1 � C� � x2 �19
x�2 � x2 �1
�3x�2
��x2 �1
�3x�2� dx � �x2 �19
x�2� dx �x3
3�
19
x�1
�1� � C �x3
3�
19x
� C �3x4 � 1
9x� C
27.
Check:ddx
��2x � C �12
�2x��1�2�2� �1
�2x
� 1
�2x dx �
12��2x��1�22 dx �
12�
�2x�1�2
1�2 � � C � �2x � C
28.
Check:ddx
��x � C �1
2�x
� 1
2�x dx �
12�x�1�2 dx �
12
x1�2
1�2� � C � �x � C
29.
Check:ddx�
25
x5�2 � 2x3�2 � 14x1�2 � C� �x2 � 3x � 7
�x
�x2 � 3x � 7�x
dx � ��x3�2 � 3x1�2 � 7x�1�2� dx �25
x5�2 � 2x3�2 � 14x1�2 � C �25�x�x2 � 5x � 35� � C
30.
Check:ddt�
23
t3�2 �45
t5�2 � C� � t1�2 � 2t3�2 �t � 2t2
�t
�t � 2t2
�t dt � ��t1�2 � 2t3�2� dt �
23
t3�2 �45
t5�2 � C �2
15t3�2�5 � 6t� � C
31.
Check:ddt�
14
t 4 � t2 � C� � t3 � 2t � t2t �2t �
�t 2t �2t � dt � ��t 3 � 2t� dt �
14
t 4 � t 2 � C
444 Chapter 5 Integration
32.
Check:ddt�
112
t4 �14t
� C� �13
t3 �1
4t2
�t3
3�
14t2� dt � �1
3t3 �
14
t�2� dt �13
t4
4� �14
t�1
�1� � C �1
12t4 �
14t
� C
33.
Check:ddy�
25
y3�2�15 � y� � C� �ddy�6y3�2 �
25
y5�2 � C� � 9y1�2 � y3�2 � �9 � y��y
��9 � y��y dy � ��9y1�2 � y3�2� dy � 923
y3�2� �25
y5�2 � C �25
y3�2�15 � y� � C
34.
Check:ddy�
4�y2
7�14 � y3�2� � C� �
ddy�2�4y2 �
27
y7�2� � C� � 16�y � 2�y5�2 � �2�y��8 � y3�2�
�2�y�8 � y3�2� dy � 2���8y � y5�2� dy � 2�4y2 �27
y7�2� � C �4�y2
7�14 � y3�2� � C
35.
� 2x2 � 4�16 � x2 � C
� 4x2
2 � � 2��16 � x2�1�2
1�2 � � C
� 4�x dx � 2��16 � x2��1�2��2x� dx
y � ��4x �4x
�16 � x2� dx 36.
�203�1 � x3 � C
�103 ��1 � x3�1�2
1�2 � � C
�103 ��1 � x3��1�2�3x2� dx
y � � 10x2
�1 � x3 dx
37.
� �1
2�x2 � 2x � 3� � C
�12�
�x2 � 2x � 3��1
�1 � � C
�12��x2 � 2x � 3��2�2x � 2� dx
y � � x � 1�x2 � 2x � 3�2 dx 38.
� �x2 � 8x � 1 � C
�12�
�x2 � 8x � 1�1�2
1�2 � � C
�12��x2 � 8x � 1��1�2�2x � 8� dx
y � � x � 4�x2 � 8x � 1
dx
39. (a)
x
y
−2 2
−1
3
(b)
−2 2
−1
2
y � �13
�4 � x2�3�2�2
2 � �13
�4 � 22�3�2 � C ⇒ C � 2�2, 2�:
� �12
�23
�4 � x2�3�2 � C � �13
�4 � x2�3�2 � C
y � �x�4 � x2 dx � �12��4 � x2�1�2��2x dx�
dydx
� x�4 � x2, �2, 2�
Section 5.5 Integration by Substitution 445
(b)
y �19
�x3 � 1�3
0 � C
�13
�x3 � 1�3
3� C �
19
�x3 � 1�3 � C
y � �x2�x3 � 1�2 dx �13��x3 � 1�2�3x2 dx�, �u � x3 � 1�
dydx
� x2�x3 � 1�2, �1, 0�40. (a)
x
y
2
−2
−2 2
41. (a)
(b)
5
6
−3
−6
y �12
sin�x2� � 1
1 �12
sin�0� � C ⇒ C � 1�0, 1�:
�12
sin�x2� � C
y � �x cos x2 dx �12�cos�x2�2x dx
dydx
� x cos x2, �0, 1�
−4 4
−4
4
x
y 42. (a)
(b)
y � �sec�2x�
�1 � �sec�0� � C ⇒ C � 0
� �sec�2x� � C
�u � 2x� y � ��2 sec�2x� tan�2x� dx,
dydx
� �2 sec�2x� tan�2x�, �0, �1�
x
y
3
−3
3−3
43. (a)
(b)
−4 8
−2
6y � �4e�x�2 � 5
�0, 1�: 1 � �4e0 � C � �4 � C ⇒ C � 5
� �4e�x�2 � C
y � �2e�x�2 dx � �4�e�x�2�12
dx�
dydx
� 2e�x�2, �0, 1�
x−2
−2
5
5
y
(0, 1)
44. (a)
(b)
y � �2.5e�0.2x2� 1
0, �32�: �
32
� �2.5e0 � C � �2.5 � C ⇒ C � 1
� �1
0.4e�0.2x2
� C � �2.5e�0.2x2� C
y � �xe�0.2x2 dx �
1�0.4
�e�0.2x2 ��0.4x�dx
dydx
� xe�0.2x2, 0, �
32�
x
y
−4
−4
4
4
446 Chapter 5 Integration
45. (a)
(b)
−6
−4
6
4
y � 3ex�3 �52
0, 12�:
12
� 3e0�3 � C ⇒ C � �52
y � �ex�3 dx � 3ex�3 � C
dydx
� ex�3, 0, 12�
−5
−5
5
5
y
x
46. (a)
(b)
y � esin x � 1
��, 2�: 2 � esin � � C � 1 � C ⇒ C � 1
y � �esin x cos x dx � esin x � C
dydx
� esin x cos x, ��, 2�
x
2
6
4
−2
y
10
47. �� sin �x dx � �cos �x � C
49. �sin 2x dx �12��sin 2x��2� dx � �
12
cos 2x � C
51. � 1� 2 cos
1�
d� � ��cos 1��
1� 2� d� � �sin
1�
� C
53. Let
� e5x�5� dx � e5x � C
du � 5 dx.u � 5x, 55.
��13
e�x3� C
� x2e�x3 dx � �
13
� e�x3��3x2� dx
48. �4x3 sin x4 dx � �sin x4�4x3� dx � �cos x4 � C
50. �cos 6x dx �16��cos 6x��6� dx �
16
sin 6x � C
52. �x sin x2 dx �12��sin x2��2x� dx � �
12
cos x2 � C
54. �e�x3��3x2� dx � e�x3�C
56. �12
ex2�2x � C � �x � 1�ex2�2x dx �12
� ex2�2x �2x � 2� dx
57. OR
OR
�sin 2x cos 2x dx �12
�2 sin 2x cos 2x dx �12
�sin 4x dx � �18
cos 4x � C2
�sin 2x cos 2x dx � �12��cos 2x���2 sin 2x� dx � �
12
�cos 2x�2
2� C1 � �
14
cos2 2x � C1
�sin 2x cos 2x dx �12��sin 2x��2 cos 2x� dx �
12
�sin 2x�2
2� C �
14
sin2 2x � C
58. �sec�2 � x� tan�2 � x� dx � ���sec�2 � x� tan�2 � x���1� dx � �sec�2 � x� � C
59. �tan4 x sec2 x dx �tan5 x
5� C �
15
tan5 x � C 60. ��tan x sec2 x dx ��tan x�3�2
3�2� C �
23
�tan x�3�2 � C
Section 5.5 Integration by Substitution 447
61.
� ��cot x��2
�2� C �
12 cot2 x
� C �12
tan2 x � C �12
�sec2 x � 1� � C �12
sec2 x � C1
�csc2 xcot3 x
dx � ���cot x��3��csc2 x� dx
63. �cot2 x dx � ��csc2 x � 1� dx � �cot x � x � C
65. � e x�e x � 1�2 dx ��e x � 1�3
3� C
67. Let
� �23
�1 � ex�3�2 � C
�ex�1 � ex dx � ���1 � ex�1�2��ex� dx
u � 1 � ex, du � �ex dx.
69.
� �52
e�2x � e�x � C
�5 � ex
e2x dx � �5e�2x dx � �e�x dx
62. � sin xcos3 x
dx � ���cos x��3��sin x� dx � ��cos x��2
�2� C �
12 cos2 x
� C �12
sec2 x � C
64. �csc2x2� dx � 2�csc2x
2�12� dx � �2 cotx
2� � C
66.
���1 � 3e x�2
6
� e x�1 � 3e x� dx ��13
� �1 � 3e x���3e x� dx
68. Let
��2
ex � e�x � C
�2ex � 2e�x
�ex � e�x�2 dx � 2��ex � e�x��2�ex � e�x� dx
u � ex � e�x, du � �ex � e�x� dx.
70.
� ex � 2x � e�x � C
� e2x � 2ex � 1
ex dx � � �ex � 2 � e�x� dx
71.
�1�
esin �x � C
�e sin �x cos �x dx �1��e sin �x�� cos �x� dx
73.
� �tan�e�x� � C
�e�x sec2�e�x� dx � ��sec2�e�x���e�x� dx
75.
�2
ln 33x�2 � C
� 3x�2 dx � 2� 3x�212� dx � 2
3x�2
ln 3� C
77.
��1
2 ln 5�5�x2 � � C
� �12�
5�x2
ln 5� C
�x 5�x2 dx � �
12�5�x2��2x� dx
72.
�12
etan 2x � C
� etan 2x sec2 2x dx �12�etan 2x�2 sec2 2x� dx
74.
� x2 � x � C
�ln�e2x�1� dx � � �2x � 1� dx
76. �4�x dx � �4�x
ln 4� C
78.
� �1
2 ln 7�7�3�x�2 � C
��3 � x�7�3�x�2 dx � �
12��2�3 � x�7�3�x�2
dx
448 Chapter 5 Integration
79.
f �x� � �13
�4 � x2�3�2 � 2
f �2� � 2 � �13
�0� � C ⇒ C � 2
� �12
�4 � x2�3�223� � C � �
13
�4 � x2�3�2 � C
f �x� � � x�4 � x2 dx � �12� �4 � x2�1�2 ��2x� dx
81.
Since Thus,
f �x� � 2 sin x2
� 3.
C � 3.f �0� � 3 � 2 sin 0 � C,
f �x� � �cos x2
dx � 2 sin x2
� C
83.
f �x� � �8e�x�4 � 9
f �0� � 1 � �8 � C ⇒ C � 9
� �8e�x�4 � C
f �x� � � 2e�x�4 dx � �8� e�x�4�14� dx
80.
y �3
ln 0.40.4x�3 �
12
�3
ln 0.4
f �0� �3
ln 0.4� C �
12 ⇒ C �
12
�3
ln 0.4
�3
ln 0.4 0.4x�3 � C
y � �0.4x�3 dx � 3�0.4x�313�dx
dydx
� 0.4x�3, 0, 12�
82.
Since Thus
f �x� � sec �x � 1.
C � �1.f �1�3� � 1 � sec���3� � C,
f �x� � �� sec �x tan �x dx � sec �x � C
84.
f �x� � �53
e�0.2x3�
196
0, 32� :
32
��53
� C ⇒ C �196
��53
e�0.2x3� C
�1
�0.6 �e�0.2x3��0.6x2� dx
f �x� � �x2e�0.2x3 dx
85.
f �x� �12
�ex � e�x�
f �0� � 1 � C2 � 1 ⇒ C2 � 0
f �x� � �12
�ex � e�x� dx �12
�ex � e�x� � C2
f��0� � C1 � 0
f��x� � � 12
�ex � e�x� dx �12
�ex � e�x� � C1 86.
f �x� � x � sin x �14
e2x
f �0� �14
� C2 �14
⇒ C2 � 0
� �sin x �14
e2x � x � C2
f �x� � � �cos x �12
e2x � 1� dx
f��x� � �cos x �12
e2x � 1
f��0� � �1 �12
� C1 �12
⇒ C1 � 1
f��x� � ��sin x � e2x� dx � �cos x �12
e2x � C1
Section 5.5 Integration by Substitution 449
87.
�2
15�x � 2�3�2�3x � 4� � C
�2
15�x � 2�3�2�3�x � 2� � 10 � C
�2u3�2
15�3u � 10� � C
�25
u5�2 �43
u3�2 � C
� ��u3�2 � 2u1�2� du
�x�x � 2 dx � ��u � 2��u du
dx � dux � u � 2,u � x � 2, 88.
�15
�2x � 3�3�2�x � 1� � C
�1
20�2x � 3�3�2�4x � 4� � C
�1
20�2x � 3�3�2�2�2x � 3� � 10 � C
�u3�2
20�2u � 10 � C
�14�
25
u5�2 � 2u3�2� � C
�14
��u3�2 � 3u1�2� du
�x�2x � 3 dx � �u � 32
�u 12
du
dx �12
dux �u � 3
2,u � 2x � 3,
89.
� �2
105�1 � x�3�2�15x2 � 12x � 8� � C
� �2
105�1 � x�3�2�35 � 42�1 � x� � 15�1 � x�2 � C
� �2u3�2
105�35 � 42u � 15u2� � C
� �23
u3�2 �45
u5�2 �27
u7�2� � C
� ���u1�2 � 2u3�2 � u5�2� du
�x2�1 � x dx � ���1 � u�2�u du
dx � �dux � 1 � u,u � 1 � x,
90.
� �25
�2 � x�3�2�x � 3� � C
� �25
�2 � x�3�2�5 � �2 � x� � C
� �2u3�2
5�5 � u� � C
� �2u3�2 �25
u5�2� � C
� ���3u1�2 � u3�2� du
��x � 1��2 � x dx � ���3 � u��u du
dx � �dux � 2 � u,u � 2 � x,
450 Chapter 5 Integration
91.
�1
15�2x � 1�3x2 � 2x � 13� � C
�1
60�2x � 1�12x2 � 8x � 52� � C
��2x � 1
60�3�2x � 1�2 � 10�2x � 1� � 45 � C
�u1�2
60�3u2 � 10u � 45� � C
�18
25
u5�2 �43
u3�2 � 6u1�2� � C
�18��u3�2 � 2u1�2 � 3u�1�2� du
�18�u�1�2��u2 � 2u � 1� � 4 du
� x2 � 1
�2x � 1 dx � ���1�2��u � 1�2 � 1
�u 12
du
dx �12
dux �12
�u � 1�,u � 2x � 1,
92. Let
�23�x � 4�2x � 13� � C
�23�x � 4�2�x � 4� � 21 � C
�23
u1�2�2u � 21� � C
�43
u3�2 � 14u1�2 � C
� ��2u1�2 � 7u�1�2� du
� 2x � 1�x � 4
dx � �2�u � 4� � 1�u
du
u � x � 4, x � u � 4, du � dx. 93.
where C1 � �1 � C.
� ��x � 2�x � 1� � C1
� �x � 2�x � 1 � 1 � C
� ��x � 1� � 2�x � 1 � C
� �u � 2�u � C
� ��u � 2u1�2� � C
� ���1 � u�1�2� du
� ����u � 1���u � 1��u��u � 1� du
� �x
�x � 1� � �x � 1 dx � ���u � 1�
u � �u du
dx � dux � u � 1,u � x � 1,
94.
�37
�t � 4�4�3�t � 3� � C
�37
�t � 4�4�3��t � 4� � 7 � C
�3u4�3
7�u � 7� � C
�37
u7�3 � 3u4�3 � C
� ��u4�3 � 4u1�3� du
�t 3�t � 4 dt � ��u � 4�u1�3 du
dt � dut � u � 4,u � t � 4, 95. Let
� �18
�x2 � 1�4�1
�1� 0
�1
�1x�x2 � 1�3 dx �
12�
1
�1�x2 � 1�3�2x� dx
du � 2x dx.u � x2 � 1,
Section 5.5 Integration by Substitution 451
96. Let
�19
��64 � 8�3 � ��8 � 8�3 � 41,472
�4
�2x2�x3 � 8�2 dx �
13�
4
�2�x3 � 8�2�3x2� dx � �1
3 �x3 � 8�3
3 �4
�2
u � x3 � 8, du � 3x2 dx.
97. Let
�49
�27 � 2�2 � 12 �89�2�
49��x3 � 1�3�2�
2
1� �2
3 �x3 � 1�3�2
3�2 �2
1 �2
12x2�x3 � 1 dx � 2 �
13�
2
1�x3 � 1�1�2�3x2� dx
u � x3 � 1, du � 3x2 dx.
98. Let
�2
0x�4 � x2 dx � �
12�
2
0�4 � x2�1�2��2x� dx � ��
13
�4 � x2�3�2�2
0� 0 �
83
�83
du � �2x dx.u � 4 � x2,
99. Let
�4
0
1
�2x � 1 dx �
12�
4
0�2x � 1��1�2�2� dx � ��2x � 1�
4
0� �9 � �1 � 2
du � 2 dx.u � 2x � 1,
100. Let
�2
0
x
�1 � 2x2 dx �
14�
2
0�1 � 2x2��1�2�4x� dx � �1
2�1 � 2x2�
2
0�
32
�12
� 1
du � 4x dx.u � 1 � 2x2,
101. �1
0 e�2x dx � �
12�
1
0 e�2x��2� dx � �
12
e�2x�1
0� �
12
e�2 �12
102. �2
1 e1�x dx � ��2
1 e1�x��1� dx � �e1�x�
2
1� �e�1 � 1
103. � �13
�e � e3� �e3
�e2 � 1�� ��13
e3�x�3
1 �3
1 e3�x
x2 dx � �13
�3
1 e3�x�
3x2� dx
104. Let
��2
0 xe�x2�2 dx � ���2
0 e�x2�2��x� dx � ��e�x2�2�
�2
0 � 1 � e�1 �e � 1
e
u ��x2
2, du � �x dx.
105. Let
�9
1
1
�x�1 � �x �2 dx � 2�9
1�1 � �x ��2 1
2�x� dx � ��2
1 � �x�9
1� �
12
� 1 �12
du �1
2�x dx.u � 1 � �x,
106. Let
�2
0x 3�4 � x2 dx �
12�
2
0�4 � x2�1�3�2x� dx � �3
8�4 � x2�4�3�
2
0�
38
�84�3 � 44�3� � 6 �32
3�4 � 3.619
du � 2x dx.u � 4 � x2,
452 Chapter 5 Integration
107.
When When
� �0
1�u3�2 � u1�2� du � �2
5u5�2 �
23
u3�2�0
1� ��2
5�
23� �
415�2
1�x � 1��2 � x dx � �0
1���2 � u� � 1�u du
u � 0.x � 2,u � 1.x � 1,
dx � �dux � 2 � u,u � 2 � x,
108. Let
When When
�163
�14�
23
�27� � 2�3�� � 23
� 2��
�14�
23
u3�2 � 2u1�2�9
1 �5
1
x�2x � 1
dx � �9
1
1�2�u � 1��u
1
2 du �
1
4�9
1�u1�2 � u�1�2� du
x � 5, u � 9.x � 1, u � 1.
u � 2x � 1, du � 2 dx, x �12
�u � 1�.
109. ���2
0cos2
3x� dx � �3
2 sin2
3x��
��2
0�
32�32 � �
3�34
111. �7
2 ln 2�
7ln 4
�1
ln 2 �4 �
12��2
�12x dx � � 2x
ln 2�2
�1
110. ���2
��3�x � cos x� dx � �x2
2� sin x�
��2
��3� �2
8� 1� � �2
18�
�32 � �
5�2
72�
2 � �32
112.
� �2x�0
�2� 4 � �0
�22 dx
�0
�2�33
� 52� dx � �0
�2�27 � 25� dx
113.
y � �2x3 � 1�3 � 3
4 � 13 � C ⇒ C � 3
y � 3�2x3 � 1�3
3� C � �2x3 � 1�3 � C
y � 3��2x3 � 1�2�6x2� dx, �u � 2x3 � 1�
dydx
� 18x2�2x3 � 1�2, �0, 4� 114.
y �8
�3x � 5�2 � 1
3 �8
�3��1� � 5�2 � C �84
� C ⇒ C � 1
�8
�3x � 5�2 � C
��16�3x � 5��2
�2� C
� ��48� 13��3x � 5��3 3 dx
y � �48��3x � 5��3 dx
dydx
��48
�3x � 5�3, ��1, 3�
115.
y � �2x2 � 1 � 3
4 � �49 � C � 7 � C ⇒ C � �3
y �12
�2x2 � 1�1� 2
1�2� C � �2x2 � 1 � C
y �12��2x2 � 1��1� 2�4x dx�, �u � 2x2 � 1�
dydx
�2x
�2x2 � 1, �5, 4�
Section 5.5 Integration by Substitution 453
116.
y � 2x2 �2
�3x3 � 1� 4
2 � 0 �21
� C ⇒ C � 4
� 2x2 �2
�3x3 � 1� C
� 2x 2 ��3x3 � 1��1� 2
��1�2� � C
y � ��4x � �3x3 � 1��3� 2 9x2� dx
dydx
� 4x �9x2
�3x3 � 1�3�2, �0, 2� 117.
When When
�120928
� 3847
� 12� � 37
�34�
� �37
u7�3 �34
u4�3�8
1
� �8
1�u4�3 � u1�3� du
Area � �7
0x 3�x � 1 dx � �8
1�u � 1� 3�u du
u � 8.x � 7,u � 1.x � 0,
dx � dux � u � 1,u � x � 1,
119. A � ��
0�2 sin x � sin 2x� dx � ��2 cos x �
12
cos 2x��
0� 4
118.
When When
� �8
0�u7�3 � 4u4�3 � 4u1�3� du � � 3
10u10�3 �
127
u7�3 � 3u4�3�8
0�
475235
Area � �6
�2x2 3�x � 2 dx � �8
0�u � 2�2 3�u du
u � 8.x � 6,u � 0.x � �2,
dx � dux � u � 2,u � x � 2,
120. A � ��
0�sin x � cos 2x� dx � ��cos x �
12
sin 2x��
0� 2
121. Area � �2��3
��2sec2x
2� dx � 2�2��3
��2sec2x
2�12� dx � �2 tanx
2��2��3
��2� 2��3 � 1�
122. Let
Area � ���4
��12csc 2x cot 2x dx �
12�
��4
��12csc 2x cot 2x�2� dx � ��
12
csc 2x���4
��12�
12
du � 2 dx.u � 2x,
123.
0 60
150
�5
0 ex dx � �ex�
5
0� e5 � 1 � 147.413
125.
−4.5
−3
4.5
3
� �2e�3�2 � 2 � 1.554
��6
0xe�x2�4 dx � ��2e�x2�4��6
0
124.
a b
�b
a
e�x dx � ��e�x�b
a� e�a � e�b
126.
−20
4
4
� �12
e�4 � 4 �12
� 4.491
�2
0 �e�2x � 2� dx � ��
12
e�2x � 2x�2
0
454 Chapter 5 Integration
127.
−1
−1 5
3
�4
0
x
�2x � 1 dx � 3.333 �
103
128.
00
2
20
�2
0x3�x � 2 dx � 7.581 129.
00 8
15
�7
3x�x � 3 dx � 28.8 �
1445
130.
1 50
50
�5
1x2�x � 1 dx � 67.505 131.
−1
−1 4
5
�3
0� � cos
�
6� d� � 7.377 132.
0
−1
2
�2
���2
0sin 2x dx � 1.0
133.
−1 2
−1
2
� �e�1 � 1 � 0.632
��2
0 xe�x2�2 dx � ��e��x2�2��
�2
0134.
−20
4
4
� �12
e�4 � 4 �12
� 4.491
�2
0 �e�2x � 2� dx � ��
12
e�2x � 2x�2
0
135.
They differ by a constant: C2 � C1 �16
��2x � 1�2 dx � ��4x2 � 4x � 1� dx �43
x3 � 2x2 � x � C2
��2x � 1�2 dx �12��2x � 1�22 dx �
16
�2x � 1�3 � C1 �43
x3 � 2x2 � x �16
� C1
136.
They differ by a constant: C2 � C1 �12
�cos2 x
2� C2 � �
�1 � sin2 x�2
� C2 �sin2 x
2�
12
� C2
�sin x cos x dx � ���cos x�1��sin x dx� � �cos2 x
2� C2
�sin x cos x dx � ��sin x�1�cos x dx� �sin2 x
2� C1
137. is even.
� 2�325
�83� �
27215
�2
�2x2�x2 � 1� dx � 2�2
0�x4 � x2� dx � 2�x5
5�
x3
3 �2
0
f �x� � x2�x2 � 1� 138. is even.
� 2�sin3 x3 �
��2
0�
23
���2
���2sin2 x cos x dx � 2���2
0sin2 x�cos x� dx
f �x� � sin2 x cos x
Section 5.5 Integration by Substitution 455
139. is odd.
�2
�2x�x2 � 1�3 dx � 0
f �x� � x�x2 � 1�3 140. is odd.
���4
���4sin x cos x dx � 0
f �x� � sin x cos x
141. the function is an even function.
(a)
(c)�2
0��x2� dx � ��2
0x2 dx � �
83
�0
�2x2 dx � �2
0x2 dx �
83
x2�2
0x2 dx � �x3
3 �2
0�
83
;
(b)
(d) �0
�23x2 dx � 3�2
0x2 dx � 8
�2
�2x2 dx � 2�2
0x2 dx �
163
142. (a) since sin x is symmetric to the origin.
(b) since cos x is symmetric to the y-axis.
(c)
(d) since and hence is symmetric to the origin.sin x cos xsin��x� cos��x� � �sin x cos x,���2
���2 sin x cos x dx � 0
���2
���2 cos x dx � 2���2
0cos x dx � �2 sin x�
��2
0� 2
���4
���4 cos x dx � 2���4
0cos x dx � �2 sin x�
��4
0� �2
���4
���4 sin x dx � 0
143. � 0 � 2�4
0�6x2 � 3� dx � 2�2x3 � 3x�
4
0� 232�4
�4�x3 � 6x2 � 2x � 3� dx � �4
�4�x3 � 2x� dx � �4
�4�6x2 � 3� dx
144. ��
��
�sin 3x � cos 3x� dx � ��
��
sin 3x dx � ��
��
cos 3x dx � 0 � 2��
0cos 3x dx � �2
3 sin 3x�
�
0� 0
145. If then and �x�5 � x2�3 dx � �12��5 � x2�3��2x� dx � �
12�u3 du.du � �2x dxu � 5 � x2,
147.
Thus,
When $250,000.Q�50� �t � 50,
Q�t� � 2�100 � t�3.
Q�0� � �k3
�100�3 � 2,000,000 ⇒ k � �6
Q�t� � �k3
�100 � t�3
Q�100� � C � 0
Q�t� � �k�100 � t�2 dt � �k3
�100 � t�3 � C
dQdt
� k�100 � t�2
146. is odd. Hence, �2
�2x�x2 � 1�2 dx � 0.f �x� � x�x2 � 1�2
148.
Solving this system yields and. Thus,
When V�4� � $340,000.t � 4,
V�t� �200,000
t � 1� 300,000.
C � 300,000k � �200,000
V�1� � �12
k � C � 400,000
V�0� � �k � C � 500,000
V�t� � � k�t � 1�2 dt � �
kt � 1
� C
dVdt
�k
�t � 1�2
456 Chapter 5 Integration
149.
(a)
Relative minimum: or June
Relative maximum: or January
(b) inches
(c) inches13�
12
9R�t� dt �
13
�13� � 4.33
�12
0R�t� dt � 37.47
�0.4, 5.5��6.4, 0.7�
0 120
8
R � 3.121 � 2.399 sin�0.524t � 1.377�
151. (a)
Maximum flow: at .is a relative maximum.
(b) thousand gallonsVolume � �24
0R�t� dt � 1272
��18.861, 61.178�t � 9.36R � 61.713
0 240
70
150.
(a) thousand units
(b) thousand units
(c) thousand units112�74.50t �
262.5�
cos �t6 �
12
0�
112894 �
262.5�
�262.5
� � � 74.5
13�74.50t �
262.5�
cos �t6 �
6
3�
13447 �
262.5�
� 223.5� � 102.352
13�74.50t �
262.5�
cos �t6 �
3
0�
13223.5 �
262.5� � � 102.352
1b � a�
b
a�74.50 � 43.75 sin
�t6 � dt �
1b � a�74.50t �
262.5�
cos �t6 �
b
a
152.
(a) amps
(b)
amps
(c) amps1
�1�30� � 0��1
30� cos�60�t� �
1120�
sin�120� t��1�30
0� 30� 1
30�� � �1
30��� � 0
�2�
�5 � 2�2� � 1.382
1�1�240� � 0��
130�
cos�60� t� �1
120� sin�120� t��
1�240
0� 240��
1
30�2��
1120�� � �
130���
1�1�60� � 0��
130�
cos�60�t� �1
120� sin�120�t��
1�60
0� 60� 1
30�� 0� � �
130��� �
4�
� 1.273
1b � a�
b
a
�2 sin�60� t� � cos�120� t� dt �1
b � a��1
30� cos�60�t� �
1120 �
sin�120�t��b
a
Section 5.5 Integration by Substitution 457
153.
When When
(a)
(b)
b � 0.586 � 58.6%
�1 � b�3�2�3b � 2� � 1
P0, b � ���1 � x�3�2
2�3x � 2��
b
0
P0.5, 0.75 � ���1 � x�3�2
2�3x � 2��
0.75
0.5� 0.353 � 35.3%
� ���1 � x�3�2
2�3x � 2��
b
a�
154 �2u3�2
15�3u � 5��
1�b
1�a �
154 �
1�b
1�a
�u3�2 � u1�2� du �154 �2
5u5�2 �
23
u3�2�1�b
1�a
Pa, b � �b
a
154
x�1 � x dx �154 �
1�b
1�a
��1 � u��u du
x � b, u � 1 � b.x � a, u � 1 � a.
dx � �dux � 1 � u,u � 1 � x,
154.
When When
(a)
(b) P0.5, 1 � �u5�2
16�105u3 � 385u2 � 495u � 231��
0
0.5� 0.736 � 73.6%
P0, 0.25 � �u5�2
16�105u3 � 385u2 � 495u � 231��
0.75
1� 0.025 � 2.5%
�115532 �2u5�2
1155�105u3 � 385u2 � 495u � 231��
1�b
1�a� �u5�2
16�105u3 � 385u2 � 495u � 231��
1�b
1�a
�115532 �1�b
1�a
�u9�2 � 3u7�2 � 3u5�2 � u3�2� du �115532 � 2
11u11�2 �
23
u9�2 �67
u7�2 �25
u5�2�1�b
1�a
Pa, b � �b
a
115532
x3�1 � x�3�2 dx �115532 �1�b
1�a
��1 � u�3u3�2 du
x � b, u � 1 � b.x � a, u � 1 � a.
dx � �dux � 1 � u,u � 1 � x,
156.
ex � 1 ≥ x ⇒ ex ≥ 1 � x for x ≥ 0
�et�x
0 ≥ �t�
x
0
�x
0 et dt ≥ �x
0 1 dt155.
Graphing utility: 0.4772 � 47.72%
0.0665�60
48 e�0.0139�t�48�2
dt
157. (a)
(b) is nonnegative because the graph of is positive atthe beginning, and generally has more positivesections than negative ones.
(c) The points on that correspond to the extrema of are points of inflection of
(d) No, some zeros of like do not correspond to an extrema of The graph of continues toincrease after because remains above the
axis.x-fx � ��2gg.
x � ��2,f,
g.fg
fg
0 9.4
−4
f
g
4 (e) The graph of h is that of g shifted two units downward.
� ���2
0f �x� dx � �t
��2f �x� dx � 2 � h�t�
g�t� � �t
0f �x� dx
0 9.4
−4
4
458 Chapter 5 Integration
158. Let
Let and use righthand endpoints.
� �1�
��1 � 1� �2�
� �1�
cos �x�1
0
� �1
0 sin �x dx
limn→
n
i�1
sin�i��n�n
� lim�x�→0
n
i�1f �ci� x
ci �in
, i � 1, 2, . . . , n
x �1n
f �x� � sin �x, 0 ≤ x ≤ 1. 159. (a) Let
(b) Let
� �1
0xb�1 � x�a dx
� �1
0ub�1 � u�a du
�1
0xa�1 � x�b dx � �0
1�1 � u�aub��du�
x � 0 ⇒ u � 1, x � 1 ⇒ u � 0
u � 1 � x, du � �dx, x � 1 � u.
� �1
0x5�1 � x�2 dx
� �1
0u5�1 � u�2 du
�1
0x2�1 � x�5 dx � �0
1�1 � u�2u5��du�
x � 0 ⇒ u � 1, x � 1 ⇒ u � 0
u � 1 � x, du � �dx, x � 1 � u.
160. (a) and
Let
(b) Let as in part (a).
� ���2
0cosn u du � ���2
0cosn x dx
���2
0sinn x dx � ���2
0cosn�
2� x� dx � �0
��2cosn u��du�
u ��
2� x
� ���2
0cos2 u du � ���2
0cos2 x dx���2
0sin2 x dx � ���2
0cos2�
2� x� dx � �0
��2cos2 u��du�
u ��
2� x, du � �dx, x �
�
2� u.
cos x � sin�
2� x�sin x � cos�
2� x�
162. False
�x�x2 � 1� dx �12��x2 � 1��2x� dx �
14
�x2 � 1�2 � C
161. False
��2x � 1�2 dx �12��2x � 1�22 dx �
16
�2x � 1�3 � C
163. True
Odd Even
�10
�10�ax3 � bx2 � cx � d� dx � �10
�10�ax3 � cx� dx � �10
�10�bx2 � d� dx � 0 � 2�10
0�bx2 � d� dx
Section 5.5 Integration by Substitution 459
165. True
4�sin x cos x dx � 2�sin 2x dx � �cos 2x � C
164. True
�b
a
sin x dx � ��cos x�b
a� �cos b � cos a � �cos�b � 2�� � cos a � �b�2�
a
sin x dx
166. False
�sin2 2x cos 2x dx �12��sin 2x�2�2 cos 2x� dx �
12
�sin 2x�3
3� C �
16
sin3 2x � C
167. Let
� �cb
ca
f �x� dx� �cb
ca
f �u� duc�b
a
f �cx� dx � c�cb
ca
f �u� duc
u � cx, du � c dx.
168. (a)
Thus,
(b) Let
(part (a))
� 2�
� 2��� cos���
� 2�sin u � u cos u��
0
� 2��
0u sin u du
��2
0sin�x dx � ��
0sin u�2u du�
u � �x, u2 � x, 2 u du � dx.
�u sin u du � sin u � u cos u � C.
� u sin ud
du�sin u � u cos u � C � cos u � cos u � u sin u
169. Because f is odd, Then
Let in the first integral. When When
� ��a
0f �u� du � �a
0f �x� dx � 0
�a
�a
f �x� dx � ��a
0f ��u���du� � �a
0f �x� dx
x � �a, u � a.x � 0, u � 0.x � �u, dx � �du
� ���a
0f �x� dx � �a
0f �x� dx. �a
�a
f �x� dx � �0
�a
f �x� dx � �a
0f �x� dx
f ��x� � �f �x�.
170. Let then When When Thus,
�b
a
f �x � h� dx � �b�h
a�h
f �u� du � �b�h
a�h
f �x� dx.
u � b � h.x � b,u � a � h.x � a,du � dx.u � x � h,
460 Chapter 5 Integration
171. Let
(Given)
By the Mean Value Theorem for Integrals, there exists in such that
Thus, the equation has at least one real zero.
0 � f �c�.
�1
0f �x� dx � f �c��1 � 0�
�0, 1�c
� a0 � a1
2�
a2
3� . . . �
an
n � 1� 0
�1
0f �x� dx � �a0 x � a1
x2
2� a2
x3
3� . . . � an
xn�1
n � 1�1
0
f �x� � a0 � a1x � a2 x2 � . . . � anxn. 172.
Adding,
Since Hence, there are no such functions.
�� � x�2 ≥ 0, f � 0.
�1
0f �x��� � x�2 dx � 0.
�1
0��2 f �x� � 2�x f �x� � x2f �x�� dx � 0
�1
0f �x�x2 dx � �2
�2��1
0f �x�x dx � �2���� � �2�2
�2�1
0f �x� dx � �2�1� � �2
Section 5.6 Numerical Integration
1. Exact:
Trapezoidal:
Simpson’s: �2
0x2 dx �
16�0 � 41
22
� 2�1�2 � 432
2
� �2�2� �83
� 2.6667
�2
0x2 dx �
14�0 � 21
22
� 2�1�2 � 232
2
� �2�2� �114
� 2.7500
�2
0x2 dx � �1
3x3�
2
0�
83
� 2.6667
2. Exact:
Trapezoidal:
Simpson’s: �1
0x2
2� 1 dx �
112�1 � 4�1�4�2
2� 1 � 2�1�2�2
2� 1 � 4�3�4�2
2� 1 � 12
2� 1� �
76
� 1.1667
�1
0x2
2� 1 dx �
18�1 � 2�1�4�2
2� 1 � 2�1�2�2
2� 1 � 2�3�4�2
2� 1 � 12
2� 1� �
7564
� 1.1719
�1
0x2
2� 1 dx � �x3
6� x�
1
0�
76
� 1.1667
3. Exact:
Trapezoidal:
Simpson’s: �2
0x3 dx �
16�0 � 41
23
� 2�1�3 � 432
3
� �2�3� �246
� 4.0000
�2
0x3 dx �
14�0 � 21
23
� 2�1�3 � 232
3
� �2�3� �174
� 4.2500
�2
0x3 dx � �x4
4 �2
0� 4.000
4. Exact:
Trapezoidal:
Simpson’s: �2
1 2x2 dx �
112�2 � 432
25 � 289 � 432
49 �12� � 1.0008
�2
1 2x2 dx �
18�2 � 232
25 � 289 � 232
49 �12� � 1.0180
�2
1 2x2 dx � ��2
x �2
1� �1 � 2 � 1
Section 5.6 Numerical Integration 461
5. Exact:
Trapezoidal:
Simpson’s: �2
0x3 dx �
112�0 � 41
43
� 224
3
� 434
3
� 2�1�3 � 454
3
� 264
3
� 474
3
� 8� � 4.0000
�2
0x3 dx �
18�0 � 21
43
� 224
3
� 234
3
� 2�1�3 � 254
3
� 264
3
� 274
3
� 8� � 4.0625
�2
0x3 dx � �1
4x4�
2
0� 4.0000
7. Exact:
Trapezoidal:
Simpson’s: �9
4
�x dx �5
24�2 � 4�378
� �21 � 4�478
� �26 � 4�578
� �31 � 4�678
� 3� � 12.6667
� 12.6640
�9
4
�x dx �5
16�2 � 2�378
� 2�214
� 2�478
� 2�264
� 2�578
� 2�314
� 2�678
� 3��9
4
�x dx � �23
x3�2�9
4� 18 �
163
�383
� 12.6667
9. Exact:
Trapezoidal:
Simpson’s:
�1
1214
�6481
�8
25�
64121
�19 � 0.1667
�2
1
1�x � 1�2 dx �
112�
14
� 4 1��5�4� � 1�2 � 2 1
��3�2� � 1�2 � 4 1��7�4� � 1�2 �
19�
�18
14
�3281
�8
25�
32121
�19 � 0.1676
�2
1
1�x � 1�2 dx �
18�
14
� 2 1��5�4� � 1�2 � 2 1
��3�2� � 1�2 � 2 1��7�4� � 1�2 �
19�
�2
1
1�x � 1�2 dx � ��
1x � 1�
2
1� �
13
�12
�16
� 0.1667
6. Exact:
Trapezoidal:
Simpson’s: �8
0 3�x dx �
13
�0 � 4 � 2 3�2 � 4 3�3 � 2 3�4 � 4 3�5 � 2 3�6 � 4 3�7 � 2� � 11.8632
�8
0 3�x dx �
12
�0 � 2 � 2 3�2 � 2 3�3 � 2 3�4 � 2 3�5 � 2 3�6 � 2 3�7 � 2� � 11.7296
�8
0
3 �x dx � �34
x4�3�8
0� 12.0000
8. Exact:
Trapezoidal:
Simpson’s: �3
1�4 � x2� dx �
16�3 � 44 �
94 � 0 � 44 �
254 � 5� � �0.6667
�3
1�4 � x2� dx �
14 3 � 2�4 � 3
22
� � 2�0� � 2�4 � 52
2
� � 5� � �0.7500
�3
1�4 � x2� dx � �4x �
x3
3 �3
1� 3 �
113
� �23
� �0.6667
10. Exact:
Trapezoidal:
Simpson’s: �2
0x�x2 � 1 dx �
16�0 � 41
2��1�2�2 � 1 � 2�1��12 � 1 � 432��3�2�2 � 1 � 2�22 � 1� � 3.392
�2
0x�x2 � 1 dx �
14�0 � 21
2��1�2�2 � 1 � 2�1��12 � 1 � 232��3�2�2 � 1 � 2�22 � 1� � 3.457
�2
0x�x2 � 1 dx �
13��x2 � 1�3�2�
2
0�
13
�53�2 � 1� � 3.393
462 Chapter 5 Integration
11. Trapezoidal:
Simpson’s:
Graphing utility: 3.241
�2
0
�1 � x3 dx �16
�1 � 4�1 � �1�8� � 2�2 � 4�1 � �27�8� � 3� � 3.240
�2
0
�1 � x3 dx �14
�1 � 2�1 � �1�8� � 2�2 � 2�1 � �27�8� � 3� � 3.283
12. Trapezoidal:
Simpson’s:
Graphing utility: 1.402
�2
0
1
�1 � x3 dx �
16�1 � 4 1
�1 � �1�2�3 � 2 1
�1 � 13 � 4 1
�1 � �3�2�3 �13� � 1.405
�2
0
1
�1 � x3 dx �
14�1 � 2 1
�1 � �1�2�3 � 2 1
�1 � 13 � 2 1
�1 � �3�2�3 �13� � 1.397
13.
Trapezoidal:
Simpson’s:
Graphing utility: 0.393
�1
0
�x�1 � x� dx �1
12�0 � 4�141 �
14 � 2�1
21 �12 � 4�3
41 �34 � � 0.372
�1
0
�x�1 � x� dx �18�0 � 2�1
41 �14 � 2�1
21 �12 � 2�3
41 �34 � � 0.342
�1
0
�x�1 � x dx � �1
0
�x�1 � x� dx
14. Trapezoidal:
Simpson’s:
Graphing utility: 1.458
��
��2
�x sin x dx ��
24 ���
2� 4�5�
8 sin5�
8 � 2�3�
4 sin3�
4 � 4�7�
8 sin7�
8 � 0� � 1.458
��
��2
�x sin x dx ��
16 ���
2�1� � 2�5�
8 sin5�
8 � 2�3�
4 sin3�
4 � 2�7�
8 sin7�
8 � 0� � 1.430
15. Trapezoidal:
Simpson’s:
Graphing utility: 0.977
� 0.978
����2
0cos�x2� dx �
���212 �cos 0 � 4 cos���2
4 2
� 2 cos���22
2
� 4 cos3���24
2
� cos��
22
�
� 0.957
����2
0cos�x2� dx �
���28 �cos 0 � 2 cos���2
4 2
� 2 cos���22
2
� 2 cos3���24
2
� cos��
22
�
16. Trapezoidal:
Simpson’s:
Graphing utility: 0.256
� 0.257
����4
0tan�x2� dx �
���412 �tan 0 � 4 tan���4
4 2
� 2 tan���42 2
� 4 tan3���44 2
� tan��
42
�
� 0.271
����4
0tan�x2� dx �
���48 �tan 0 � 2 tan���4
4 2
� 2 tan���42 2
� 2 tan3���44 2
� tan��
42
�
Section 5.6 Numerical Integration 463
17. Trapezoidal:
Simpson’s:
Graphing utility: 0.089
�1.1
1sin x2 dx �
1120
�sin�1� � 4 sin�1.025�2 � 2 sin�1.05�2 � 4 sin�1.075�2 � sin�1.1�2� � 0.089
�1.1
1sin x2 dx �
180
�sin�1� � 2 sin�1.025�2 � 2 sin�1.05�2 � 2 sin�1.075�2 � sin�1.1�2� � 0.089
18. Trapezoidal:
Simpson’s:
Graphing utility: 1.910
���2
0
�1 � cos2 x dx ��
24��2 � 4�1 � cos2���8� � 2�1 � cos2���4� � 4�1 � cos2�3��8� � 1� � 1.910
���2
0
�1 � cos2 x dx ��
16��2 � 2�1 � cos2���8� � 2�1 � cos2���4� � 2�1 � cos2�3��8� � 1� � 1.910
19.
Trapezoidal: 1.684
Simpson’s: 1.649
Graphing utility: 1.648
�2
0 x ln�x � 1� dx
20. Trapezoidal:
Simpson’s:
Graphing utility: 1.296
�3
1 ln x dx �
16
�0 � 4 ln�1.5� � 2 ln 2 � 4 ln�2.5� � ln 3� �7.7719
6� 1.295
�3
1 ln x dx �
14
�0 � 2 ln�1.5� � 2 ln 2 � 2 ln�2.5� � ln 3� �5.1284
4� 1.282
21. Trapezoidal:
Simpson’s:
Graphing utility: 0.186
���4
0x tan x dx �
�
48�0 � 4 �
16 tan �
16 � 22�
16 tan2�
16 � 43�
16 tan3�
16 ��
4� � 0.186
���4
0x tan x dx �
�
32�0 � 2 �
16 tan �
16 � 22�
16 tan2�
16 � 23�
16 tan3�
16 ��
4� � 0.194
22. Trapezoidal:
Simpson’s:
Graphing utility: 1.852
��
0 sin x
x dx �
�
12�1 �4 sin���4�
��4�
2 sin���2���2
�4 sin�3��4�
3��4� 0� � 1.852
��
0 sin x
x dx �
�
8�1 �2 sin���4�
��4�
2 sin���2���2
�2 sin�3��4�
3��4� 0� � 1.836
23. Trapezoidal:
Simpson’s:
Graphing utility: 92.744
�4
0 �xex dx �
13
�0 � 4e1 � 2�2e2 � 4�3e3 � 2e 4� � 93.375
�4
0 �xe x dx �
12
�0 � 2e1 � 2�2e2 � 2�3e3 � 2e4� � 102.555
24. Trapezoidal:
Simpson’s:
Graphing utility: 0.594
�2
0 2xe�x dx �
16
�0 � 2e�1�2 � 2e�1 � 6e�3�2 � 2e�2� �3.5583
6� 0.5930
�2
0 xe�x dx �
14
�0 � e�1�2 � 2e�1 � 3e�3�2 � 2e�2� �2.2824
4� 0.5706
464 Chapter 5 Integration
25.
The Trapezoidal Rule overestimates the area if the graphof the integrand is concave up.
xa b
y f x= ( )
y 26. Trapezoidal: Linear polynomials
Simpson’s: Quadratic polynomials
27.
(a) Trapezoidal: since is maximum in when
(b) Simpson’s: since f �4��x� � 0.Error ≤ �2 � 0�5
180�44� �0� � 0
x � 2.�0, 2�� f� �x��Error ≤ �2 � 0�3
12�42� �12� � 0.5
f �4��x� � 0
f���x� � 6
f � �x� � 6x
f��x� � 3x2
f �x� � x3
28.
The error is 0 for both rules.
f � �x� � 0
f��x� � 2
f �x� � 2x � 3
29.
(a) Trapezoidal: because is at most 1 on
(b) Simpson’s: because is at most 1 on �0, ��.� f �4��x��Error ≤ �� � 0�5
180�44� �1� ��5
46,080� 0.006641
�0, ��.� f � �x��Error ≤ �� � 0�3
12�42� �1� ��3
192� 0.1615
f �4��x� � cos x
f�� �x� � sin x
f � �x� � �cos x
f��x� � �sin x
f �x� � cos x
30.
(a) Trapezoidal: since on
(b) Simpson’s: since on �0, 1�.� f �4� �x�� ≤ � 4Error ≤ �1 � 0�5
180�44� � 4 �� 4
46,080� 0.0021
�0, 1�.� f � �x�� ≤ � 2Error ≤ �1 � 0�3
12�42� � 2 ��2
192� 0.0514
f �4��x� � � 4 sin��x�
f�� �x� � �� 3 cos��x�
f � �x� � �� 2 sin��x�
f��x� � � cos��x�
f �x� � sin��x�
Section 5.6 Numerical Integration 465
31.
(a) Maximum of is
Trapezoidal:
(b) Maximum of is
Simpson’s:
n ≥ 6.2. Let n � 8 (even).
n4 ≥32�15��2180�256� 105 �
�296
105
Error ≤ 25
180n415�2256 ≤ 0.00001
15�2256
� 0.0829.
� f �4� �x�� � � �1516�x � 2�7�2�
n ≥ 76.8. Let n � 77.
n2 ≥8�2
12�16�105 ��224
105
Error ≤ �2 � 0�3
12n2 �216 ≤ 0.00001
�216
� 0.0884.� f � �x�� � � �14�x � 2�3�2�
f �4��x� ��1516
�x � 2��7�2
f�� �x� �38
�x � 2��5�2
f � �x� � �14
�x � 2��3�2
f��x� �12
�x � 2��1�2
f �x� � �x � 2�1�2, 0 ≤ x ≤ 2 32.
(a) Maximum of is on
Trapezoidal:
(b) Maximum of is on
Simpson’s:
n ≥ 18.5. Let n � 20 (even).
n4 ≥76
� 105
Error ≤ 25
180n410516 ≤ 0.00001
�1, 3�.10516� f �4� �x�� � � 105
16x9�2� n ≥ 223.6. Let n � 224.
n2 ≥12
� 105
Error ≤ �3 � 1�3
12n2 34 ≤ 0.00001
�1, 3�.34� f � �x�� � � 3
4x5�2�f �4��x� �
10516
x�9�2
f���x� ��15
8x�7�2
f��x� �34
x�5�2
f��x� ��12
x�3�2
1 ≤ x ≤ 3 f �x� � x�1�2,
33.
(a) Maximum of is
Trapezoidal:
(b) Maximum of is
Simpson’s:
n ≥ 15.3. Let n � 16.
n4 ≥�4
180� 105
Error ≤ 1
180n4 � 4 ≤ 0.00001
� 4.� f �4� �x�� � �� 4cos��x��
n ≥ 286.8. Let n � 287.
n2 ≥�2
12� 105
Error ≤ �1 � 0�3
12n2 �2 ≤ 0.00001
� 2.� f � �x�� � ��� 2 cos��x��f �4��x� � � 4 cos��x�
f�� �x� � � 3 sin��x�
f � �x� � �� 2 cos��x�
f��x� � �� sin��x�
f �x� � cos��x�, 0 ≤ x ≤ 1 34.
All derivatives are bounded by 1.
(a) Trapezoidal:
(b) Simpson’s:
n ≥ 8.5. Let n � 10 (even).
n4 ≥�5
5760� 105
Error ≤ ���2�5
180n4 �1� ≤ 0.00001
n ≥ 179.7. Let n � 180.
n2 ≥�3
96� 105
Error ≤ ���2�3
12n2 �1� ≤ 0.00001
f �4��x� � sin x
f�� �x� � �cos x
f � �x� � �sin x
f��x� � cos x
f �x� � sin x, 0 ≤ x ≤�
2
466 Chapter 5 Integration
35.
(a) in
is maximum when and
Trapezoidal: Error ≤ < 0.00001, > 16,666.67, n > 129.10; let
(b) in
is maximum when and
Simpson’s: Error ≤ < 0.00001, > 16,666.67, n > 11.36; let n � 12.n432180n415
16� f �4��0�� �
1516
.x � 0� f �4��x��
�0, 2�f �4��x� ��15
16�1 � x�7�2
n � 130.n2812n21
4� f � �0�� �
14
.x � 0� f � �x��
�0, 2�f� �x� � �1
4�1 � x�3�2
f �x� � �1 � x
36.
(a) in
is maximum when and
Trapezoidal: Error ≤ < 0.00001, > 14,814.81, n > 121.72; let
(b) in
is maximum when and
Simpson’s: Error ≤ < 0.00001, > 12,290.81, n > 10.53; let (In Simpson’s Rule n must be even.)n � 12.n432180n456
81� f �4��0�� �
5681
.x � 0� f �4��x��
�0, 2�f �4��x� � �56
81�x � 1�10�3
n � 122.n2812n42
9�f � �0�� �
29
.x � 0� f� �x��
�0, 2�f � �x� � �2
9�x � 1�4�3
f �x� � �x � 1�2�3
38.
(a) in
is maximum when and
Trapezoidal: Error ≤ < 0.00001, > 19,044.17, n > 138.00; let
(b) in
is maximum when and
Simpson’s: Error ≤ < 0.00001, > 15,793.61, n > 11.21; let n � 12.n4�1 � 0�5
180n4 �28.4285�
� f �4��0.852�� � 28.4285.x � 0.852� f �4��x���0, 1�f �4��x� � �16x4 � 12� sin�x2� � 48x2 cos�x2�
n � 139.n2�1 � 0�3
12n2 �2.2853�
�f ��1�� � 2.2853.x � 1�f ��x���0, 1�f � �x� � 2��2x2 sin�x2� � cos�x2��
f �x� � sin�x2�
37.
(a) in
is maximum when and
Trapezoidal: Error ≤ < 0.00001, > 412,754.17, n > 642.46; let
(b) in
is maximum when and
Simpson’s: Error ≤ < 0.00001, > 5,102,485.22, n > 47.53; let n � 48.n4�1 � 0�5
180n4 �9184.4734�
� f �4��1�� � 9184.4734.x � 1� f �4��x���0, 1�f �4��x� � 8 sec2�x2��12x2 � �3 � 32x4� tan�x2� � 36x2 tan2�x2� � 48x4 tan3�x2��
n � 643.n2�1 � 0�3
12n2 �49.5305�
� f ��1�� � 49.5305.x � 1� f ��x���0, 1�f ��x� � 2 sec2�x2��1 � 4x2 tan�x2��
f �x� � tan�x2�
Section 5.6 Numerical Integration 467
39. (a)
(b)
�773
� 25.67
�4
0f �x� dx �
412
�3 � 4�7� � 2�9� � 4�7� � 0�
�12
�49� �492
� 24.5
�4
0f �x� dx �
48
�3 � 2�7� � 2�9� � 2�7� � 0�
b � a � 4 � 0 � 4, n � 4 40.
(a)
(b)
�13
�134� �1343
� 4�10� � 2�9� � 4�6� � 0�
�8
0f �x� dx �
824
�0 � 4�1.5� � 2�3� � 4�5.5� � 2�9�
�12
�88� � 44
� 2�10� � 2�9� � 2�6� � 0�
�8
0f �x� dx �
816
�0 � 2�1.5� � 2�3� � 2�5.5� � 2�9�
n � 8, b � a � 8 � 0 � 8
41. The program will vary depending upon the computer or programmable calculator that you use.
42. on �0, 4�f �x� � �2 � 3x2
n
4 12.7771 15.3965 18.4340 15.6055 15.4845
8 14.0868 15.4480 16.9152 15.5010 15.4662
10 14.3569 15.4544 16.6197 15.4883 15.4658
12 14.5386 15.4578 16.4242 15.4814 15.4657
16 14.7674 15.4613 16.1816 15.4745 15.4657
20 14.9056 15.4628 16.0370 15.4713 15.4657
S�n�T�n�R�n�M�n�L�n�
43. on �0, 1�f �x� � �1 � x2
n
4 0.8739 0.7960 0.6239 0.7489 0.7709
8 0.8350 0.7892 0.7100 0.7725 0.7803
10 0.8261 0.7881 0.7261 0.7761 0.7818
12 0.8200 0.7875 0.7367 0.7783 0.7826
16 0.8121 0.7867 0.7496 0.7808 0.7836
20 0.8071 0.7864 0.7571 0.7821 0.7841
S�n�T�n�R�n�M�n�L�n�
44. on �0, 4�f �x� � sin�x
n
4 2.8163 3.5456 3.7256 3.2709 3.3996
8 3.1809 3.5053 3.6356 3.4083 3.4541
10 3.2478 3.4990 3.6115 3.4296 3.4624
12 3.2909 3.4952 3.5940 3.4425 3.4674
16 3.3431 3.4910 3.5704 3.4568 3.4730
20 3.3734 3.4888 3.5552 3.4643 3.4759
S�n�T�n�R�n�M�n�L�n�
48.
Simpson’s Rule:
� 4001512�125 � 15
123
� . . . � 0� � 10,233.58 ft � lb
�5
0100x�125 � x3 dx �
53�12��0 � 400 5
12�125 � 512
3
� 2001012�125 � 10
123
n � 12
W � �5
0100x�125 � x3 dx
468 Chapter 5 Integration
45. on �1, 2�f �x� �sin x
x
n
4 0.7070 0.6597 0.6103 0.6586 0.6593
8 0.6833 0.6594 0.6350 0.6592 0.6593
10 0.6786 0.6594 0.6399 0.6592 0.6593
12 0.6754 0.6594 0.6431 0.6593 0.6593
16 0.6714 0.6594 0.6472 0.6593 0.6593
20 0.6690 0.6593 0.6496 0.6593 0.6593
S�n�T�n�R�n�M�n�L�n�
46. on �0, 2�f �x� � 6e�x2�2
n
4 8.4410 7.1945 5.8470 7.1440 7.1770
8 7.8178 7.1820 6.5208 7.1693 7.1777
10 7.6911 7.1804 6.6535 7.1723 7.1777
12 7.6063 7.1796 6.7416 7.1740 7.1777
16 7.4999 7.1788 6.8514 7.1756 7.1777
20 7.4358 7.1784 6.9170 7.1764 7.1777
S�n�T�n�R�n�M�n�L�n�
47. on �0, 3�f �x� � �x ln�x � 1�
n
4 2.5311 3.3953 4.3320 3.4316 3.4140
8 2.9632 3.4026 3.8637 3.4135 3.4074
10 3.0508 3.4037 3.7711 3.4109 3.4068
12 3.1094 3.4044 3.7097 3.4095 3.4065
16 3.1829 3.4050 3.6331 3.4050 3.4062
20 3.2273 3.4054 3.5874 3.4073 3.4061
S�n�T�n�R�n�M�n�L�n�
Section 5.6 Numerical Integration 469
49. (a) Trapezoidal:
Simpson’s:
�2
0f �x� dx �
23�8��4.32 � 4�4.36� � 2�4.58� � 4�5.79� � 2�6.14� � 4�7.25� � 2�7.64� � 4�8.08� � 8.14� � 12.592
�2
0f �x� dx �
22�8��4.32 � 2�4.36� � 2�4.58� � 2�5.79� � 2�6.14� � 2�7.25� � 2�7.64� � 2�8.08� � 8.14� � 12.518
(b) Using a graphing utility,
Integrating, �2
0y dx � 12.53.
y � �1.3727x3 � 4.0092x2 � 0.6202x � 4.2844.
50. Simpson’s Rule,
�1
36�113.098� � 3.1416
� �1
2� 0
3�6� �6 � 4�6.0209� � 2�6.0851� � 4�6.1968� � 2�6.3640� � 4�6.6002� � 6.9282�
n � 6�1�2
0
6
�1 � x2 dx,
52. Area �10002�10��125 � 2�125� � 2�120� � 2�112� � 2�90� � 2�90� � 2�95� � 2�88� � 2�75� � 2�35�� � 89,250 sq m
51. Simpson’s Rule:
� 3.14159
� � 4�1
0
11 � x2 dx �
43�6��1 �
41 � �1�6�2 �
21 � �2�6�2 �
41 � �3�6�2 �
21 � �46�2 �
41 � �5�6�2 �
12�
n � 6
53.
� 7435 sq m
Area �120
2�12��75 � 2�81� � 2�84� � 2�76� � 2�67� � 2�68� � 2�69� � 2�72� � 2�68� � 2�56� � 2�42� � 2�23� � 0�
55.
By trial and error, we obtain t � 2.477.
n � 10�t
0sin�x dx � 2,
56. The quadratic polynomial
passes through the three points.
p�x� ��x � x2��x � x3�
�x1 � x2��x1 � x3� y1 �
�x � x1��x � x3��x2 � x1��x2 � x3�
y2 ��x � x1��x � x2�
�x3 � x1��x3 � x2� y3
54. Let Then
Simpson’s: Error ≤
Therefore, Simpson’s Rule is exact when approximatingthe integral of a cubic polynomial.
Example:
This is the exact value of the integral.
�1
0x3 dx �
16�0 � 41
23
� 1� �14
�b � a�5
180n4 �0� � 0
f �4��x� � 0.f �x� � Ax3 � Bx2 � Cx � D.
17.
�x3
3� 2x �
12
ln�x2 � 2� � C
�x4 � x � 4x2 � 2
dx � ��x2 � 2 �x
x2 � 2� dx 18.
� �3x �x2
2�
12
ln�x2 � 3� � C
�x3 � 3x2 � 4x � 9x2 � 3
dx � ���3 � x �x
x2 � 3� dx
470 Chapter 5 Integration
Section 5.7 The Natural Logarithmic Function: Integration
1. �5
x dx � 5�1
x dx � 5 ln�x� � C 2. �10
x dx � 10�1
x dx � 10 ln�x� � C 3.
� 1x � 1
dx � ln�x � 1� � C
du � dxu � x � 1,
4.
� 1x � 5
dx � ln�x � 5� � C
du � dxu � x � 5, 5.
� �12
ln�3 � 2x� � C
� 13 � 2x
dx � �12� 1
3 � 2x��2� dx
du � �2 dxu � 3 � 2x, 6.
�13
ln�3x � 2� � C
� 13x � 2
dx �13� 1
3x � 2�3� dx
7.
� ln�x2 � 1 � C
�12
ln�x2 � 1� � C
� xx2 � 1
dx �12� 1
x2 � 1�2x� dx
du � 2x dxu � x2 � 1, 8.
� �13
ln�3 � x3� � C
� x2
3 � x3 dx � �13� 1
3 � x3��3x2� dx
du � �3x2 dxu � 3 � x3, 9.
�x2
2� 4 ln�x� � C
�x2 � 4x
dx � ��x �4x� dx
10.
� ��9 � x2 � C
� x
�9 � x2 dx � �
12��9 � x2��1�2��2x� dx
du � �2x dxu � 9 � x2, 11.
�13
ln�x3 � 3x2 � 9x� � C
� x2 � 2x � 3x3 � 3x2 � 9x
dx �13�3�x2 � 2x � 3�
x3 � 3x2 � 9x dx
du � 3�x2 � 2x � 3� dxu � x3 � 3x2 � 9x,
12.
�13
ln�x3 � 3x2 � 4� � C
� x�x � 2�x3 � 3x2 � 4
dx �13� 3x2 � 6x
x3 � 3x2 � 4 dx, �u � x3 � 3x2 � 4�
13.
�x2
2� 4x � 6 ln�x � 1� � C
�x2 � 3x � 2x � 1
dx � ��x � 4 �6
x � 1� dx 14.
� x2 � 11x � 19 ln�x � 2� � C
�2x2 � 7x � 3x � 2
dx � ��2x � 11 �19
x � 2� dx
15.
�x3
3� 5 ln�x � 3� � C
�x3 � 3x2 � 5x � 3
dx � ��x2 �5
x � 3� dx 16.
�x3
3�
5x2
2� 19x � 115 ln�x � 5� � C
�x3 � 6x � 20x � 5
dx � ��x2 � 5x � 19 �115
x � 5� dx
Section 5.7 The Natural Logarithmic Function: Integration 471
19.
��ln x�2
x dx �
13
�ln x�3 � C
du �1x dxu � ln x, 20.
�13
ln�ln�x�� � C
� 1x ln�x3� dx �
13� 1
ln x�
1x dx
21.
� 2�x � 1 � C
� 2�x � 1�1�2 � C
� 1
�x � 1 dx � ��x � 1��1�2 dx
du � dxu � x � 1, 22.
� 3 ln�1 � x1�3� � C
� 1x2�3�1 � x1�3� dx � 3� 1
1 � x1�3� 13x2�3� dx
du �1
3x2�3 dxu � 1 � x1�3,
23.
� 2 ln�x � 1� �2
�x � 1� � C
� 2� 1x � 1
dx � 2� 1�x � 1�2 dx
� �2�x � 1��x � 1�2 dx � 2� 1
�x � 1�2 dx
� 2x�x � 1�2 dx � �2x � 2 � 2
�x � 1�2 dx 24.
� ln�x � 1� �1
2�x � 1�2 � C
� � 1x � 1
dx � � 1�x � 1�3 dx
� ��x � 1�2
�x � 1�3 dx � � 1�x � 1�3 dx
�x�x � 2��x � 1�3 dx � �x2 � 2x � 1 � 1
�x � 1�3 dx
25.
where C � C1 � 1.
� �2x � ln�1 � �2x � � C
� �1 � �2x � � ln�1 � �2x� � C1
� u � ln�u� � C1
� 1
1 � �2x dx � ��u � 1�
u du � ��1 �
1u� du
du �1
�2x dx ⇒ �u � 1� du � dxu � 1 � �2x, 26.
�23�3x �
23
ln�1 � �3x� � C1
�23
1 � �3x � ln�1 � �3x� � C
�23
u � ln�u� � C
�23��1 �
1u� du
� 1
1 � �3x dx � �1
u 2
3�u � 1� du
u � 1 � �3x, du �3
2�3x dx ⇒ dx �
2
3�u � 1� du
27.
where C � C1 � 27.
� x � 6�x � 18 ln��x � 3� � C
� ��x � 3�2� 12��x � 3� � 18 ln��x � 3� � C1
� u2 � 12u � 18 ln�u� � C1
� 2�u2
2� 6u � 9 ln�u�� � C1
� 2�u2 � 6u � 9u
du � 2��u � 6 �9u� du
� �x
�x � 3 dx � 2��u � 3�2
u du
du �1
2�x dx ⇒ 2�u � 3� du � dxu � �x � 3,
37.
� ln�cos�e�x�� � C
� ���ln�cos�e�x��� � C
�e�x tan�e�x� dx � �� tan�e�x���e�x� dx 38.
� tan t � sec t � C
�sec t�sec t � tan t� dt � �sec2 t dt � �sec t tan t dt
472 Chapter 5 Integration
28.
� 3 ln�x1�3 � 1� �3x2�3
2� 3x1�3 � x � C1
� 3��x1�3 � 1�3
3�
3�x1�3 � 1�2
2� 3�x1�3 � 1� � ln�x1�3 � 1�� � C
� 3�u3
3�
3u2
2� 3u � ln�u�� � C
� 3��u2 � 3u � 3 �1u� du
� 3�u � 1u
�u2 � 2u � 1� du
� 3�x3�x � 1
dx � �u � 1u
3�u � 1�2 du
u � x1�3 � 1, du �1
3x2�3 dx ⇒ dx � 3�u � 1�2 du
29.
�u � sin �, du � cos � d��
�cos �sin �
d� � ln�sin �� � C 30.
� �15
ln�cos 5�� � C
�tan 5� d� �15�5 sin 5�
cos 5� d�
31.
� �12
ln�csc 2x � cot 2x� � C
�csc 2x dx �12��csc 2x��2� dx 32.
� 2 ln�sec x2
� tan x2� � C
�sec x2
dx � 2�sec x2�
12� dx
33. � cos t1 � sin t
dt � ln�1 � sin t� � C 34.
�csc2 tcot t
dt � �ln�cot t� � C
u � cot t, du � �csc2 t dt
35. �sec x tan xsec x � 1
dx � ln�sec x � 1� � C 36.
� ln�sec t�sec t � tan t�� � C
� ln�sec t � tan tcos t � � C
��sec t � tan t� dt � ln�sec t � tan t� � ln�cos t� � C
Section 5.7 The Natural Logarithmic Function: Integration 473
39.
y � �3 ln�x � 2�0 � �3 ln�1 � 2� � C ⇒ C � 0�1, 0�:
� �3 ln�x � 2� � C
� �3� 1x � 2
dx−10 10
−10
(1, 0)
10y � � 3
2 � x dx 40.
−9 9
−4
(0, 4)
8y � ln�x2 � 9� � 4 � ln 9
4 � ln�0 � 9� � C ⇒ C � 4 � ln 9�0, 4�:
� ln�x2 � 9� � C
y � � 2xx2 � 9
dx
41.
s � �12
ln�cos 2�� � 2
2 � �12
ln�cos�0�� � C ⇒ C � 2�0, 2�:
� �12
ln�cos 2�� � C
�12�tan�2���2 d�� −3 3
−3
(0, 2) 4s � �tan�2�� d� 42.
−8 8
− 2
( , 4)π
10r � ln�tan t � 1� � 4
4 � ln�0 � 1� � C ⇒ C � 4��, 4�:
� ln�tan t � 1� � C
r � � sec2 ttan t � 1
dt
43.
f �x� � �2 ln x � 3x � 2
f �1� � 1 � �2�0� � 3 � C1 ⇒ C1 � �2
f �x� � �2 ln x � 3x � C1
f ��x� ��2x
� 3
f ��1� � 1 � �2 � C ⇒ C � 3
f ��x� ��2x
� C
f � �x� �2x2 � 2x�2, x > 0 44.
f �x� � 4 ln�x � 1� � x2 � 7
f �2� � 3 � 4�0� � 4 � C1 ⇒ C1 � 7
f �x� � 4 ln�x � 1� � x2 � C1
f ��x� �4
x � 1� 2x
f ��2� � 0 � 4 � 4 � C ⇒ C � 0
f ��x� �4
�x � 1� � 2x � C
f � �x� ��4
�x � 1�2 � 2 � �4�x � 1��2 � 2, x > 1
45.
(a)
x
y
3
−3
4−2
(0, 1)
�0, 1�dydx
�1
x � 2,
(b)
Hence, y � ln�x � 2� � 1 � ln 2 � ln�x � 22 � � 1.
y�0� � 1 ⇒ 1 � ln 2 � C ⇒ C � 1 � ln 2 −3 6
−3
3
y � � 1x � 2
dx � ln�x � 2� � C
474 Chapter 5 Integration
46.
(a)
(b)
Hence, y ��ln x�2
2� 2.
y�1� � �2 ⇒ �2 ��ln 1�2
2� C ⇒ C � �2
y � �ln xx
dx ��ln x�2
2� C
x
−1
−2
1
2
y
4
�1, �2�dydx
�ln x
x, 47. (a)
(b)
8
−1
−1
8
y � x � ln x � 3
4 � 1 � 0 � C ⇒ C � 3
y � x � ln x � C
dydx
� 1 �1x, �1, 4�
8−1
−2
−3
1
2
3
4
5
x
y(1, 4)
48. (a)
−4
4
x
y
2π
2π−
(b)
y � ln�sec x � tan x� � 1
1 � ln�1 � 0� � C ⇒ C � 1
y � ln�sec x � tan x� � C
5
−3
−�2
�2
dydx
� sec x, �0, 1�
49.
�53
ln 13 4.275
�4
0
53x � 1
dx � �53
ln�3x � 1��4
050.
� ln 3 � ln 1 � ln 3
�1
�1
1x � 2
dx � �ln�x � 2��1
�151.
�73
�e
1 �1 � ln x�2
x dx � �1
3�1 � ln x�3�
e
1
du �1x dxu � 1 � ln x,
52.
� �ln�ln�x���e2
e� ln 2 �e2
e
1x ln x
dx � �e2
e� 1
ln x�1x dx
du �1x dxu � ln x, 53.
� �12
x2 � x � ln�x � 1��2
0� �ln 3
�2
0 x2 � 2x � 1
dx � �2
0�x � 1 �
1x � 1� dx
54.
� �x � 2 ln�x � 1��1
0� 1 � 2 ln 2
�1
0 x � 1x � 1
dx � �1
01 dx � �1
0
�2x � 1
dx 55.
� ln�2 � sin 21 � sin 1� 1.929
�2
1 1 � cos �� � sin �
d� � �ln�� � sin ���2
1
56.
� ��cot 2� � csc 2� � ��0.2
0.1 0.0024 � �0.2
0.1�2 csc2 2� � 2 csc 2� cot 2� � 1� d�
�0.2
0.1�csc 2� � cot 2��2 d� � �0.2
0.1�csc2 2� � 2 csc 2� cot 2� � cot2 2�� d�
Section 5.7 The Natural Logarithmic Function: Integration 475
57.
where C � C1 � 2. � 2�x � ln�1 � �x � � C
� 1
1 � �x dx � 2�1 � �x � � 2 ln�1 � �x � � C1
58.
� 4�x � x � 4 ln�1 � �x � � C where C � C1 � 5.
�1 � �x
1 � �x dx � ��1 � �x �2 � 6�1 � �x � � 4 ln�1 � �x � � C1
59. � �xx � 1
dx � ln��x � 1�x � 1� � 2�x � C 60. � x2
x � 1 dx � ln�x � 1� �
x2
2� x � C
61.
� ln��2 � 1� ��22
0.174
���2
��4�csc x � sin x� dx � ��ln�csc x � cot x� � cos x�
��2
��4
62.
� ln��2 � 1
�2 � 1� � 2�2 �1.066
���4
���4
sin2 x � cos2 xcos x
dx � �ln�sec x � tan x� � 2 sin x���4
���4
Note: In Exercises 63–66, you can use the Second Fundamental Theorem of Calculus to integrate the function.
63.
F��x� �1x
F �x� � �x
1 1t dt 64.
F� �x� � tan x
F �x� � �x
0tan t dt
65.
(by Second Fundamental Theorem of Calculus)
Alternate Solution:
F� �x� �13x
�3� �1x
F�x� � �3x
1 1t dt � �ln�t��
3x
1� ln�3x�
F� �x� �13x
�3� �1x
F�x� � �3x
1 1t dt 66.
F��x� �2xx2 �
2x
F �x� � �x2
1 1t dt
67.
Matches (d).A 1.25;
x1−1
2
12
12
−
y 68.
Matches (a).A 3;
x1 2 3 4
1
2
−1
−2
y 69. A � �3
1 4x dx � 4 ln�x��
3
1� 4 ln 3
476 Chapter 5 Integration
70.
� 2 ln�2 ln 2ln 2 � � 2 ln 2
� 2ln�ln 4� � ln�ln 2�
� 2 ln�ln x��4
2
A � �4
2
2x ln x
dx � 2�4
2
1ln x
1x dx 71.
� ln�2 �ln 2
2 0.3466
� �ln �22
� 0
A � ���4
0tan x dx � �ln�cos x��
��4
0
72.
� ln�3 � 2�2 �
� ln�2 � �2
2 � �2� � �ln�1 �
�22 � � ln�1 �
�22 �
� �ln�1 � cos x��3��4
��4
A � �3��4
��4
sin x1 � cos x
dx 73.
0 60
10
�152
� 8 ln 2 13.045 square units
� �x2
2� 4 ln x�
4
1� �8 � 4 ln 4� �
12
A � �4
1 x2 � 4
x dx � �4
1�x �
4x� dx
74.
00
9
6
� 3 � 4 ln 4 8.5452
� 4 � 4 ln 4 � 1
� �x � 4 ln x�4
1
A � �4
1 x � 4
x dx � �4
1�1 �
4x� dx 75.
0 40
10
�12�
ln�2 � �3 � 5.03041
�12�
ln�sec �
3� tan
�
3 � �12�
ln�1 � 0�
� �12�
ln�sec �x6
� tan �x6 ��2
0
�2
02 sec
�x6
dx �12��2
0sec��x
6 ��
6 dx
76.
0 5
−2
8
� �16 �103
ln cos�1.2�� � �1 �103
ln cos�0.3�� 11.7686
�4
1�2x � tan�0.3x�� dx � �x2 �
103
ln�cos�0.3x���4
1
Section 5.7 The Natural Logarithmic Function: Integration 477
77.
Trapezoid:
Simpson:
Calculator:
Exact: 12 ln 5
�5
1 12x
dx 19.3133
43�4� f �1� � 4f �2� � 2f �3� � 4f �4� � f �5� �
13
12 � 24 � 8 � 12 � 2.4 19.4667
42�4� f �1� � 2f �2� � 2f �3� � 2f �4� � f �5� �
12
12 � 12 � 8 � 6 � 2.4 � 20.2
f �x� �12x
, b � a � 5 � 1 � 4, n � 4
78.
Trapezoid:
Simpson:
Calculator:
Exact: 4 ln 5
�4
0
8xx2 � 4
dx 6.438
43�4� f �0� � 4f �1� � 2f �2� � 4f �3� � f �4� 6.4615
42�4� f �0� � 2f �1� � 2f �2� � 2f �3� � f �4� �
12
0 � 3.2 � 4 � 3.6923 � 1.6 6.2462
f �x� �8x
x2 � 4, b � a � 4 � 0 � 4, n � 4
79.
Trapezoid:
Simpson:
Calculator: �6
2 ln x dx 5.3643
43�4� f �2� � 4f �3� � 2f �4� � 4f �5� � f �6� 5.3632
42�4� f �2� � 2f �3� � 2f �4� � 2f �5� � f �6� �
12
0.6931 � 2.1972 � 2.7726 � 3.2189 � 1.7918 5.3368
f �x� � ln x, b � a � 6 � 2 � 4, n � 4
80.
Trapezoid:
Simpson:
Calculator: ���3
���3sec x dx 2.6339
2��33�4� � f ��
�
3� � 4f ���
6� � 2f �0� � 4f ��
6� � f ��
3�� 2.6595
2��32�4� � f ��
�
3� � 2f ���
6� � 2f �0� � 2f ��
6� � f ��
3�� �
122 � 2.3094 � 2 � 2.3094 � 2 2.780
f �x� � sec x, b � a ��
3� ��
�
3� �2�
3, n � 4
81. Power Rule 82. Substitution: andPower Rule
�u � x2 � 4� 83. Substitution: and Log Rule
�u � x2 � 4�
84. Substitution: and Log Rule
�u � tan x� 85.
� ln�sec x� � C
�ln�cos x� � C � ln� 1cos x� � C 86.
� �ln�csc x� � C
ln�sin x� � C � ln� 1csc x� � C
87.
� ln� 1sec x � tan x� � C � �ln�sec x � tan x� � C
� ln�sec2 x � tan2 xsec x � tan x � � Cln�sec x � tan x� � C � ln��sec x � tan x��sec x � tan x�
�sec x � tan x� � � C
478 Chapter 5 Integration
88.
� �ln� 1csc x � cot x� � C � ln�csc x � cot x� � C
� �ln�csc2 x � cot2 xcsc x � cot x � � C�ln�csc x � cot x� � C � �ln��csc x � cot x��csc x � cot x�
�csc x � cot x� � � C
89.
� �4�14
�12� � 1
� ��41x�
4
2
� 4�4
2x�2 dx
Average value �1
4 � 2 �4
2 8x2 dx 90.
� 2�ln 2 �14� � ln 4 �
12
1.8863
� 2�ln 4 �14
� ln 2 �12�
� 2�ln x �1x�
4
2
� 2�4
2�1
x�
1x2� dx
Average value �1
4 � 2 �4
2 4�x � 1�
x2 dx
91.
�1
2e � 2 0.291
�1
e � 1�12�
�1
e � 1��ln x�2
2 �e
1
Average value �1
e � 1�e
1
ln xx
dx 92.
�3�
ln�2 � �3�
�3�
ln�2 � �3� � ln�1 � 0�
� �12�
6�� ln�sec
�x6
� tan �x6 ��2
0
Average value �1
2 � 0�2
0sec
�x6
dx
93.
P�3� � 100012�ln 1.75� � 1 7715
� 100012 ln�1 � 0.25t� � 1
P�t� � 12,000 ln�1 � 0.25t� � 1000
C � 1000
P�0� � 12,000 ln�1 � 0.25�0�� � C � 1000
� 12,000 ln�1 � 0.25t� � C
P�t� � � 30001 � 0.25t
dt � �3000��4�� 0.251 � 0.25t
dt 94.
�10
ln 2�ln�43�� 4.1504 units of time
�10
ln 2�ln�T � 100��300
250�
10ln 2
ln 200 � ln 150
t �10
ln 2 �300
250
1T � 100
dT
95.
$168.27
1
50 � 40�50
40
90,000400 � 3x
dx � �3000 ln�400 � 3x��50
4096.
Solving this system yields and Thus,
S�t� �100 ln t
ln 2� 100 � 100� ln t
ln 2� 1�.
C � 100.k � 100�ln 2
S�4� � k ln 4 � C � 300
S�2� � k ln 2 � C � 200
S�t� � � kt
dt � k ln�t� � C � k ln t � C since t > 1.
dSdt
�kt
Section 5.7 The Natural Logarithmic Function: Integration 479
97. (a)
(c) In part (a):
Using a graphing utility the graphs intersect at The slopes are 3.295 and
respectively.�0.304 � ��1��3.295,�2.214, 1.344�.
y� �2xy
4x � 2yy� � 0
2x2 � y2 � 8
−10 10
−10
10
y2 � ��2x2 � 8
y1 � �2x2 � 8
y2 � 2x2 � 8
2x2 � y2 � 8 (b)
Let and graph
−10
−10 10
10
�y1 �2�x
, y2 � �2�x�y2 �
4x.k � 4
y2 � e���1�x� dx � e�ln x�C � eln�1�x��eC � �1x
k
98.
limk→0�
fk�x� � ln x
k � 0.1: f0.1�x� �10�x � 1
0.1� 10�10�x � 1�
k � 0.5: f0.5�x� ��x � 1
0.5� 2��x � 1�
x2 4 6 8 10
2
4
6
8
10f1
f0.5
f0.1
yk � 1: f1�x� � x � 1
99. False
�ln x�1�2
12
�ln x� � ln�x1�2�
100. False
ddx
ln x �1x
101. True
� ln�Cx�, C 0
� ln�x� � ln�C�
�1x
dx � ln�x� � C1
102. False; the integrandhas a nonremovablediscontinuity at x � 0.
103.
—CONTINUED—
5 10
0.5
1
y
x
f �x� �x
1 � x2(a) intersects
�12
ln 2 �14
� �12
ln�x2 � 1� �x2
4 �1
0
A � �1
0 �� x
1 � x2� �12
x� dx
x � 1
1 � x2 � 2
12
x �x
1 � x2
f �x� �x
1 � x2:y �12
x (b)
Hence, for the graphs of and enclose a finite region.y � mx
f0 < m < 1,
f ��0� � 1
f ��x� ��1 � x2� � x�2x�
�1 � x2�2 �1 � x2
�1 � x2�2
In part (b):
y� ��2yx2 �
�2yy2x 2 �
�2y4x
��y2x
2yy� ��4x2
y2 �4x
� 4x�1
480 Chapter 5 Integration
103. —CONTINUED—
(c)
�12
[m � ln�m� � 1�
�12
ln� 1m� �
12
�1 � m�
�12
ln�1 �1 � m
m � �12
m�1 � mm �
� �12
ln�1 � x2� �mx2
2 ���1�m�m
0
A � ��1�m�m
0 � x
1 � x2 � mx� dx, 0 < m < 1
x ��1 � mm
, intersection point
x2 �1 � m
m
1 � m � mx2
1 − mm
0.5
x
y
f (x) =
y = mx
xx2 + 1
x
1 � x2 � mx
104.
Alternate Solution:
� ln 2
� ln 2 � ln x � ln x
F�x� � ln t�2x
x� ln�2x� � ln x
F� �x� �12x
�2� �1x
� 0 ⇒ F is constant on �0, ��.
F�x� � 2x
x
1t dt, x > 0
Section 5.8 Inverse Trigonometric Functions: Integration
1. 5
�9 � x2 dx � 5 arcsin�x
3� � C 2. 3
�1 � 4x2 dx �
3
2 2
�1 � 4x2 dx �
3
2 arcsin�2x� � C
3. 716 � x2 dx �
74
arctan�x4� � C 4. 4
1 � 9x2 dx �43 3
1 � 9x2 dx �43
arctan�3x� � C
5. 1
x�4x2 � 1 dx � 2
2x��2x�2 � 1 dx � arcsec�2x� � C 6. 1
4 � �x � 1�2 dx �12
arctan�x � 12 � � C
7. (Use long division.) x3
x2 � 1 dx � �x �
xx2 � 1� dx � x dx �
12 2x
x2 � 1 dx �
12
x2 �12
ln�x2 � 1� � C
8. x4 � 1x2 � 1
dx � �x2 � 1� dx �13
x3 � x � C 9. 1
�1 � �x � 1�2 dx � arcsin�x � 1� � C
105. implies that
The second formula follows by the Chain Rule.
1x dx � ln�x� � C.
ddx
ln�x� �1x
Section 5.8 Inverse Trigonometric Functions: Integration 481
10. Let
tt 4 � 16
dt �12 1
�4�2 � �t 2�2�2t� dt �18
arctan t2
4� C
u � t 2, du � 2t dt. 11. Let
t
�1 � t4 dt �
12 1
�1 � �t2�2�2t� dt �
12
arcsin�t 2� � C
u � t 2, du � 2t dt.
12. Let
�14
arcsec x2
2� C
1
x�x4 � 4 dx �
12 1
x2��x2�2 � 22�2x� dx
u � x2, du � 2x dx. 13. Let
e2x
4 � e4x dx �
12 2e2x
4 � �e2x�2 dx �
14
arctan e2x
2� C
u � e2x, du � 2e2x dx.
14.
�1�3
arctan�x � 2�3 � � C
1
3 � �x � 2�2 dx � 1
��3 �2 � �x � 2�2 dx 15.
� 2 arcsin �x � C
1
u�1 � u2�2u du� � 2 du
�1 � u2� 2 arcsin u � C
1�x �1 � x
dx, u � �x, x � u2, dx � 2u du
16.
� 3 arctan�x � C
32 2u du
u�1 � u2� � 3 du1 � u2 � 3 arctan u � C
3
2�x�1 � x� dx, u � �x, du �
1
2�x dx, dx � 2u du 17.
�12
ln�x2 � 1� � 3 arctan x � C
x � 3x2 � 1
dx �12 2x
x2 � 1 dx � 3 1
x2 � 1 dx
18. � �4�1 � x2 � 3 arcsin x � C 4x � 3�1 � x2
dx � ��2� �2x�1 � x2
dx � 3 1�1 � x2
dx
19.
� ��6x � x2 � 8 arcsin�x3
� 1� � C � ��9 � �x � 3�2 � 8 arcsin�x � 33 � � C
x � 5�9 � �x � 3�2
dx � �x � 3��9 � �x � 3�2
dx � 8�9 � �x � 3�2
dx
20.
�12
ln�x2 � 2x � 5� �32
arctan�x � 12 � � C
x � 2�x � 1�2 � 4
dx �12 2x � 2
�x � 1�2 � 4 dx � 3
�x � 1�2 � 4 dx
21. Let .
� �13
arcsin�3x��16
0�
�
18
16
0
1
�1 � 9x2 dx �
1316
0
1
�1 � �3x�2�3� dx
du � 3 dxu � 3x, 22. 1
0
1
�4 � x2 dx � �arcsin
x2�
1
0�
�
6
23. Let
� �12
arctan�2x���32
0�
�
6
�32
0
11 � 4x2 dx �
12
�32
0
21 � �2x�2 dx
du � 2 dx.u � 2x, 24. 3
�3
19 � x2 dx � �1
3 arctan
x3�
3
�3�
�
36
482 Chapter 5 Integration
25. Let
1�2
0
arcsin x
�1 � x2 dx � �1
2 arcsin2 x�
1�2
0�
�2
32� 0.308
u � arcsin x, du �1
�1 � x2 dx. 26. Let
� ��12
arccos2 x�1�2
0�
3� 2
32� 0.925
1�2
0
arccos x
�1 � x2 dx � �1�2
0 �arccos x
�1 � x2 dx
u � arccos x, du � �1
�1 � x2 dx.
27. Let
� �0.134
� ���1 � x2�0
�12�
�3 � 22
0
�12
x
�1 � x2 dx � �
120
�12 �1 � x2��12��2x� dx
u � 1 � x2, du � �2x dx. 28. Let
� �12
ln�1 � x2��0
��3� �ln 2
0
��3
x1 � x2 dx �
120
��3
11 � x2�2x� dx
u � 1 � x2, du � 2x dx.
29. Let
� ��arctan�cos x���
�2�
�
4
�
�2
sin x1 � cos2 x
dx � ��
�2
�sin x1 � cos2 x
dx
u � cos x, du � �sin x dx. 30. �2
0
cos x1 � sin2 x
dx � arctan�sin x���2
0�
�
4
31.
� �arctan�x � 1��2
0�
�
2
2
0
dxx2 � 2x � 2
� 2
0
11 � �x � 1�2 dx 32.
�13
arctan�43�
� �13
arctan�x � 23 ��
2
�2
2
�2
dxx2 � 4x � 13
� 2
�2
dx�x � 2�2 � 9
33.
� ln�x2 � 6x � 13� � 3 arctan�x � 32 � � C
2xx2 � 6x � 13
dx � 2x � 6x2 � 6x � 13
dx � 6 1x2 � 6x � 13
dx � 2x � 6x2 � 6x � 13
dx � 6 14 � �x � 3�2 dx
34. 2x � 5x2 � 2x � 2
dx � 2x � 2x2 � 2x � 2
dx � 7 11 � �x � 1�2 dx � ln�x2 � 2x � 2� � 7 arctan�x � 1� � C
35.
� arcsin�x � 22 � � C
1
��x2 � 4x dx � 1
�4 � �x � 2�2 dx 36.
� 2 arcsin�x � 22 � � C
� 2�4 � �x � 2�2
dx
2��x2 � 4x
dx � 2�4 � �x2 � 4x � 4�
dx
37. Let
� ���x2 � 4x � C
x � 2
��x2 � 4x dx � �
12��x2 � 4x��12��2x � 4� dx
u � �x2 � 4x, du � ��2x � 4� dx. 38. Let
� �x2 � 2x � C
x � 1
�x2 � 2x dx �
12�x2 � 2x��12�2x � 2� dx
u � x2 � 2x, du � �2x � 2� dx.
Section 5.8 Inverse Trigonometric Functions: Integration 483
39.
� ��2�4x � x2 � arcsin�x � 22 ��
3
2� 4 � 2�3 �
�
6� 1.059
3
2
2x � 3
�4x � x2 dx � 3
2
2x � 4
�4x � x2 dx � 3
2
1
�4x � x2 dx � �3
2�4x � x2��12�4 � 2x� dx � 3
2
1
�4 � �x � 2�2 dx
40.
� arcsec�x � 1� � C
1
�x � 1��x2 � 2x dx � 1
�x � 1���x � 1�2 � 1 dx 41. Let
�12
arctan�x2 � 1� � C
xx 4 � 2x2 � 2
dx �12 2x
�x2 � 1�2 � 1 dx
u � x2 � 1, du � 2x dx.
42. Let
x
�9 � 8x2 � x 4 dx �
12 2x
�25 � �x2 � 4�2 dx �
12
arcsin� x2 � 45 � � C
u � x2 � 4, du � 2x dx.
43. Let Then and
� 2u � 2�3 arctan u
�3� C � 2�et � 3 � 2�3 arctan �et � 3
3� C
�et � 3 dt � 2u2
u2 � 3 du � 2 du � 6
1u2 � 3
du
2u duu2 � 3
� dt.u2 � 3 � et, 2u du � et dt,u � �et � 3.
44. Let
� 2u �6
�3 arctan
u
�3� C � 2�x � 2 � 2�3 arctan �x � 2
3� C
�x � 2x � 1
dx � 2u2
u2 � 3 du � 2u2 � 6 � 6
u2 � 3 du � 2du � 6 1
u2 � 3 du
2u du � dx.u2 � 2 � x,u � �x � 2,
45.
Let
� 2��
3�
�
4� ��
6
� 2 arctan�u���3
1
�3
1
2u duu�1 � u2� � �3
1
21 � u2 du
u � �x, u2 � x, 2u du � dx, 1 � x � 1 � u2.
3
1
dx�x �1 � x�
46.
Let
��
4�
�
6�
�
12
� arcsin��22 � � arcsin�1
2�
� arcsin�u2��
�2
1
�2
1
2u du
2�4 � u2 u� �2
1
du�4 � u2
�3 � x � �4 � u2.2u du � dx,u2 � x � 1,u � �x � 1,
1
0
dx
2�3 � x�x � 1
47. (a)
(b)
(c) cannot be evaluated using the basic
integration rules.
1
x�1 � x2 dx
x
�1 � x2 dx � ��1 � x2 � C, u � 1 � x2
1
�1 � x2 dx � arcsin x � C, u � x 48. (a) cannot be evaluated using the basic
integration rules.
(b)
(c) 1x2 e1x dx � �e1x � C, u �
1x
xex2 dx �
12
ex2� C, u � x2
ex2 dx
484 Chapter 5 Integration
49. (a)
(b) Let Then and
(c) Let Then and
Note: In (b) and (c), substitution was necessary before the basic integration rules could be used.
�23
u�u2 � 3� � C �23�x � 1�x � 2� � C x
�x � 1 dx � u2 � 1
u�2u� du � 2�u2 � 1� du � 2�u3
3� u� � C
dx � 2u du.x � u2 � 1u � �x � 1.
�2
15 u3�3u2 � 5� � C �
215
�x � 1�32 3�x � 1� � 5� � C �2
15�x � 1�32�3x � 2� � C
x�x � 1 dx � �u2 � 1��u��2u� du � 2�u4 � u2� du � 2�u5
5�
u3
3 � � C
dx � 2u du.x � u2 � 1u � �x � 1.
�x � 1 dx �23
�x � 1�32 � C, u � x � 1
50. (a) cannot be evaluated using the basic
integration rules.
(b)
(c)
�14
ln�1 � x4� � C, u � 1 � x4
x3
1 � x4 dx �14 4x3
1 � x 4 dx
�12
arctan�x2� � C, u � x2
x1 � x4 dx �
12 2x
1 � �x2�2 dx
11 � x4 dx 51.
Matches (c).
x1−1
2
32
12
12
−
yArea � �1��1� � 1
52. No. This integral does not correspond to any of the basic differentiation rules.
53.
y � arcsin�x2� � �
y�0� � � � C
y � 1
�4 � x2 dx � arcsin�x
2� � C
y� �1
�4 � x2, �0, �) 54.
y �12
arctan�x2� �
7�
8
��
8� C ⇒ C �
7�
8
� �12
arctan�22� � C
y � 1
4 � x2 dx �12
arctan x2
� C
y� �1
4 � x2, �2, ��
55. (a)
x
y
5
5
−5
−5(0, 0)
(b)
y � 3 arctan x
�0, 0�: 0 � 3 arctan�0� � C ⇒ C � 0
y � 3 dx
1 � x2 � 3 arctan x � C 8−8
32�−
32�
dydx
�3
1 � x2, �0, 0�
Section 5.8 Inverse Trigonometric Functions: Integration 485
56. (a)
4
−4
4
x
y (b)
y �23
arctan�x3� � 2
2 � C
y � 2
9 � x2 dx �23
arctan�x3� � C
5
4
−1
−4
y� �2
9 � x2, �0, 2�
57. (a)
−1 1 4
−2
−3
−4
1
2
3
4
x
y (b)
y �12
arcsec x2
� 1, x ≥ 2
1 �12
arcsec�1� � C � C
y � 1
x�x2 � 4 dx �
12
arcsec �x�2
� C
4
4
−4
−4
y� �1
x�x2 � 4, �2, 1�
58. (a)
x
y
−4 4
−4
4
(b)
y � 2 arcsin�x5�
� � 2 arcsin�1� � C ⇒ C � 0
y � 2
�25 � x2 dx � 2 arcsin�x
5� � C
5
−5
−5 5
y� �2
�25 � x2, �5, ��
59.
−6 12
−8
4
dydx
�10
x�x2 � 1, y�3� � 0 60.
4
6
−4
−6
dydx
�1
12 � x2, y�4� � 2
61.
−3
−1
3
3
dy
dx�
2y�16 � x2
, y�0� � 2 62.
7
60
−1
dydx
��y
1 � x2, y�0� � 4
486 Chapter 5 Integration
63.
��
8
�12
arctan�1� �12
arctan�0�
�12
arctan�x � 12 ��
3
1
A � 3
1
1x2 � 2x � 5
dx � 3
1
1�x � 1�2 � 4
dx 64.
��
4
� arctan�1� � arctan�0�
� arctan�x � 22 ��
0
�2
Area � 0
�2
2x2 � 4x � 8
dx � 0
�2
2�x � 2�2 � 4
dx
65.
��
6
� arcsin�12� � arcsin�0�
� arcsin�x2��
1
0
x
1
112
y
Area � 1
0
1�4 � x2
dx 66.
��
3�
�
4�
�
12
� arcsec�2� � arcsec� 2�2�
� arcsec x�2
2�2
Area � 2
2�2
1
x�x2 � 1 dx
67.
�3�
4�
3�
4�
3�
2
� 3 arctan�1� � 3 arctan��1�
� 3 arctan�sin x���2
��2
Area � �2
��2
3 cos x1 � sin2 x
dx � 3�2
��2
11 � sin2 x
�cos x dx� 68.
��
3�
�
4�
�
12
� arctan��3 � � arctan�1�
� arctan�ex��ln
�3
0
A � ln��3 �
0
ex
1 � e2x dx, �u � ex�
69. (a)
Thus,
(b)
�12
ln 3 �12
ln 2 ���3
9�
�
4� 0.3835
� �ln �3 �12
ln�4� �arctan �3
�3 � � ��12
ln 2 � arctan�1��
� �ln x �12
ln�1 � x2� �arctan x
x ��3
1
A � �3
1 arctan x
x2 dx
arctan x
x2 dx � ln x �12
ln�1 � x2� �arctan x
x� C.
�1 � x2 � x2
x�1 � x2� �1
x�1 � x2� �arctan x
x2 �arctan x
x2
ddx�ln x �
12
ln�1 � x2� �arctan x
x� C� �
1x
�x
1 � x2 � � x 1�1 � x2�� � arctan x x2 �
70. (a)
(b) �� 2
4� 2 � 0.4674� ���
2�2
� 2� � �0�� �x�arcsin x�2 � 2x � 2�1 � x2 arcsin x�1
0 A � 1
0�arcsin x�2 dx
� �arcsin x�2 � 2x�arcsin x� 1�1 � x2
� 2 �2x
�1 � x2 arcsin x � 2�1 � x2 1
�1 � x2� �arcsin x�2
ddx
x�arcsin x�2 � 2x � 2�1 � x2 arcsin x � C�
Section 5.8 Inverse Trigonometric Functions: Integration 487
71. (a)
Shaded area is given by
(b) 1
0 arcsin x dx � 0.5708
1
0arcsin x dx.
1 2
1
2
2
y
x
π
(c) Divide the rectangle into two regions.
Hence, 1
0arcsin x dx �
�
2� 1, ��0.5708�.
� 1
0arcsin x dx � 1
�
2� 1
0arcsin x dx � ��cos y��
�2
0
Area rectangle � 1
0arcsin x dx � �2
0 sin y dy
Area rectangle � �base��height� � 1��
2� ��
2
72. (a)
(b) Let
(c) 3.1415927
41
0
11 � x2 dx � 4� 1
18��1 �4
1 � �136� �2
1 � �19� �4
1 � �14� �2
1 � �49� �4
1 � �2536� �12� � 3.1415918
n � 6.
1
0
41 � x2 dx � �4 arctan x�
1
0� 4 arctan 1 � 4 arctan 0 � 4��
4� � 4�0� � �
73.
(a) represents the average value of over the interval Maximum at since the graph is greatest on
(b)
when x � �1.F��x� �1
1 � �x � 2�2 �1
1 � x2 ��1 � x2� � �x2 � 4x � 5�
�x2 � 1��x2 � 4x � 5� ��4�x � 1�
�x2 � 1��x2 � 4x � 5� � 0
F�x� � �arctan t�x�2
x� arctan�x � 2� � arctan x
�1, 1�.x � �1, x, x � 2�.f �x�F�x�
F�x� �12x�2
x
2t 2 � 1
dt
74.
(a)
(b)
� 2 arcsin� u
�6� � C � 2 arcsin��x
�6� � C
1
�6u2 � u4�2u du� � 2
�6 � u2 du
u � �x, u2 � x, 2u du � dx
1
�6x � x2 dx � dx
�9 � �x � 3�2� arcsin�x � 3
3 � � C
6x � x2 � 9 � �x2 � 6x � 9� � 9 � �x � 3�2
1
�6x � x2 dx
(c)
The antiderivatives differ by a constant,
Domain: 0, 6�
�2.
7
−2
−1
y1
y2
4
75. False, dx
3x�9x2 � 16�
112
arcsec �3x�4
� C 76. False, dx25 � x2 dx �
15
arctan x5
� C
77. True
ddx��arccos
x2
� C� �12
�1 � �x2�2�
1�4 � x2
78. False. Use substitution: u � 9 � e2x, du � �2e2x dx
488 Chapter 5 Integration
79.
Thus, du
�a2 � u2� arcsin�u
a� � C.
ddx�arcsin�u
a� � C� �1
�1 � �u2a2��u�
a � �u�
�a2 � u280.
Thus, dua2 � u2 � u�
a2 � u2 dx �1a
arctan ua
� C.
�1a2� u�
�a2 � u2�a2� �u�
a2 � u2
ddx�
1a
arctan ua
� C� �1a�
u�a1 � �ua�2�
81. Assume
The case is handled in a similar manner. Thus,
du
u�u2 � a2� u�
u�u2 � a2 dx �
1a
arcsec �u�a
� C.
u < 0
ddx�
1a
arcsec ua
� C� �1a�
u�a
�ua���ua�2 � 1� �1a�
u�
u��u2 � a2�a2� �u�
u�u2 � a2.
u > 0.
82. (a)
(b) Trapezoidal Rule:
(c) Because
you can use the Trapezoidal Rule to approximate and hence, For example, using you obtain
� � 4�0.785397� � 3.141588.
n � 200,�.�4,
1
0
11 � x2 dx � arctan x�
1
0�
�
4,
A � 0.7847
n � 8, b � a � 1 � 0 � 1
A � 1
0
11 � x2 dx
83. (a)
(c)
When andwe have
—CONTINUED—
v�t� ��32k
tan�arctan�500� k32� � �32k t�.
C � arctan�500�k32 �,v � 500,t � 0,
v ��32k
tan�C � �32k t�
� k32
v � tan�C � �32k t�
arctan�� k32
v� � ��32k t � C
1
�32k arctan�� k
32v� � �t � C1
132 � kv2 dv � � dt
0 200
550
v�t� � �32t � 500 (b)
When the object reaches its maximum height,
(d) When
v�t� � 0 when t0 � 6.86 sec.
00
7
500
v(t� ��32,000 tan arctan�500�0.00003125� ��0.032 t�k � 0.001:
� 3906.25 ft �Maximum height�
s�15.625� � �16�15.625�2 � 500�15.625�
t � 15.625
�32t � �500
v�t� � �32t � 500 � 0
v�t� � 0.
s�t� � �16t2 � 500t
s�0� � �16�0� � 500�0� � C � 0 ⇒ C � 0
� �16t2 � 500t � C
s�t� � v�t� dt � ��32t � 500� dt
Section 5.9 Hyperbolic Functions 489
83. —CONTINUED—
(e)
Simpson’s Rule: feet
(f) Air resistance lowers the maximum height.
h � 1088n � 10;
h � �6.86
0
�32,000 tan�arctan�500�0.00003125 � � �0.032 t� dt
84.
Since and is increasing for
for Thus,
Since and g is increasing for for Thus, Therefore,x
1 � x2 < arctan x < x.
x > arctan x.x > 0.x � arctan x > 0x > 0,g�0� � 0
g� �x� � 1 �1
1 � x2 �x2
1 � x2 > 0 for x > 0.
Let g�x� � x � arctan x
arctan x > x
1 � x2.x > 0.arctan x �x
1 � x2 > 0
x > 0,ff �0� � 0
f � �x� �1
1 � x2 �1 � x2
�1 � x2�2 �2x2
�1 � x2� > 0 for x > 0.
x2 4 6 8 10
1
3
5
4
2
y
y
y3
2
1
yLet f �x� � arctan x �x
1 � x2
Section 5.9 Hyperbolic Functions
1. (a)
(b) tanh��2� �sinh��2�cosh��2� �
e�2 � e2
e�2 � e2 � �0.964
sinh 3 �e3 � e�3
2� 10.018 2. (a)
(b) sech 1 �2
e � e�1 � 0.648
cosh 0 �e0 � e0
2� 1
3. (a)
(b)
�5 � �1�5�5 � �1�5� �
1312
coth�ln 5� �cosh�ln 5�sinh�ln 5� �
eln 5 � e�ln 5
eln 5 � e�ln 5
csch�ln 2� �2
eln 2 � e�ln 2 �2
2 � �1�2� �43
4. (a)
(b) tanh�1 0 � 0
sinh�1 0 � 0
5. (a)
(b) sech�1 23
� ln1 � �1 � �4�9�2�3 � 0.962
cosh�1 2 � ln�2 � �3 � � 1.317 6. (a)
(b) coth�1 3 �12
ln42 � 0.347
csch�1 2 � ln1 � �52 � 0.481
7. �e2x � 2 � e�2x � 4
�ex � e�x�2 �e2x � 2 � e�2x
e2x � 2 � e�2x � 1tanh2 x � sech2 x � ex � e�x
ex � e�x2
� 2ex � e�x
2
8.1 � cosh 2x
2�
1 � �e2x � e�2x��22
�e2x � 2 � e�2x
4� ex � e�x
2 2
� cosh2 x
490 Chapter 5 Integration
9.
�14
�2�ex�y � e��x�y��� �e�x�y� � e��x�y�
2� sinh�x � y�
�14
�ex�y � e�x�y � ex�y � e��x�y� � ex�y � e�x�y � ex�y � e��x�y��
sinh x cosh y � cosh x sinh y � ex � e�x
2 ey � e�y
2 � ex � e�x
2 ey � e�y
2
10. 2 sinh x cosh x � 2ex � e�x
2 ex � e�x
2 �e2x � e�2x
2� sinh 2x
11.
�12
�e3x � e�x � ex � ex � e�3x � e�x� �e3x � e�3x
2� sinh�3x�
� ex � e�x
2 �3 � e2x � 2 � e�2x� �12
�ex � e�x��e2x � e�2x � 1�
3 sinh x � 4 sinh3 x � sinh x�3 � 4 sinh2 x� � ex � e�x
2 �3 � 4ex � e�x
2 2
�
12.
� cosh x � cosh y
� 2�ex � ey � e�y � e�x
4 � �ex � e�x
2�
ey � e�y
2
2 cosh x � y
2 cosh
x � y2
� 2�e�x�y��2 � e��x�y��2
2 ��e�x�y��2 � e��x�y��2
2 �
13.
coth x �1
3��13�
�133
sech x �1
�13�2�
2�1313
csch x �1
3�2�
23
tanh x �3�2
�13�2�
3�1313
cosh2 x � 32
2
� 1 ⇒ cosh2 x �134
⇒ cosh x ��13
2
sinh x �32
14.
csch x �1
�3�3� �3
sinh x � tanh x cosh x � 122�3
3 ��33
coth x �1
1�2� 2
cosh x �1
�3�2�
2�33
12
2
� sech2 x � 1 ⇒ sech2 x �34
⇒ sech x ��32
tanh x �12
Putting these in order:
tanh x �12
coth x � 2
cosh x �2�3
3 sech x �
�32
sinh x ��33
csch x � �3
Section 5.9 Hyperbolic Functions 491
21.
h� �x� �12
cosh�2x� �12
�cosh�2x� � 1
2� sinh2 x
h�x� �14
sinh 2x �x2
22.
h� �t� � 1 � csch2 t � coth2 t
h�t� � t � coth t
23.
f � �t� �1
1 � sinh2 t�cosh t� �
cosh tcosh2 t
� sech t
f �t� � arctan�sinh t� 24.
� �6 sech2 3x tanh 3x
g� �x� � �2 sech�3x� sech�3x� tanh�3x��3�
g�x� � sech2 3x
15.
y� � �sech�x � 1� tanh�x � 1�
y � sech�x � 1� 16.
y� � �3 csch2�3x�
y � coth 3x 17.
f � �x� �1
sinh x�cosh x� � coth x
f �x� � ln�sinh x�
18.
� tanh x
g� �x� �1
cosh x�sinh x�
g�x� � ln�cosh x� 19.
�1
sinh x� csch x
�1
2 sinh�x�2� cosh�x�2�
y� �1�2
tanh�x�2� sech2x2
y � lntanh x2 20.
� x sinh x
y� � x sinh x � cosh x � cosh x
y � x cosh x � sinh x
25.
Tangent line:
y � �2x � 2
y � 0 � �2�x � 1�
y� �1� � �2
y� � cosh�1 � x2���2x�
y � sinh�1 � x2�, �1, 0� 26.
At
Tangent line:
Note: cosh�1� � 1.5431
y � cosh�1�x � cosh�1� � 1
y � 1 � cosh�1��x � 1�
�1, 1�, y� � cosh�1�.
y�
y�
cosh xx
� sinh x ln x
ln y � cosh x ln x
y � x cosh x, �1, 1�
27.
At
Tangent line:
y � �2x � 1
y � 1 � �2�x � 0�
�0, 1�, y� � 2�1���1� � �2.
y� � 2�cosh x � sinh x��sinh x � cosh x�
y � �cosh x � sinh x�2, �0, 1� 28.
Tangent line:
y � x � 1
y � 1 � 1�x � 0�
y� �0� � e0�1� � 1
y� � esinh x cosh x
y � esinh x, �0, 1�
29.
Relative maxima:
Relative minimum: �0, �1�
�±�, cosh ��
� 2 sin x cosh x � 0 when x � 0, ±�.
f � �x� � sin x cosh x � cos x sinh x � cos x sinh x � sin x cosh x
�−2
− 2
(0, 1)−
( , cosh )− ππ ( , cosh )π π12
�2
f �x� � sin x sinh x � cos x cosh x, �4 ≤ x ≤ 4
492 Chapter 5 Integration
30.
By the First Derivative Test,is a relative minimum.
�0, �cosh��1�� � �0, �1.543�
f � �x� � 0 for x � 0.
f � �x� � x cosh�x � 1� � sinh�x � 1� � sinh�x � 1� � x cosh�x � 1�
(0, −1.543)
−6
−2
6
6 f �x� � x sinh�x � 1� � cosh�x � 1�
31.
Using a graphing utility,
By the First Derivative Test, is a relative maximum and is a relative minimum.
−1
(1.20, 0.66)
( 1.20, 0.66)− −
1
��−
��1.1997, �0.6627��1.1997, 0.6627�
x � ±1.1997.
x tanh x � 1
� sech x�1 � x tanh x� � 0
g� �x� � sech x � x sech x tanh x
g�x� � x sech x 32.
Using a graphing utility,
From the First Derivative Test, is a relative maximum and is a relativeminimum.
−3 3
−2
(0.88, 0.53)
( 0.88, 0.53)− −
2
��0.8814, �0.5328��0.8814, 0.5328�
x � 0.8814.
sech2 x �12
h� �x� � 2 sech2 x � 1 � 0
h�x� � 2 tanh x � x
35.
f
P1
P2
3
−2
−3
2
P2�x� � f �0� � f ��0��x � 0� �12 f � �0��x � 0�2 � x
P1�x� � f �0� � f ��0��x � 0� � x
f � �x� � �2 sech2 x tanh x, f � �0� � 0
f ��x� � sech2 x, f ��0� � 1
f �x� � tanh x, f �0� � 0 36.
−2 20
f
P1
P2
3
P2�x� � 1 �12x2
P1�x� � f �0� � f � �0��x � 0� � 1
f � �x� � cosh x f � �0� � cosh�0� � 1
f � �x� � sinh x f � �0� � sinh�0� � 0
f �x� � cosh x f �0� � cosh�0� � 1
37. (a)
2010
10
20
30
−10
y
x
y � 10 � 15 cosh x
15, �15 ≤ x ≤ 15
33.
Therefore, y��� � y� � 0.
y��� � a cosh x
y� � a sinh x
y� � a cosh x
y � a sinh x 34.
Therefore, y� � y � 0.
y� � a cosh x
y� � a sinh x
y � a cosh x
(b) At
At
(c) At x � 15, y� � sinh�1� � 1.175.y� � sinh x
15.
x � 0, y � 10 � 15 cosh�0� � 25.
x � ±15, y � 10 � 15 cosh�1� � 33.146.
Section 5.9 Hyperbolic Functions 493
38. (a)
−25
−10
80
25
y � 18 � 25 cosh x
25, �25 ≤ x ≤ 25 (b) At
At
(c) At x � 25, y� � sinh�1� � 1.175.y� � sinh x
25.
x � 0, y � 18 � 25 � 43.
x � ±25, y � 18 � 25 cosh�1� � 56.577.
39. Let
� �12
cosh�1 � 2x� � C
�sinh�1 � 2x� dx � �12�sinh�1 � 2x���2� dx
u � 1 � 2x, du � �2 dx. 40. Let
�cosh �x
�x dx � 2�cosh �x 1
2�x dx � 2 sinh�x � C
u � �x, du �1
2�x dx.
41. Let
�cosh2�x � 1� sinh�x � 1� dx �13
cosh3�x � 1� � C
u � cosh�x � 1�, du � sinh�x � 1� dx. 42. Let
� �sech x � C
� sinh1 � sinh2 x
dx � � sinh xcosh2 x
dx ��1
cosh x� C
u � cosh x, du � sinh x dx.
45. Let
�x csch2 x2
2 dx � �csch2
x2
2 x dx � �coth x2
2� C
u �x2
2, du � x dx. 46. Let
� �13
sech3 x � C
�sech3 x tanh x dx � ��sech2 x��sech x tanh x� dx
u � sech x, du � �sech x tanh x dx.
47. Let
� csch 1x
� C
�csch�1�x� coth�1�x�x2 dx � ��csch
1x coth
1x�
1x2 dx
u �1x, du � �
1x2 dx. 48. Let
� arcsinex � e�x
6 � C
� cosh x
�9 � sinh2 x dx � arcsinsinh x
3 � C
u � sinh x, du � cosh x dx.
49. Let
� xx4 � 1
dx �12� 2x
�x2�2 � 1 dx �
12
arctan�x2� � C
u � x2, du � 2x dx. 50.
� 2 ln�1 � �1 � 4x2
2x � C
� 2
x�1 � 4x2 dx � 2� 1
�2x��1 � �2x�2�2� dx
43. Let
�cosh xsinh x
dx � ln sinh x � C
u � sinh x, du � cosh x dx. 44. Let
�12
tanh�2x � 1� � C
�sech2�2x � 1� dx �12�sech2�2x � 1��2� dx
u � 2x � 1, du � 2 dx.
494 Chapter 5 Integration
51.
Note: cosh�ln 2� �eln 2 � e�ln 2
2�
2 � �1�2�2
�54
� ln54 � 0 � ln5
4 � ln�cosh�ln 2� � ln�cosh�0��
� ln�cosh x��ln 2
0
�ln 2
0tanh x dx � �ln 2
0 sinh xcosh x
dx, �u � cosh x� 52.
�12
�12
sinh�1� cosh�1�
�12�1 �
12
sinh�2��
�12�x �
12
sinh�2x��1
0
�1
0 cosh2 x dx � �1
0 1 � cosh�2x�
2 dx
53.
� � 110
ln 5 � x5 � x �4
0�
110
ln 9 �15
ln 3
�4
0
125 � x2 dx �
110� 1
5 � x dx �
110� 1
5 � x dx 54. �4
0
1
�25 � x2 dx � �arcsin
x5�
4
0� arcsin
45
55. Let
� �arcsin�2x���2�4
0�
�
4
��2�4
0
2�1 � 4x2
dx � ��2�4
0
1�1 � �2x�2
�2� dx
u � 2x, du � 2 dx. 56.
�38
� ln 2
� �ln 2 �12
14� � �0 �
12�
� �x �12
e�2x�ln 2
0
�ln 2
02e�x cosh x dx � �ln 2
0�1 � e�2x� dx
2e�x cosh x � 2e�x�ex � e�x
2 � � 1 � e�2x
57.
y� �3
�9x2 � 1
y � cosh�1�3x� 58.
y� �1
1 � �x�2�212 �
24 � x2
y � tanh�1 x2 59.
y� �1
�tan2 x � 1�sec2 x� � sec x
y � sinh�1�tan x�
60.
since for 0 < x < ��4.sin 2x ≥ 0
y� ��1
cos 2x�1 � cos2 2x��2 sin 2x� �
2 sin 2xcos 2x sin 2x �
2cos 2x
� 2 sec 2x,
y � sech�1�cos 2x�, 0 < x < �
4
61.
y� �1
1 � sin2 2x�2 cos 2x� � 2 sec 2x
y � tanh�1�sin 2x� 62.
y� � 2 csch�1 x �1
x �1 � x2 ��2 csch�1 x
x �1 � x2
y � �csch�1 x�2
63.
� 2 sinh�1�2x�
y� � 2x 2
�1 � 4x2 � 2 sinh�1�2x� �4x
�1 � 4x2
y � 2x sinh�1�2x� � �1 � 4x2 64.
y� � x 11 � x2 � tanh�1 x �
�x1 � x2 � tanh�1 x
y � x tanh�1 x � ln�1 � x2 � x tanh�1 x �12
ln�1 � x2�
Section 5.9 Hyperbolic Functions 495
65. Answers will vary. 66. See the definitions and graphs inthe textbook.
67. limx→�
sinh x � �
68. limx→�
tanh x � 1 69. limx→�
sech x � 0 70. limx→��
csch x � 0
71. limx→0
sinh x
x� lim
x→0 ex � e�x
2x� 1 72. does not exist.
�coth x →� for x → 0�, coth x → �� for x → 0��
limx→0
coth x
73.
� �ln1 � �1 � e2x
ex � C
� 1
�1 � e2x dx � � ex
ex�1 � �ex�2 dx � �csch�1�ex� � C 74.
� �1
12 ln 3 � x2
3 � x2 � C
� �12
16 ln 3 � x2
3 � x2 � C
� x9 � x4 dx � �
12� �2x
9 � �x2�2 dx
77. � �14x � x2 dx � � 1
�x � 2�2 � 4 dx �
14
ln �x � 2� � 2�x � 2� � 2 �
14
ln x � 4x � C
78. � �12
ln2 � ��x � 2�2 � 4
x � 2 � C � dx
�x � 2��x2 � 4x � 8� � dx
�x � 2���x � 2�2 � 4
79.
�1
2�6 ln �2 �x � 1� � �3
�2 �x � 1� � �3 � C �1�2
�1
2�3 ln �3 � �2 �x � 1�
�3 � �2 �x � 1� � C
�1�2
� �2
��3 �2 � ��2�x � 1��2 dx
� 11 � 4x � 2x2 dx � � 1
3 � 2�x � 1�2 dx
80.
�1�2
� 1
�x � 1���x � 1�2 � ��3 �2dx � �
1
�6 ln�3 � ��x � 1�2 � 3
x � 1 � C
� 1
�x � 1��2x2 � 4x � 8 dx � � 1
�x � 1��2�x � 1�2 � 6 dx
75. Let
� 1
�x�1 � x dx � 2� 1
�1 � ��x �2 1
2�x dx � 2 sinh�1�x � C � 2 ln��x � �1 � x � � C
u � �x, du �1
2�x dx.
76. Let
� �x
�1 � x3 dx �
23� 1
�1 � �x3�2�232�x dx �
23 sinh�1�x3�2� � C �
23
ln�x3�2 � �1 � x3 � � C
u � x3�2, du �32�x dx.
81. Let
�14� 4
�81 � �4x � 1�2 dx �
14
arcsin4x � 19 � C y � � 1
�80 � 8x � 16x2 dx
u � 4x � 1, du � 4 dx.
496 Chapter 5 Integration
82. Let
� �1
�3 ln �3 � ��4x2 � 8x � 1
2�x � 1� � C � � 2
2�x � 1����3 �2 � �2�x � 1��2 dx
y � � 1
�x � 1���4x2 � 8x � 1 dx
u � 2�x � 1�, du � 2 dx.
83.
��x2
2� 4x �
103
ln 5 � xx � 1 � C
� �x2
2� 4x �
103
ln 1 � x5 � x � C
� �x2
2� 4x �
206
ln 3 � �x � 2�3 � �x � 2� � C
� ���x � 4� dx � 20� 132 � �x � 2�2 dx
y � � x3 � 21x5 � 4x � x2 dx � ��x � 4 �
205 � 4x � x2 dx 84.
� ln 4x � x2 �34
ln x � 4x � C
� ln 4x � x2 �34
ln �x � 2� � 2�x � 2� � 2 � C
y � � 1 � 2x4x � x2 dx � � 4 � 2x
4x � x2 dx � 3� 1�x � 2�2 � 4
dx
85.
� 8 arctan�e2� � 2� � 5.207
� �8 arctan�ex�2��4
0
� 4�4
0
ex�2
�ex�2�2 � 1 dx
� 2�4
0
2ex�2 � e�x�2 dx
A � 2�4
0sech
x2
dx 86.
� ln�e4 � e�4
2� 1.654
� �12
ln�e2x � e�2x��2
0�
12
ln�e4 � e�4� �12
ln 2
�12�2
0
1e2x � e�2x�2��e2x � e�2x� dx
� �2
0 e2x � e�2x
e2x � e�2x dxA � �2
0tanh 2x dx
87.
�52
ln�4 � �17 � � 5.237
� �52
ln�x2 � �x4 � 1 ��2
0
�52�2
0
2x
��x2�2 � 1 dx
A � �2
0
5x
�x4 � 1 dx 88.
� 6 ln5 � �21
3 � �5 � 3.626
� 6 ln�5 � �21 � � 6 ln�3 � �5 �
� �6 ln�x � �x2 � 4 ��5
3
A � �5
3
6
�x2 � 4 dx
89. (a)
(b) ��3
0
dx�x2 � 1
� sinh�1 x��3
0� sinh�1��3 � � 1.317
� ln��3 � 2� � 1.317
��3
0
dx�x2 � 1
� ln�x � �x2 � 1 ���3
090. (a)
(b)
�ln 3
2� �
ln 32 � ln 3
�1�2
�1�2
dx1 � x2 � �tanh�1 x�
1�2
�1�2
� ln 32
� ln 12
� ln 3
�12�ln
32
� ln 12
� ln 12
� ln 32�
� �12
ln 1 � x �12
ln 1 � x �1�2
�1�2
�1�2
�1�2
dx1 � x2 � �1�2
�1�2 � 1�2
1 � x�
1�21 � x� dx
Section 5.9 Hyperbolic Functions 497
91.
3kt16
� � 1�x � 6�2 � 4
dx �1
2�2� ln �x � 6� � 2�x � 6� � 2 � C �
14
ln x � 8x � 4 � C
�3k16
dt � � 1x2 � 12x � 32
dx
When
C � �14
ln�2�
t � 0
x � 0: When
k �2
15 ln7
6
30k16
�14
ln �7�3 �
14
ln�2� �14
ln76
t � 10
x � 1: When
x �10462
�5231
� 1.677 kg
62x � 104
4936
�x � 8
2x � 8
ln76
2
� ln x � 8
2x � 8
316
215 ln7
6�20� �14
ln x � 8
2x � 8
t � 20:
92. (a)
(b)
(c)
Let then
Since
� ��32
�k tanh��32k t�
��32�k ���e�32k t � e��32k t�
e�32k t � e��32k t � v �
�32�e�2�32k t � 1��k�e�2�32k t � 1� �
e�32k t
e�32k t
v��k � �k e�2�32k t� � �32�e�2�32k t � 1�
�32 � �k v � e�2�32k t��32 � �k v�
�32 � �k v
�32 � �k v� e�2�32k t
ln �32 � �k v
�32 � �k v � �2�32k t
v�0� � 0, C � 0.
1
�k�
1
2�32 ln �32 � �k v
�32 � �k v � �t � C
du � �k dv.u � �k v,
� dv32 � kv2 � ��dt
� dvkv2 � 32
� �dt
dvdt
� �32 � kv2
s�t� � �16t 2 � 400
s�0� � �16�0�2 � C � 400 ⇒ C � 400
s�t� � �v�t� dt � ���32t� dt � �16t2 � C
v�t� � �32t (d)
The velocity is bounded by
(e) Since tanh ln cosh (which can be verified by differentiation), then
When
When
when seconds
when seconds
When air resistance is not neglected, it takesapproximately 3.3 more seconds to reach the ground.
(f) As increases, the time required for the object to reachthe ground increases.
k
t � 8.3s2�t� � 0
t � 5s1�t� � 0
s1�t� � �16t2 � 400
s2�t� � 400 � 100 ln�cosh�0.32 t�k � 0.01:
� 400 ⇒ 400 � �1�k� ln�cosh��32k t��.s�0� � C
t � 0,
� �1k ln�cosh��32k t�� � C.
� ��32
�k
1
�32k ln�cosh��32k t�� � C
s�t� � ���32
�k tanh��32k t� dt
�ct��ct� dt � �1�c��
��32��k.
limt→�
���32
�k tanh��32k t� � �
�32
�k
498 Chapter 5 Integration
93.
dydx
��1
�x�a��1 � �x2�a2��
x
�a2 � x2�
�a2
x�a2 � x2�
x
�a2 � x2�
x2 � a2
x�a2 � x2�
��a2 � x2
x
y � a sech�1 xa
� �a2 � x2, a > 0
94. Equation of tangent line through
When
Hence, Q is the point
Distance from P to Q: d � �x02 � ���a2 � x0
2 �2 � a
�0, a sech�1�x0�a��.
y � a sech�1 x0
a� �a2 � x0
2 � �a2 � x02 � a sech�1
x0
a.
x � 0,
y � a sech�1 x0
a� �a2 � x0
2 � ��a2 � x0
2
x0�x � x0�
( , 0)aP
Q
L
x
a
yP � �x0, y0�:
95. Let
u �12
ln1 � x1 � x, �1 < x < 1
2u � ln1 � x1 � x
e2u �1 � x1 � x
e2u�1 � x� � 1 � x
e2u � 1 � xe2u � x
eu � e�u � xeu � xe�u
sinh ucosh u
�eu � e�u
eu � e�u � x
tanh u � x.u � tanh�1 x, �1 < x < 1, 96. Let Then,
and
Thus, Therefore,
arctan�sinh x� � arcsin�tanh x�.
y � arctan�sinh x�.
tan y �ex � e�x
2� sinh x.
sin y � tanh x �ex � e�x
ex � e�x
y
2
e e−e e+
x
x
−x
−x
y � arcsin�tanh x�.
97.
�2x sinh�xb�
�2x �
exb � e�xb
2 �
�exb
x�
e�xb
x
�b
�b
ext dt �ext
x �b
�b98.
y� �ex � e�x
2� sinh x
y � cosh x �ex � e�x
2
99.
��1
�sech y��1 � sech2 y�
�1
x�1 � x2
y� ��1
�sech y��tanh y�
��sech y��tanh y�y� � 1
sech y � x
y � sech�1 x 100.
y� �1
sinh y�
1
�cosh2 y � 1�
1
�x2 � 1
�sinh y�� y� � � 1
cosh y � x
y � cosh�1 x
Review Exercises for Chapter 5 499
Review Exercises for Chapter 5
1.
x
f
f
y 2.
x
f ′
f
y
103.
Let be a point on the catenary.
The slope at is The equation of line is
When The length of is
the ordinate of the point P.y1
�c2 sinh2�x1
c � � c2 � c � cosh x1
c� y1,
Lc �x
sinh�x1�c� ⇒ x � c sinh�x1
c �.y � 0,
y � c ��1
sinh�x1�c� �x � 0�.
Lsinh�x1�c�.P
y� � sinh xc
P�x1, y1�
x
y
P(x1, y1)
(0, c) L
y � c cosh xc
104. There is no such common normal. To see this, assume there is a common normal.
Normal line at is
Similarly, is normal at Also,
The slope between the points is Therefore,
for all Hence, But,
a contradiction.�a � c
cosh a � sinh c< 0,
c < a.x ⇒ sinh c < cosh c � sinh a < cosh a.sinh x < cosh x
cosh c > 0 ⇒ a > 0
�a � c
cosh a � sinh c� cosh c � sinh a.
sinh c � cosh ac � a
.
�1sinh a
��1
cosh c ⇒ cosh c � sinh a.
�c, sinh c�.y � sinh c ��1
cosh c�x � c�
y � cosh a ��1
sinh a�x � a�.�a, cosh a�
y � cosh x ⇒ y� � sinh x.
x
y
y = sinh x
y = cosh x
(a, cosh a)
(c, sinh c)
101.
y� �1
cosh y�
1
�sinh2 y � 1�
1
�x2 � 1
�cosh y�y� � 1
sinh y � x
y � sinh�1 x 102.
� � �2ex � e�x��ex � e�x
ex � e�x� � �sech x tanh x
y� � �2�ex � e�x��2�ex � e�x�
y � sech x �2
ex � e�x
15.
v�30� � 8�30� � 240 ft�sec
a �2�3600�
�30�2 � 8 ft�sec2.
s�30� �a2
�30�2 � 3600 or
s�t� �a2
t2
s�0� � 0 � C2 � 0 when C2 � 0.
s�t� � �at dt �a2
t2 � C2
v�t� � at
v�0� � 0 � C1 � 0 when C1 � 0.
v�t� � �a dt � at � C1
a�t� � a 16.
Solving the system
we obtain and We now solveand get Thus,
Stopping distance from 30 mph to rest is
475.2 � 264 � 211.2 ft.
s�725 � � �
55�122 �72
5 �2
� 66�725 � � 475.2 ft.
t � 72�5.��55�12�t � 66 � 0a � 55�12.t � 24�5
s(t� � �a2
t2 � 66t � 264
v�t� � �at � 66 � 44
s�t� � �a2
t2 � 66t since s�0� � 0.
v�t� � �at � 66 since v�0� � 66 ft�sec.
a�t� � �a
30 mph � 44 ft�sec
45 mph � 66 ft�sec
500 Chapter 5 Integration
3. ��2x2 � x � 1� dx �23
x3 �12
x2 � x � C 4.
� 23�3x
dx �23��3x��1�3�3� dx � �3x�2�3 � C
du � 3 dx
u � 3x
5. �x3 � 1x2 dx � ��x �
1x2� dx �
12
x2 �1x
� C
7. ��4x � 3 sin x� dx � 2x2 � 3 cos x � C
9. ��5 � ex� dx � 5x � ex � C
11. �5x dx � 5 lnx � C
13.
When
y � 2 � x2
C � 2
y � �1 � C � 1
x � �1:
f �x� � ��2x dx � �x2 � C
��1, 1�f��x� � �2x,
6.
�12
x2 � 2x �1x
� C
�x3 � 2x2 � 1x2 dx � ��x � 2 � x�2� dx
8. ��5 cos x � 2 sec2 x� dx � 5 sin x � 2 tan x � C
10. ��t � et� dt �t2
2� et � C
12. �10x
dx � 10 lnx � C
14.
Since the slope of the tangent line at is 3,
f �x� � 2ex � x � 1
f �0� � 1 � 2 � C2 ⇒ C2 � �1
f �x� � ��2ex � 1� dx � 2ex � x � C2
f��x� � 2ex � 1
⇒ C1 � 1.f��0� � 2 � C1 � 3�0, 1�
f��x� � �2ex dx � 2ex � C1
f ��x� � 2ex, �0, 1�
Review Exercises for Chapter 5 501
17.
(a) when t sec.
(b) ft
(c) when sec.
(d) fts�32� � �16�9
4� � 96�32� � 108
t �32
v�t� � �32t � 96 �962
s�3� � �144 � 288 � 144
� 3v�t� � �32t � 96 � 0
s�t� � �16t2 � 96t
v�t� � �32t � 96
a�t� � �32 18.
(a) when sec.
(b)
(c) when sec.
(d) s�2.04� � 61.2 m
t �209.8
� 2.04v�t� � �9.8t � 40 � 20
s�4.08� � 81.63 m
t �409.8
� 4.08v�t� � �9.8t � 40 � 0
�s�0� � 0�s�t� � �4.9t2 � 40t,
v�t� � �9.8t � v0 � �9.8t � 40
a�t� � �9.8 m�sec2
19. (a)
(b)
(c) 10
i�1�4i � 2�
n
i�1i3
10
i�1�2i � 1�
20.
(a)
(b)
(c)
(d) 5
i�2�xi � xi�1� � ��1 � 2� � �5 � ��1�� � �3 � 5� � �7 � 3� � 5
5
i�1�2xi � xi
2� � �2�2� � �2�2� � �2��1� � ��1�2� � �2�5� � �5�2� � �2�3� � �3�2� � �2�7� � �7�2� � �56
5
i�1 1xi
�12
� 1 �15
�13
�17
�37
210
15
5
i�1xi �
15
�2 � 1 � 5 � 3 � 7� �165
x5 � 7x4 � 3,x3 � 5,x2 � �1,x1 � 2,
22.
7 < Area of region < 14
s�3� � 1�20 � 21 � 22� � 7
S�3� � 1�21 � 22 � 23� � 14
y � 2x, �x � 1, n � 3
21.
9.0385 < Area of region < 13.0385
s�n� � s�4� �12
10�1�2�2 � 1
�10
1 � 1�
10�3�2�2 � 1
�10
22 � 1� � 9.0385
S�n� � S�4� �12
101
�10
�1�2�2 � 1�
10�1�2 � 1
�10
�3�2�2 � 1� � 13.0385
y �10
x2 � 1, �x �
1
2, n � 4
23.
� 24 � 8 � 16
� limn→�
24 � 8 n � 1
n �
� limn→�
4n 6n �
4n
n�n � 1�
2 �
� limn→�
n
i�1 �6 �
4in �
4n
Area � limn→�
n
i�1 f �ci� �x
x86
y
4
6
8
2
−2
2−2 4
y � 6 � x, �x �4n
, right endpoints
502 Chapter 5 Integration
26.
� 4 � 6 � 4 � 1 � 15
� limn→�
4n n �
3n
n�n � 1�
2�
3n2
n�n � 1��2n � 1�6
�1n3
n2�n � 1�2
4 �
� limn→�
4n
n
i�1 1 �
3in
�3i2
n2 �i3
n3�
� limn→�
12n
n
i�1 8 �
24in
�24i2
n2 �8i3
n3 �
� limn→�
n
i�1 14�2 �
2in �
3
�2n�
Area � limn→�
n
i�1 f �ci� �x
1
5
10
15
20
2 3 4
y
x
y �14
x3, �x �2n
24. right endpoints
�83
� 6 �263
� limn→�
43
�n � 1��2n � 1�
n2 � 6�
� limn→�
2n
4n2
n�n � 1��2n � 1�6
� 3n�
� limn→�
2n
n
i�1 4i2
n2 � 3�
� limn→�
n
i�1 �2i
n �2
� 3��2n�
Area � limn→�
n
i�1 f �ci� �x
1
2
4
6
8
10
12
2
y
x
y � x2 � 3, �x �2n
,
25.
� 3 � 18 � 9 � 12
� limn→�
3 � 18 n � 1
n�
92
�n � 1��2n � 1�
n2 �
� limn→�
3n n �
12n
n�n � 1�
2�
9n2
n�n � 1��2n � 1�6 �
� limn→�
3n
n
i�1 1 �
12in
�9i2
n2 �
� limn→�
n
i�1 5 � ��2 �
3in �
2
��3n�
Area � limn→�
n
i�1 f �ci� �x
x4
1
3
−4
y
4
−2
2
6
1 2−1−3 3
y � 5 � x2, �x �3n
Review Exercises for Chapter 5 503
27.
� 18 �92
� 9� �272
� limn→�
3n 6n �
3n
n�n � 1�
2�
9n2
n�n � 1��2n � 1�6 �
� limn→�
3n
n
i�1 6 �
3in
�9i2
n2 �
� limn→�
3n
n
i�1 10 �
15in
� 4 � 12in
�9i2
n2 �
Area � limn→�
n
i�1 5�2 �
3in � � �2 �
3in �
2
��3n�
1
1 2 3 4 5 6
2
3
4
6
y
x
x � 5y � y2, 2 ≤ y ≤ 5, �y �3n
28. (a)
(b)
(c) Area � limn→�
mb2�n � 1�
2n� lim
n→� mb2�n � 1�
2n�
12
mb2 �12
�b��mb� �12
�base��height�
s�n� � n�1
i�0 f �bi
n ��bn� �
n�1
i�0m�bi
n ��bn� � m�b
n�2
n�1
i�0i �
mb2
n2 ��n � 1�n2 � �
mb2�n � 1�2n
S�n� � n
i�1 f �bi
n ��bn� �
n
i�1�mbi
n ��bn� � m�b
n�2
n
i�1i �
mb2
n2 �n�n � 1�2 � �
mb2�n � 1�2n
s � m�0��b4� � m�b
4��b4� � m�2b
4 ��b4� � m�3b
4 ��b4� �
mb2
16�1 � 2 � 3� �
3mb2
8
xx b=
y mx=
yS � m�b
4��b4� � m�2b
4 ��b4� � m�3b
4 ��b4� � m�4b
4 ��b4� �
mb2
16�1 � 2 � 3 � 4� �
5mb2
8
29. lim��� →�
n
i�1�2ci � 3� �xi � �6
4�2x � 3� dx 30. lim
��� →�
n
i�13ci�9 � ci
2� �xi � �3
13x�9 � x2� dx
31.
(triangle)
�5
0�5 � x � 5� dx � �5
0�5 � �5 � x�� dx � �5
0x dx �
252
x9
y
6
9
12
3
−3
3−3 6
Triangle
32.
(semicircle)�4
�4
�16 � x2 dx �12
�4�2 � 8
1−1
1
2
3
5
6
−2
−2−3−4 2 3 4
y
x
33. (a)
(b)
(c)
(d) �6
25f �x� dx � 5�6
2f �x� dx � 5�10� � 50
�6
2�2f �x� � 3g�x�� dx � 2�6
2f �x� dx � 3�6
2g�x� dx � 2�10� � 3�3� � 11
�6
2� f �x� � g�x�� dx � �6
2f �x� dx � �6
2g�x� dx � 10 � 3 � 7
�6
2� f �x� � g�x�� dx � �6
2f �x� dx � �6
2g�x� dx � 10 � 3 � 13
504 Chapter 5 Integration
42. �2
1� 1
x2 �1x3� dx � �2
1�x�2 � x�3� dx � �
1x
�1
2x2�2
1� ��
12
�18� � ��1 �
12� �
18
43.
� 1 ��22
��2 � 2
2
�3�4
0sin d � �cos �
3�4
0� ���
�22 � � 1
45. �2
0�x � ex� dx � x2
2� ex�
2
0� 2 � e2 � 1 � 1 � e2
47.
−1−2 1 2 3 4 5 6−1
1
2
3
4
5
6
x
y
�3
1�2x � 1� dx � x2
� x�3
1� 6
44. ��4
��4sec2 t dt � tan t�
�4
��4� 1 � ��1� � 2
46. �6
1 3x dx � 3 lnx�
6
1� 3 ln 6
48.
x1 2 3 4 5−1−2
1
2
4
5
6
7
y
�2
0�x � 4� dx � x2
2� 4x�
2
0� 10
34. (a)
(b)
(c)
(d) �6
3�10 f �x� dx � �10�6
3f �x� dx � �10��1� � 10
�4
4f �x� dx � 0
�3
6f �x� dx � ��6
3f �x� dx � ���1� � 1
�6
0f �x� dx � �3
0f �x� dx � �6
3f �x� dx � 4 � ��1� � 3
35. (c)�734
,�8
1� 3�x � 1� dx � 3
4x4�3 � x�
8
1� 3
4�16� � 8� � 3
4� 1�
36. (d)�3
1 12x3 dx � 12x�2
�2 �3
1� �6
x2 �3
1�
�69
� 6 �163
, 37. �4
0�2 � x� dx � 2x �
x2
2 �4
0� 8 �
162
� 16
38. �1
�1�t2 � 2� dt � t3
3� 2t�
1
�1�
143
39. �1
�1�4t3 � 2t� dt � t4 � t2�
1
�1� 0
40.
�5215
� �325
�163
� 10� � ��325
�163
� 10�
�2
�2�x4 � 2x2 � 5� dx � x5
5�
2x3
3� 5x�
2
�2
41. �9
4x�x dx � �9
4x3�2 dx � 2
5x5�2�
9
4�
25
���9 �5� ��4 �5� �
25
�243 � 32� �4225
Review Exercises for Chapter 5 505
49.
1 2 4 5 6 7 8
1
2
3
4
5
6
7
8
x
y
� �643
� 36� � �9 � 27� �103
�4
3�x2 � 9� dx � x3
3� 9x�
4
350.
x1 3−2
2
3
−1
−2
y
�103
�76
�92
� ��83
� 2 � 4� � �13
�12
� 2�
�2
�1��x2 � x � 2� dx � �
x3
3�
x2
2� 2x�
2
�1
51.
x
1
1
1
y
�12
�14
�14�1
0�x � x3� dx � x2
2�
x4
4 �1
052.
x1
1
y
�23
�25
�4
15
� 23
x3�2 �25
x5�2�1
0
�1
0
�x�1 � x� dx � �x1�2 � x3�2� dx
53.
2−2−2
2
4
6
8
10
4 6 8 10x
y
� 8�3 � 1� � 16� 4x1�2
�1�2��9
1 Area � �9
1
4�x
dx 54.
2
3
4
5
π6
π3
y
x
� tan x��3
0� �3 Area � ��3
0sec2 x dx
55.
−1−1
1
2
3
4
5
1 2 3 4 5
y
x
Area � �3
1 2x dx � 2 ln x�
3
1� 2 ln 3 � 2 ln 1 � ln 9
y �2x
56.
4
6
8
10
1 2
y
x
� 2 � e2 � 1 � 1 � e2 � 8.3891
� x � ex�2
0 Area � �2
0�1 � ex� dx
506 Chapter 5 Integration
67. ,
�x�1 � 3x2�4 dx � �16��1 � 3x2�4��6x dx� � �
130
�1 � 3x2�5 � C �1
30�3x2 � 1�5 � C
du � �6x dxu � 1 � 3x2
68. ,
� x � 3�x2 � 6x � 5�2 dx �
12� 2x � 6
�x2 � 6x � 5�2 dx ��12
�x2 � 6x � 5��1 � C ��1
2�x2 � 6x � 5� � C
du � �2x � 6� dxu � x2 � 6x � 5
69. �sin3 x cos x dx �14
sin4 x � C 70. �x sin 3x2 dx �16��sin 3x2��6x� dx � �
16
cos 3x2 � C
71. � sin
�1 � cos d � ��1 � cos ��1�2 sin d � 2�1 � cos �1�2 � C � 2�1 � cos � C
72. � cos x
�sin x dx � ��sin x��1�2 cos x dx � 2�sin x�1�2 � C � 2�sin x � C
66. ,
�x2�x3 � 3 dx �13��x3 � 3�1�23x2 dx �
29
�x3 � 3�3�2 � C
du � 3x2 dxu � x3 � 3
57.
x �254
�x �52
1
2 4 6 8 10
2
2255,4
x
y 2
5�
1�x
�25
�3 � 2� �25
, Average value
1
9 � 4 �9
4
1�x
dx � 1
5 2�x�
9
458.
x � 3�2
x1 2
2
4
6
8
( 2 , 2)3
yx3 � 2
12 � 0
�2
0x3 dx � x4
8 �2
0� 2
59. F��x� � x2�1 � x3 60. F��x� �1x2 61. F��x� � x2 � 3x � 2 62. F��x� � csc2 x
63.
�x7
7�
35
x5 � x3 � x � C
��x2 � 1�3 dx � ��x6 � 3x 4 � 3x2 � 1� dx 64.
�x3
3� 2x �
1x
� C
��x �1x�
2
dx � ��x2 � 2 � x�2� dx
65. ,
� x2
�x3 � 3 dx � ��x3 � 3��1�2x2 dx �
13��x3 � 3��1�23x2 dx �
23
�x3 � 3�1�2 � C
du � 3x2 dxu � x3 � 3
Review Exercises for Chapter 5 507
73. �tann x sec2 x dx �tann�1 xn � 1
� C, n � �1 74. �sec 2x tan 2x dx �12��sec 2x tan 2x��2� dx �
12
sec 2x � C
75. ��1 � sec x�2 sec x tan x dx �1��1 � sec x�2� sec x tan x� dx �
13
�1 � sec x�3 � C
77. �xe�3x2 dx � �
16�e�3x2��6x� dx � �
16
e�3x 2� C
79.
�1
2 ln 55�x�1�2
� C
��x � 1�5�x�1�2 dx �
12�5�x�1�2
2�x � 1� dx
76. �cot4 � csc2 � d� � ���cot ��4��csc2 �� d� � �15
cot 5� � C
78. �e1 x
x2 dx � ��e1 x��
1x2� dx � �e1 x � C
80. �1t2
2�1 t dt � �2�1 t�t�2� dt �1
ln 2 2�1 t � C
81. �2
�1x�x2 � 4� dx �
12�
2
�1�x2 � 4��2x� dx �
12
�x2 � 4�2
2 �2
�1�
14
�0 � 9� � �94
82. �1
0x2�x3 � 1�3 dx �
13
�1
0�x3 � 1�3�3x2� dx �
112 �x3 � 1�4�
1
0�
112
�16 � 1� �54
83. �3
0
1
�1 � x dx � �3
0�1 � x��1�2 dx � 2�1 � x�1�2�
3
0� 4 � 2 � 2
85.
When When
� 2�0
1�u3�2 � 2u1�2� du � 2 2
5u5�2 �
43
u3�2�0
1�
28
15
2�1
0�y � 1��1 � y dy � 2�0
1���1 � u� � 1��u du
u � 0.y � 1,u � 1.y � 0,
dy � �duy � 1 � u,u � 1 � y,
86.
When When
� 2�1
0�u5�2 � 2u3�2 � u1�2� du � 2 2
7u7�2 �
45
u5�2 �23
u3�2�1
0�
32
105
2�0
�1 x2�x � 1 dx � 2�1
0�u � 1�2�u du
u � 1.x � 0,u � 0.x � �1,
dx � dux � u � 1,u � x � 1,
84. �6
3
x
3�x2 � 8 dx �
16
�6
3�x2 � 8��1�2�2x� dx � 1
3�x2 � 8�1�2�
6
3�
13
�2�7 � 1�
87. �
0cos�x
2� dx � 2�
0cos�x
2� 12
dx � 2 sin�x2��
0� 2 88. since sin 2x is an odd function.��4
��4 sin 2x dx � 0
508 Chapter 5 Integration
93. Trapezoidal Rule
Simpson’s Rule 0.685
Graphing utility: 0.704
�n � 4�:
��2
0
�x cos x dx � 0.637�n � 4�:
95. Trapezoidal Rule
Simpson’s Rule
Graphing utility: 1.494
�n � 4� � 1.494
�1
�1e�x2
dx � 1.463�n � 4�:
97.
� 17x � 2
dx �17� 1
7x � 2�7� dx �
17
ln7x � 2 � C
u � 7x � 2, du � 7 dx
94. Trapezoidal Rule
Simpson’s Rule 3.820
Graphing utility: 3.820
�n � 4�:
�
0
�1 � sin2 x dx � 3.820�n � 4�:
96. (a)
(b)
�13
4 � 4�2� � 2�1� � 4�12� �
14� � 5.417
S�4� �4 � 03�4� � f �0� � 4 f �1� � 2 f �2� � 4 f �3� � f �4��
R < I < T < L
98.
� xx2 � 1
dx �12� 2x
x2 � 1 dx �
12
lnx2 � 1 � C
u � x2 � 1, du � 2x dx
99.
� �ln1 � cos x � C
� sin x1 � cos x
dx � �� �sin x1 � cos x
dx 100.
�ln�xx
dx �12��ln x��1
x� dx �14
�ln x�2� C
u � ln x, du �1x dx
89.
(a) 2000 corresponds to
�15,000
M 1.20t � 0.02t2�11
10�
24,300M
C �15,000
M �11
10�1.20 � 0.04t� dt
t � 10.
C �15,000
M �t�1
t
p ds
p � 1.20 � 0.04t
(b) 2005 corresponds to
C �15,000
M 1.20t � 0.02t2�16
15�
27,300M
t � 15.
90. liters
Increase is liters.7
�5.1
�1.9
� 0.6048
�2
01.75 sin
t2
dt � �2 1.75 cos
t2 �
2
0� �
2
�1.75���1 � 1� �7
� 2.2282
91. Trapezoidal Rule
Simpson’s Rule
Graphing utility: 0.254
�2
1
11 � x3 dx �
112
11 � 13 �
41 � �1.25�3 �
21 � �1.5�3 �
41 � �1.75�3 �
11 � 23� � 0.254�n � 4�:
�2
1
11 � x3 dx �
18
11 � 13 �
21 � �1.25�3 �
21 � �1.5�3 �
21 � �1.75�3 �
11 � 23� � 0.257�n � 4�:
92. Trapezoidal Rule
Simpson’s Rule
Graphing utility: 0.166
�1
0
x3�2
3 � x2 dx �1
12 0 �4�1�4�3�2
3 � �1�4�2 �2�1�2�3�2
3 � �1�2�2 �4�3�4�3�2
3 � �3�4�2 �12� � 0.166�n � 4�:
�1
0
x3�2
3 � x2 dx �18 0 �
2�1�4�3�2
3 � �1�4�2 �2�1�2�3�2
3 � �1�2�2 �2�3�4�3�2
3 � �3�4�2 �12� � 0.172�n � 4�:
Review Exercises for Chapter 5 509
101. � 3 � ln 4�4
1
x � 1x
dx � �4
1�1 �
1x� dx � x � lnx�
4
1
103. ��3
0sec d � lnsec � tan �
�3
0� ln�2 � �3 �
105. Let Then
�e2x � e�2x
e2x � e�2x dx �12
ln�e2x � e�2x� � C
du � 2�e2x � e�2x� dx.
u � e2x � e�2x.�e2x � e�2x
e2x � e�2x dx.
102. �e
1 ln x
x dx � �e
1�ln x�1�1
x� dx � 12
�ln x�2�e
1�
12
104.
� 0 � ln� 1
�2� �
12
ln 2
��4
0tan�
4� x� dx � lncos�
4� x���4
0
106. � e2x
e2x � 1 dx �
12
ln�e2x � 1� � C
107. Let
�12
arctan�e2x� � C
�12� 1
1 � �e2x�2�2e2x� dx
� 1e2x � e�2x dx � � e2x
1 � e4x dx
u � e2x, du � 2e2x dx.
109. Let
� x
�1 � x4 dx �
12� 1
�1 � �x2�2�2x� dx �
12
arcsin x2 � C
u � x2, du � 2x dx.
108. Let
�1
5�3 arctan
5x
�3� C
� 13 � 25x2 dx �
15� 1
��3 �2 � �5x�2�5� dx
u � 5x, du � 5 dx.
110. � 116 � x2 dx �
14
arctan x4
� C
111. Let
� x16 � x2 dx �
12� 1
16 � x2�2x� dx �12
ln�16 � x2� � C
u � 16 � x2, du � 2x dx.
113. Let
�arctan�x�2�4 � x2 dx �
12��arctan
x2��
24 � x2� dx �
14�arctan
x2�
2
� C
u � arctan�x2�, du �
24 � x2 dx.
112. � 4 � x
�4 � x2 dx � 4� 1
�4 � x2 dx �
12��4 � x2��1�2��2x� dx � 4 arcsin
x2
� �4 � x2 � C
114. Let
� arcsin x
�1 � x2 dx �
12
�arcsin x�2 � C
u � arcsin x, du �1
�1 � x2 dx.
510 Chapter 5 Integration
Problem Solving for Chapter 5
1. (a)
(b) by the Second Fundamental Theorem of Calculus.
(c) for
(Note: The exact value of x is e, the base of the natural logarithm function.)
(d) We first show that To see this, let and
Then Now,
� L�x1� � L�x2�.
� �x1
1
1u
du � �x2
1 1u
du
� �1
1�x1
1u
du � �x2
1 1u
du
�using u �t
x1� L�x1x2� � �x1x2
1 1t dt � �x2
1�x1
1u
du
�x1
1 1t dt � �1
1�x1
1ux1
�x1 du� � �1
1�x1
1u
du � �1
1�x1
1t dt.
du �1x1
dt.u �t
x1�x1
1 1t dt � �1
1�x1
1t dt.
�2.718
1 1t dt � 0.999896
x � 2.718L�x� � 1 � �x
1 1t dt
L��1� � 1
L��x� �1x
L�1� � �1
1
1
t dt � 0
116.
Since the area of region A is
the shaded area is �1
0arcsin x dx �
�
2� 1 � 0.571.
1����2
0sin y dy�,
10.25 0.750.5
π2
π4
y
x
y = arcsin x
117.
y� � 2 �1
2�x�sinh �x� � 2 �
sinh�x
2�x
y � 2x � cosh�x 118.
y� � x� 21 � 4x2� � tanh�1 2x �
2x1 � 4x2 � tanh�1 2x
y � x tanh�1 2x
119. Let
�12
ln�x2 � �x4 � 1 � � C
� x
�x4 � 1 dx �
12� 1
��x2�2 � 1�2x� dx
u � x2, du � 2x dx. 120. Let
�x2�sech x3�2 dx �13��sech x3�2�3x2� dx �
13
tanh x3 � C
u � x3, du � 3x2 dx.
115.
Since when you have Thus,
y � A sin�� km
t�.
sin�� km
t� �yA
C � 0.t � 0,y � 0
arcsin�yA� �� k
m t � C
� dy
�A2 � y2� �� k
m dt
Problem Solving for Chapter 5 511
3.
(a)
x
y
1
1 3
2
−2
−1
S�x� � �x
0sin��t2
2 � dt
(b)
The zeros of correspond to the relative extrema of S�x�.y � sin �x2
2
x
3
y
2
1
1 2 32 3 5 6 7 2 2
(c)
Relative maximum at and
Relative minimum at and
(d)
Points of inflection at x � 1, �3, �5, and �7
S� �x� � cos��x2
2 ���x� � 0 ⇒ �x2
2�
�
2� n� ⇒ x2 � 1 � 2n ⇒ x � �1 � 2n, n integer
x � �8 � 2.8284x � 2
x � �6 � 2.4495x � �2 � 1.4142
S��x� � sin �x2
2� 0 ⇒ �x2
2� n� ⇒ x2 � 2n ⇒ x � �2n, n integer
2. (a)
(b)
(c)
Since
�Note: sin 4 � �0.7568�
F��x� � sin x2, F��2� � sin 4 � limx→2
G�x�.
� limx→2
G�x�
� limx→2
1
x � 2 �x
2sin t2 dt
F��2� � limx→2
F�x� � F�2�
x � 2
limx→2
G�x� � �0.75
G�x� �1
x � 2 �x
2sin t2 dt
F�x� � �x
2sin t2 dt
x 0 1.0 1.5 1.9 2.0 2.1 2.5 3.0 4.0 5.0
0.0611 0 �0.2769�0.0576�0.0312�0.3743�0.0867�0.0265�0.4945�0.8048F�x�
x 1.9 1.95 1.99 2.01 2.05 2.1
�0.8671�0.8174�0.7697�0.7436�0.6873�0.6106G�x�
512 Chapter 5 Integration
(d)
is a point of inflection, whereas is not. ( is not continuous at )x � 6.f �x � 6x � 2
0 < x < 2
2 < x < 6
6 < x < 8F� �x� � f��x� �
�1,
1,
1,2
6. (a)
(b) v is increasing (positive acceleration) on and
(c) Average acceleration
(d) This integral is the total distance traveled in miles.
(e) One approximation is
(other answers possible)a�0.8� �v�0.9� � v�0.8�
0.9 � 0.8�
50 � 400.1
� 100 mi�hr2.
�1
0v�t� dt �
110
0 � 2�20� � 2�60� � 2�40� � 2�40� � 65� �38510
� 38.5 miles
�v�0.4� � v�0�
0.4 � 0�
60 � 00.4
� 150 mi�hr2
�0.7, 1.0�.�0, 0.4�
0.2 0.4 0.6 0.8 1.0
20
40
60
80
100
v
t
4. Let d be the distance traversed and a be the uniform acceleration.We can assume that and Then
when
The highest speed is The lowest speed is The mean speed is
The time necessary to traverse the distance d at the mean speed is
which is the same as the time calculated above.
t �d
�ad�2��2d
a
12
��2ad � 0� ��ad2
.v � 0.v � a�2da
� �2ad.
t ��2da
.s�t� � d
s�t� �12
at2.
v�t� � at
a�t� � a
s�0� � 0.v�0� � 0
5. (a)
(b)
x
y
1
2 4 5 6 7 8 9
2345
−2−3−4−5
−1
f(0, 0)
(6, 2)(8, 3)
(2, −2)
x 0 1 2 3 4 5 6 7 8
0 314
�2�72
�4�72
�2�12
F�x�
(c)
F is decreasing on and increasingon Therefore, the minimum is at andthe maximum is 3 at x � 8.
x � 4,�4�4, 8�.�0, 4�F��x� � f �x�.
0 ≤ x < 2
2 ≤ x < 6
6 ≤ x ≤ 8��x2�2�
�x2�2��1�4�x2
� 4x
� x
,
� 4,
� 5,
F�x� � �x
0f �t� dt �
0 ≤ x < 2
2 ≤ x < 6
6 ≤ x ≤ 8f �x� �
�x,
x � 4,
1x � 1,
2
Problem Solving for Chapter 5 513
9. Consider Thus,
�12
f �b�2 � f �a�2�.
�12
F�b� � F�a��
� �12
F�x� b
a
�b
a
f �x� f��x� dx � �b
a
12
F��x� dx
F�x� � f �x��2 ⇒ F��x� � 2f �x� f��x�.
11. Consider
The corresponding Riemann Sum using right endpoints is
Thus, limn→�
S�n� � limn→�
15 � 25 � . . . � n5
n6 �16
.
�1n615 � 25 � . . . � n5�. S�n� �
1n��
1n�
5
� �2n�
5
� . . . � �nn�
5
�1
0x5 dx �
x6
6 1
0�
16
.
10. Consider The corresponding
Riemann Sum using right endpoints is
Thus, limn→�
�1 � �2 � . . . � �nn3�2 �
23
.
�1
n3�2 �1 � �2 � . . . � �n�.
S�n� �1n��1
n��2
n� . . . ��n
n
�1
0
�x dx �23
x3�2 1
0�
23
.
7. (a)
Error:
(b)
(Note: Exact answer is )
(c) Let
�2b3
� 2d
p��1�3� � p� 1
�3� � �b
3� d� � �b
3� d�
�2b3
� 2d
�1
�1p�x� dx � �ax4
4�
bx3
3�
cx2
2� dx
1
�1
p�x� � ax3 � bx2 � cx � d.
��2 � 1.5708
�1
�1
11 � x2 dx �
11 � �1�3� �
11 � �1�3� �
32
�1.6829 � 1.6758� � 0.0071
�1
�1cos x dx � sin x
1
�1� 2 sin�1� � 1.6829
� 2 cos� 1�3� � 1.6758
�1
�1cos x dx � cos��
1�3� � cos� 1
�3� 8.
Thus,
Differentiating the other integral,
Thus, the two original integrals have equal derivatives,
Letting we see that C � 0.x � 0,
�x
0f �t��x � t� dt � �x
0��t
0f �v� dv� dt � C.
ddx�
x
0��x
0f �v� dv� dt � �x
0f �v� dv.
ddx�
x
0f �t��x � t� dt � x f �x� � �x
0f �t� dt � x f �x� � �x
0f �t� dt.
� x�x
0f �t� dt � �x
0t f �t� dt
�x
0f �t��x � t� dt � �x
0x f �t� dt � �x
0t f �t� dt
12. (a)
—CONTINUED—
� 227 � 9� � 36
� 2�9x �x3
3 3
0
−4 −2 −1 1
12345678
10
2 4 5
y
x
Area � �3
�3�9 � x2� dx � 2�3
0�9 � x2� dx
514 Chapter 5 Integration
14. (a)
(b)
Hence,
(c)
⇒ �n
i�1i2 �
n�n � 1��2n � 1�6
�n�n � 1��2n � 1�
2
�2n3 � 3n2 � n
2
�2n3 � 6n2 � 6n � 3n2 � 3n � 2n
2
⇒ �n
i�13i2 � n3 � 3n2 � 3n �
3n�n � 1�2
� n
�n � 1�3 � 1 � �n
i�1�3i2 � 3i � 1� � �
n
i�13i2 �
3�n��n � 1�2
� n
�n � 1�3 � �n
i�1�3i2 � 3i � 1� � 1.
� �23 � 13� � �33 � 23� � . . . � ��n � 1�3 � n3�� � �n � 1�3 � 1
�n
i�1�3i2 � 3i � 1� � �
n
i�1�i � 1�3 � i3�
3i2 � 3i � 1 � �i � 1�3 � i3
�1 � i�3 � 1 � 3i � 3i2 � i3 ⇒ �1 � i�3 � i3 � 3i2 � 3i � 1
15. Since
��b
a� f �x�� dx ≤ �b
a
f �x� dx ≤ �b
a� f �x�� dx ⇒ ��b
af �x� dx� ≤ �b
a� f �x�� dx.
��f �x�� ≤ f �x� ≤ �f �x��,
13. By Theorem 5.8,
Similarly,
Thus, On the interval and
Thus, �Note: �1
0
�1 � x4 dx � 1.0894�1 ≤ �1
0
�1 � x4 dx ≤ �2.
b � a � 1.0, 1�, 1 ≤ �1 � x4 ≤ �2m�b � a� ≤ �b
a
f �x� dx ≤ M�b � a�.
m ≤ f �x� ⇒ m�b � a� � �b
a
m dx ≤ �b
a
f �x� dx.
0 < f �x� ≤ M ⇒ �b
a
f �x� dx ≤ �b
a
M dx � M�b � a�.
12. —CONTINUED—
(b)
(c) Let the parabola be given by
Archimedes’ Formula: Area �23�
2ba ��b2� �
43
b3
a
Base �2ba
, height � b2
� 2�b3
a�
13
b3
a �43
b3
a � 2�b2�b
a� �a2
3 �ba�
3
− ba
ba
b2
y
x
� 2�b2x � a2 x3
3 b�a
0 Area � 2�b�a
0�b2 � a2x2� dx
y � b2 � a2x2, a, b > 0.
area �23
bh �23
�6��9� � 36Base � 6, height � 9,
Problem Solving for Chapter 5 515
17. Let
� 0.8109
� 2 ln 3 � 2 ln 2 � 2 ln�32�
� �2 ln u 3
2
� �3
2
2u
du
� �3
2
2�u � 1�u2 � u
du
Area � �4
1
1�x � x
dx � �3
2
2u � 2
�u � 1� � �u2 � 2u � 1� du
dx � �2u � 2� du.
u � 1 � �x, �x � u � 1, x � u2 � 2u � 1, 18. Let
�12
arctan�12�
� �12
arctan�u2�
1
0
� �1
0
duu2 � 4
Area � ���4
0
1sin2 x � 4 cos2 x
dx � ���4
0
sec2 xtan2 x � 4
dx
u � tan x, du � sec2 x dx.
16. (a)
Area B � �12��
�
6� ��
12� 0.2618
� ��22
��32
��3 � �2
2� 0.1589
Area A � ���4
��6sin y � dy � �cos y
�4
��6
y � f �x� � arcsin x, sin y � x
(c)
(d)
��
12�4�3 � 3� �
12
ln 2 � 0.6818
Area C � ��3
1arctan x dx � ��
3���3� �12
ln 2 � ��
4��1�
� �ln 12
� ln �22
� ln�2 �12
ln 2 A
CB
π4
π3
1 3
y
x
y = arctan x � �ln�cos y�
��3
��4
Area A � ���3
��4tan y dy
tan y � x
Area B � �3
1ln x dx � 3�ln 3� � A � 3 ln 3 � 2 � ln 27 � 2 � 1.2958
1 2 3
AB
ln 3
y
x
y = ln x
ey = x
Area A � �ln 3
0ey dy � ey
ln 3
0� 3 � 1 � 2
(b)
� 0.1346
� ���28
�112� �
�2 � �32
���2
8�
�3 � �22
��
12
��2�2
1�2arcsin x dx � Area�C� � ��
4���22 � � A � B
516 Chapter 5 Integration
19. (a) (i)
−2 2
−1
y
y1
4
y1 � 1 � x
y � ex (ii)
−2 2
−1
y
y2
4
y2 � 1 � x � �x2
2 � y � ex (iii)
−2 2
−1
y y3
4
y3 � 1 � x �x2
2�
x3
6
y � ex
(b) th term is in polynomial:
(c) Conjecture: ex � 1 � x �x2
2!�
x3
3!� . . .
y4 � 1 � x �x2
2!�
x3
3!�
x4
4!xn�n!n
20. (a)
Then,
Thus,
(b) b � 1 ⇒ �1
0
sin xsin�1 � x� � sin x
dx �12
A �b2
.
� �b
01 dx � b.
2A � �b
0
f �x�f �x� � f �b � x� dx � �b
0
f �b � x�f �b � x� � f �x� dx
� �b
0
f �b � x�f �b � x� � f �x� dx
� �b
0
f �b � u�f �b � u� � f �u� du
A � �0
b
f �b � u�
f �b � u� � f �u���du�
Let u � b � x, du � �dx.
Let A � �b
0
f �x�f �x� � f �b � x� dx.