Thomas Calculus 13th [Solutions]

85. (a) Wrong:
(b)
(c)
d
dx
3
(2x
1)
3
C
3
Wrong: d ((2x
1) C)
dx
3
Right: d ((2x
1) C)
dx
2
3(2 x 1) (2)
2(2x
1) 2 (2x
1)
2
3
3(2x
2
1) (2)
6(2x
2
1)
2 2
6(2x
1) 3(2x
1)
2 1/2
86. (a) Wrong: d
dx ( x x C)
1 2 1/2
( ) (2 1)
2 x x C x 2x
1 2x
1
2
2 x x C
2 1/2
(b) Wrong: d ( x x)
C 1 2 1/2
( ) (2 1)
dx
2 x x x 2x
1 2x
1
2
2 x x
3
(c) Right: d 1 2x
1 C d 1 3/2
(2x
1) C 3 1/2
(2 1) (2)
dx 3
dx 3 6 x 2x
1
Section 4.8 Antiderivatives 313
87. Right: d
dx
x
x
3
2
3
2
x 3 ( x 2) 1 ( x 3) 1
x 2 ( x 2)
C 3
2
2
( 3)
3 x 5
( x 2) ( x 2)
2 2
4
2
15( x 3)
( x 2)
88. Wrong:
d
dx
2
sin( x )
x
C
2 2
x cos( x )(2 x) sin( x ) 1
x
2
2 2 2
2x cos( x ) sin( x )
x
2
2 2
xcos( x ) sin( x )
x
2
dy
89. Graph (b), because 2x
dx
2
y x C . Then y(1) 4 C 3.
90. Graph (b), because
dy
dx
1 2
x y x C . Then y( 1) 1 C 3 .
2
2
dy
91. 2x
7
dx
2
2
y x 7 x C ; at x 2 and y 0 we have 0 2 7(2) C C 10
y
2
x
7x
10
dy
92. 10
dx
x
2
y 10 x x C;
at x 0 and y 1 we have
2
1 10(0) 0 C
2
2
C 1 y 10x
x 1
2
2
dy
93. 1 2
x x x y
dx 2
x
1
2
y x x 1 or y
2 2
2
1 x
1
x C ; at x 2 and y 1 we have 1 2 2 2
C C 1
2
2
2
1
2
x 1
x 2 2
94.
dy
dx
2
9x
4x
5
C 10 y
3 2
y 3x 2x 5 x C ; at x 1
3 2
and y 0 we have 0 3( 1) 2( 1) 5( 1) C
3 2
3x 2x 5x
10
95.
dy
dx
y
1/3
x
1
3
2/3 3
1/3
3x y C 9 y 9 x C ; at x 1 and y 5 we have
1/3
9x
4
1/3
5 9( 1) C C 4
96.
dy
dx
1 1 x
2 x 2
1/2
1/2
1/2
y x C ; at x 4 and y 0 we have 0 4 C C
1/2
2 y x 2
97. ds 1 cost s t sin t C ; at t 0 and s 4 we have 4 0 sin 0 C C 4 s t sin t 4
dt
98. ds cos t sin t s sin t cos t C ; at t and s 1 we have 1 sin cos C C 0
dt
s sin t cost
99. dr sin
d
r cos ( ) C ; at r 0 and 0 we have 0 cos ( 0) C C 1 r cos ( ) 1
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