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6 ex
FL
x2
’ sin x cos x dx = ’
(b) xe
2 2
2
2
AM

e36 sin2 6 e4 sin2 2
= ’ ’ ’ .
2 2 2 2
4
’ cos x 2
ln2 x
4
TE



ln x
+ x sin x 2 dx = +
(c)
2 2
x
1
0

ln2 4 ’ cos 42 ln2 1 ’ cos 12
= + ’ +
2 2 2 2

ln2 4 cos 16 cos 1
= ’ + .
2 2 2
2
2 sin x 3
tan x ’ x cos x dx = ’ ln | cos x| ’
2 3
(d)
3
1 1
sin 23 sin 13
= ’ ln | cos 2| ’ ’ ’ ln | cos 1| ’
3 3
sin 8 sin 1
= ’ ln | cos 2| ’ + ln | cos 1| + .
3 3
e
ln2 x 2 ln2 e2 ln2 12
ln x 2
e
dx = = ’ = 1 ’ 0 = 1.
(e)
4 4 4
x
1
1
289
Chapter 4

8
sin2 x 3
8
x · cos x sin x dx =
2 3 3
(f)
6
4
4

sin2 83 sin2 43 sin2 512 sin2 64
= ’ = ’ .
6 6 6 6
5
5 x3 x2
Area = x + x + 6 dx = + + 6x
2
5. (a)
3 2
2 2
53 52 2 3 22 405
= + +6·5 ’ + +6·2 = .
3 2 3 2 6
π/4
sin2 x
π/4
Area = sin x cos x dx =
(b)
2
0
0

sin2 π/4 sin2 0 1
= ’ =.
2 2 4
2
2 2 2
2 e2 e1 e4
ex e
x2
Area = dx = = ’ = ’.
(c) xe
2 2 2 2 2
1
1
e
ln2 x ln2 e ln2 1
e ln x 10 1
Area = dx = = ’ =’=.
(d)
2 2 2 22 2
x
1
1
6.


y




1
2

π 2π x

positive for x µ(kπ/6, (k + 1) π/6), k even
negative for x µ(kπ/6, (k + 1) π/6), k odd




Fig. S4.6(b)
Fig. S4.6(a)
290 Solutions to Exercises




Fig. S4.6(c)

Fig. S4.6(d)


0 2
Area = ’(x + 3x) dx + x 3 + 3x dx
3
7. (a)
’2 0
0 2
x4 3x 2 x 4 3x 2
=’ + + +
4 2 4 2
’2 0

0 0 (’2)4 3 · (’2)2
=’’+ +
42 4 2

24 3 · 2 2 0 0
+ + ’’ = 20.
4 2 42

11 (j +1)π/6
Area = (’1)j sin 3x cos 3x dx
(b)
j =’12 j π/6
11 (j +1)π/6 1
= (’1)j sin 6x dx
2
j =’12 j π/6
(j +1)π/6
11
(’1)j cos 6x
= ’
2 6 j π/6
j =’12
291
Chapter 4

11
1 1
= ’’
6 6
j =’12
= 8.
1 e
ln x ln x
Area = ’ dx +
(c) dx
x x
1/2 1
1 e
ln2 x ln2 x
=’ +
2 2
1/2 1

ln2 (1/2) 12 02
0
=’’’ + ’
2 2 2 2
≈ 0.7404.
0 3
3 x4 4
Area = ’x e dx + x 3 ex dx
(d)
’3 0
0 3
4 4
ex ex
=’ +
4 4
’3 0

e81 e81 1 e81 1
1
=’’’ + ’ = ’.
4 4 4 4 2 2
2
2 x 4 3x 2
Signed Area = x + 3x dx = +3
8. (a)
4 2
’2 ’2

24 3 · 22 (’2)4 3 · (’2)2
= + ’ + = 0.
4 2 4 2
2π 2π 1
Signed Area = sin 3x cos 3x dx =
(b) sin 6x dx
2
’2π ’2π

1 cos 6x 1 1
= ’ =’ ’ ’ = 0.
2 6 6 6
’2π
e
ln2 x
e ln x
Signed Area = dx =
(c)
2
x
1/2
1/2
12 ln2 (1/2)
= ’
2 2
≈ 0.2598.
292 Solutions to Exercises

3
4
3 e81 e81
ex
3 x4
Signed Area = dx = = ’ = 0.
(d) xe
4 4 4
’3
’3
1 1
Area = [’3x + 10] ’ [2x ’ 4] dx = ’5x 2 + 14 dx
2 2
9. (a)
’1 ’1
1
’5x 3 ’5 5 74
= + 14x = + 14 ’ ’ 14 = .
3 3 3 3
’1
1
1 x3 x4
Area = x ’ x dx = ’
2 3
(b)
3 40
0
11 00 1
= ’ ’ ’ =.
34 34 12
1
1 x3
Area = [’x + 3] ’ 2x dx = ’ + 3x ’ x 2
2
(c)
3
’3 ’3
’27
1
= ’ + 3 · 1 ’ 12 ’ ’ + 3 · (’3) ’ (’3)2
3 3
32
=.
3
e
x2
e
Area = x ’ ln x dx = ’ [x ln x ’ x]
(d)
2
1 1
e2 ’ 3
e2 12
= ’ [e · 1 ’ e] ’ ’ [1 · 0 ’ 1] = .
2 2 2
π/4
x2
π/4
Area = x ’ sin x dx = + cos x
(e)
2
0 0
√ √
π2 π2
2 0 2
= + ’ +1 = + ’ 1.
32 2 2 32 2
3
3 x2
Area = e ’ x dx = e ’
x x
(f)
20
0
9 0 11
= e3 ’ ’ e0 ’ = e3 ’ .
2 2 2
1
1 x2 x3 11 00
Area = x ’ x dx = ’ = ’ ’ ’
2
10. (a)
2 3 23 23
0 0
1
=.
6
293
Chapter 5

1
1√ x 3/2 x 3
Area = x ’ x dx = ’
2
(b)
3/2 3
0 0
21 00 1
= ’ ’ ’ =.
33 33 3



√ 3 ( 3)5
3
3 x5
Area = √ 3x 2 ’x 4 dx = x 3 ’ = ( 3) ’
(c)
5 ’√3 5
’3


√ ( ’3)5 12 3
’ ( ’3)3 ’ = .
5 5
1
’2x 3
1 x5
Area = [’2x + 3] ’ x dx = + 3x ’
2 4
(d)
3 5
’1 ’1
’2 1 2 64
(’1)
= +3’ ’ ’3’ = .
3 5 3 5 15
21/4
21/4 2x 5
Area = [’x + 2] ’ [x ’ 2] dx = 4x ’
4 4
(e)
5
’21/4 ’21/4

2(21/4 )5 2(’21/4 )5
= 4·2 ’ ’ 4 · (’2 )’
1/4 1/4
5 5
32 · 21/4
= .
5
1
1 x3
Area = [’x 2 + 3] ’ 2x dx = ’ + 3x ’ x 2
(f)
3
’3 ’3
’27
1
= ’ + 3 · 1 ’ 12 ’ ’ + 3 · (’3) ’ (’3)2
3 3
32

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