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sin3 x cos2 x dx = sin x(1 ’ cos2 x) cos2 x dx.

Let u = cos x, du = ’ sin x dx. Then the integral becomes
u3 u5
’ (1 ’ u )u du = ’ + + C.
2 2
3 5
Resubstituting x, we obtain the ¬nal answer
cos3 x cos5 x
sin x cos x dx = ’ + + C.
3 2
3 5
Let u = tan x, du = sec2 x dx. Then the integral becomes
(c)
u4
u du = + C.
3
4
Resubstituting x, we obtain the ¬nal answer
tan4 x
tan x sec x dx = + C.
3 2
4
Let u = sec x, du = sec x tan x. Then the integral becomes
(d)
u3
u2 du = + C.
3
Resubstituting x, we obtain the ¬nal answer
sec3 x
tan x sec x dx = + C.
3
3
5. We do (a), (b), (c), (d).
(a) Use integration by parts twice:
1
1 1
e sin x dx = sin x ·e ’
x x
ex cosx dx
0 0
0
1 1
= [e·sin 1’0]’ cosxe ’
x
ex (’sin x)dx
0
0
1
= e·sin 1’e·cos1+1’ ex sin x dx.
0
307
Chapter 7

We may now solve for the desired integral:
1 1
ex sin x dx = [e · sin 1 ’ e · cos 1] .
2
0

Integrate by parts with u = ln x, dv = x 2 dx. Thus
(b)
e
x3 x3 1
e e
x ln x dx = ln x · ’ · dx
2
3 3x
1 1
1
e
e3 13 x 3
=1· ’0· ’
3 3 6 1
e3 e3 13
= ’ + .
3 9 9
(c) We write
2x + 1 ’1
1 1
= + 2+ .
x+1
x 2 (x + 1) x x
Thus
(2x + 1) dx ’1
4 4 4 4
1 1
= dx + dx + dx
x+1
x3 + x2 x2
x
2 2 2 2

’1 ’1
= [ln 4 ’ ln 2] + ’ + [ln 3 ’ ln 5]
4 2
61
= ln +.
54
(d) We write
π π
1
sin x cos x dx =
2 2
sin2 2x dx
4
0 0
1 ’ cos 4x
π
1
= dx
4 2
0
π
1 sin 4x
= x’
8 4 0
1
= [(π ’ 0) ’ (0 ’ 0)]
8
π
=.
8
308 Solutions to Exercises

Chapter 8

1. At position x in the base circle, the y-coordinate is 1 ’ x 2 . Therefore the

half-disk slice has radius 1 ’ x 2 and area π(1 ’ x 2 )/2. The volume of
the solid is then
π(1 ’ x 2 )
1
V= dx
2
’1
1
x3
π
= x’
2 3 ’1

’1
1
π
= 1’ ’ (’1) ’
2 3 3

=
.
3



Y
2. We calculate the volume of half the solid, and then double the answer.

For 0√ x ¤ 1/ 2, at position x in the base square, the y-coordinate
¤ √
FL
is 1/ 2 ’ x. Thus the disk slice has radius (1/ 2 ’ x) and area

π(1/ 2 ’ x)2 . Thus the volume of the solid is

AM


1/ 2
V =2 π(1/ 2 ’ x)2 dx
0

1/ 2
3
TE



2π 1
=’ √ ’x
3 2 0
3
2π 3 1
=’ 0’ √
3 2
π
= √.
32
5
π [x 2 ]2 dx
3. (a)
2
3
π [y 2 ]2 dy
(b)
1
2
π [x 3/2 + 1]2 dx
(c)
0
2
π [5 ’ (x + 3)]2 dx
(d)
’1
309
Chapter 8

6
π [y 2 + 2]2 dy
(e)
2
π/2
π[sin x]2 dx
(f )
0
4
1 + [(2/3)x ’1/3 ]2 dx
2π · x 2/3 ·
4. (a)
0

3
2π · y 2 · 1 + [2y]2 dy
(b)
0

3
2π · [x 2 ’ 2] · 1 + [2x]2 dx
(c)
0
π
2π · sin x · 1 + [cos x]2 dx
(d)
0
2
2π · [y 2 + 2] · 1 + [2y]2 dy
(e)
1
1
2π · x 3 · 1 + [3x 2 ]2 dx
(f )
0
5. The depth of points in the window ranges from 6 to 10 feet. At depth x

in this range, the window has chord of length 2 16x ’ x 2 ’ 60. Thus the
total pressure on the lower half of the window is
10
P= 62.4 · x · 2 16x ’ x 2 ’ 60 dx .
8

6. At depth x, the corresponding subtriangle has side-length 2(5 ’ x/ 3).
Therefore the total pressure on one end of the pool is


53
P= 62.4 · x · 2(5 ’ x/ 3) dx.
0
7. Let t = 0 be the moment when the climb begins. The weight of the sack at
time t is then 100 ’ t pounds. Then the work performed during the climb is
8
W= (100 ’ t) · 5 dt.
0
Thus
8
5t 2 320 0
W = 500t ’ = 4000 ’ ’ 100 · 0 ’ = 3840 ft lbs.
2 2 2
0
310 Solutions to Exercises

8. The work performed is
100
W= [3x 2 + 4x + 6] dx
3
100
= x 3 + 2x 2 + 6x
3
= (1000000 + 20000 + 600) ’ (27 + 18 + 18)
= 1020547 ft lbs.
π
1 + [cos x]2 dx
9. (a)
0
8
1 + [(2/3)x ’1/3 ]2 dx
(b)
1
π/2
1 + [’ sin y]2 dy
(c)
0
4
1 + [2x]2 dx
(d)
1
5
1
sin2 x dx
10. (a)
3 2
π/4
1
(b) tan x dx
π/4 0
2
1 x
(c) dx
x+1
4 ’2

1 sin x
(d) dx
2 + cos x
3π ’π
6
2
e’(2j/3) ·
2
11. (a)
3
j =1
10
2
sin(e’2+2j/5 ) ·
(b)
5
j =1
5
2
cos(’2 + 2j/5)2 ·
(c)
5
j =1
12
ej/3 1
·
(d)
2 + sin(j/3) 3
j =1
311
Chapter 8

12. We do (a) and (b).
2/3 ’02
+ 2 · e’(2/3) + 2 · e’(4/3) + 2 · e’(6/3)
2 2 2
(a) e
2
+2 · e’(8/3) + 2 · e’(10/3) + e’(12/3)
2 2 2


2/5
sin(e’10/5 ) + 2 · sin(e’8/5 ) + 2 · sin(e’6/5 )
(b)
2
+ 2 · sin(e’4/5 ) + 2 · sin(e’2/5 ) + 2 · sin(e0/5 )
+ 2 · sin(e2/5 ) + 2 · sin(e4/5 ) + 2 · sin(e6/5 )
+ 2 · sin(e8/5 ) + · sin(e10/5 )
13. We do (a) and (b)
2/3 ’02
+ 4e’(2/3) + 2 · e’(4/3) + 4 · e’(6/3)
2 2 2
(a) e
3
+ 2 · e’(8/3) + 4 · e’(10/3) + e’(12/3)
2 2 2


2/5
sin(e’10/5 ) + 4 · sin(e’8/5 ) + 2 · sin(e’6/5 )
(b)
3
+ 4 · sin(e’4/5 ) + 2 · sin(e’2/5 ) + 4 · sin(e0/5 )
+ 2 · sin(e2/5 ) + 4 · sin(e4/5 ) + 2 · sin(e6/5 )
+ 4 · sin(e8/5 ) + sin(e10/5 )
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FINAL EXAM




1. The operations that preserve rational numbers are
(a) addition, multiplication, subtraction, division
(b) addition and multiplication
(c) multiplication and division
(d) square roots and logarithm
(e) sine and cosine
2. The number 3.157575757 . . . , expressed as a rational fraction, is
3
(a)
2
2976
(b)
355
1563
(c)
495
2
(d)
3
111
(e)
222
√ √2
The number ( 3 + 2) is
3.
(a) rational
(b) irrational

313

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