Ln 2 0 ln 2 0 ln 2 ln2 0 0 ln1 2 2 2 1 2 t t t t e e

( )
( )
( )
ln 2
0
ln 2
0
ln 2 ln2 0 0
ln(1 2)
2 2
1
2
t t
t t
e e dt
e e
e e e e
e
−
−
−
−
= +
= + − +
= + −
=
∫
18.
( )
( )
6 6
2 3 2
1 1
6
4 3
1
1
4 3
3025
12
x x dx x x dx
x x
− = −
⎛ ⎞
= −
⎜ ⎟
⎜ ⎟
⎝ ⎠
=
∫ ∫
20.
( )
( ) ( )
( )
0
4
3
0
4
3
0
5
3
2 6
1
2 6 2 6
2
2 6
1
2 5
3,888
5
x dx
x d x
x
−
−
−
+
= + +
+
=
=
∫
∫
22.
2
2
3 2
1
2
3 2 3
1
2
3 1
1
( 1)
1
( 1) ( 1)
3
1 ( 1)
3 1
7
54
x
dx
x
x d x
x
−
−
+
= + +
+
=
−
=
⌠
⎮
⌡
∫
24.
1 1
2 2
0 0
1
2
0
6 (2 )
3
1 1
3ln( 1)
3ln 2
t t dt
dt
t t
t
=
+ +
= +
=
⌠ ⌠
⎮ ⎮
⌡ ⌡

200 Chapter 5. Integration
26.
2
6
1
( 1)( 2) t t dt + −
∫
Let 2 u t = − so 1 3 t u + = + , du dt = .
Further when 1 t = , 1 u = − and 2 t = ,
0 u = . Thus
2 0
6 6
1 1
0
7 6
1
0
8 7
1
( 1)( 2) ( 3)
( 3 )
3
8 7
17
56
t t dt u u du
u u du
u u
−
−
−
+ − = +
= +
⎛ ⎞
= + ⎜ ⎟
⎜ ⎟
⎝ ⎠
=
∫ ∫
∫
28.
2
1
ln
e
e
dx
x x
⌠
⎮
⌡
Let ln , .
dx
u x du
x
= =
If
2
, 1, and if , 2. x e u x e u = = = =
2
2
1
2
1
1
ln
ln
ln 2
0.693
e
e
du
dx
x x u
u
=
=
=
=
⌠ ⌠
⎮ ⎮
⌡ ⌡
30.
4
3/ 2
1
( 1) x
dx
x
− ⌠
⎮
⌡
Let 1, .
2
dx
u x du
x
= − =
If 1, 0, and if 4, 1. x u x u = = = =
4
3/ 2
1
3/ 2
0
1
1
5/ 2
0
( 1)
2
4
5
4
5
x
dx u du
x
u
−
=
=
=
⌠
⎮
⌡
∫
32.
1
3
1 1
3 3
1 1
3 3
[4 ( ) 3 ( )]
4 ( ) 3 ( )
4 ( ) 3 ( )
4(0) 3(4)
12
−
− −
− −
−
= −
= −
= −
= −
∫
∫ ∫
∫ ∫
f x g x dx
f x dx g x dx
f x dx g x dx
34.
3 2
2 3
( ) ( ) 5 f x dx f x dx
−
−
= − = −
∫ ∫
36. Since
2 1 2
3 3 1
( ) ( ) ( ) g x dx g x dx g x dx
− −
= +
∫ ∫ ∫
we
have
2 2 1
1 3 3
( ) ( ) ( )
2 4
6 .
g x dx g x dx g x dx
− −
= −
= − −
= −
∫ ∫ ∫
38.
[ ]
1
3
1 1
3 3
2 ( ) 3 ( )
2 ( ) 3 ( )
2(0) 3(4)
12
f x g x dx
f x dx g x dx
−
− −
+
= +
= +
=
∫
∫ ∫
40.
4
0
4
3/ 2 1/ 2
0
4
5/ 2 3/ 2
0
Area ( 1)
( )
2 2
5 3
64 16
5 3
272
15
x x dx
x x dx
x x
= +
= +
= +
= +
=
∫
∫
42.
0
8
3
Area
9 2
−
=
−
∫
dx
x
Let u = 9 − 2 x , then du = − 2 dx . If x = − 8,
u = 25 and if x = 0, u = 9, so

Chapter 5. Integration 201
( )
( )
0
8
0
8
9
25
9
25
3
Area
9 2
3 1
( 2 )
2
9 2
3 1
2
3
2
2
3 9 25
3( 2)
6
−
−
=
−
= − −
−
= −
= −
= − −
= − −
=
∫
∫
∫
dx
x
dx
x
du
u
u
44.
2
3
0
Area
x
xe dx
−
=
∫
Let
2
u x = then 2 du xdx = . If 0, 0 x u = =
and if 3, 9 x u = = so
2
2
3
0
3
0
9
0
9
0
9
Area
1
(2 )
2
1
2
1
2
1 1
2 2
0.499938
x
x
u
u
xe dx
e x dx
e du
e
e
−
−
−
−
−
=
=
=
= −
= − +
≈
∫
∫
∫
46.
2
2
1
1
2
3
Area
3ln
3ln 3ln1
6
e
e
dx
x
x
e
=
=
= −
=
⌠
⎮
⌡
48. At 11:00
A
.
M
., 2 t = , and at 1:00
P
.
M
.,
4 t = . The number of people who enter
the fair between 11:00
A
.
M
. and 1:00
P
.
M
.
is
4
2
'( ) N t dt
∫
.
50. Let ( ) P t denote the revenue generated by
the oil after t months. The price per barrel
after t months is $95 plus $0.30 for each
month or 95 + 0.30 t . At 400 barrels per
month, the revenue is changing at the rate
of ( ) 400(95 0.30 )
′
= + P t t
for 0 24 t ≤ ≤ (months). The total revenue
will be
24
0
24
2
0
( ) 400 (95 0.30 )
0.30
400 95
2
$946,560.
= +
⎛ ⎞
= +
⎜ ⎟
⎜ ⎟
⎝ ⎠
=
∫
P t t dt
t
t
52. Let A ( t ) denote the amount of pollution in
the lake after t years. Let P ( t ) denote the
population in thousands of people. Then
we are given
2
0.6 0.2 0.5
dP
t t
dt
= + + thousand
people/yr and 5
dA
dP
= units/thousand
people.
By the chain rule
2
2
5(0.6 0.2 0.5)
3 2.5 units/year.