Thomas Calculus 13th [Solutions]

624 Chapter 8 Techniques of Integration
(b)
dx ; [ t ln x ] dt and this integral is essentially the same as in Exercise 65(a): it converges
2
p
x(ln x)
ln 2
p
t
for p 1 and diverges for p 1
2x dx
b
2 2 2
66. lim ln x 1 lim ln b 1 0 lim ln b 1 the integral 2x
dx
0
2
2
x 1 b 0 b b
x 1
b 2x dx
b
2 2 2
2
diverges. But lim lim ln x 1 lim ln b 1 ln b 1 lim ln b 1
2 2
b x 1 b b b b b 1
lim (ln1) 0
b
67.
A
x
x
b
e dx lim e
0 b
0
lim
b
e
0
e 0 1 1
b
68.
69.
70.
x 1 x x x
b
b b
0 0
xe dx lim xe e lim be e 0 e e
A 0 b
0 b
0 1 1;
y
2
1 x 1 2x 1 1 2x b
1 1 2b
1 1 2 0
e dx e dx lim e lim e e 0 1 1
2A
0 2 0 b
2 2 0 b
2 2 2 2 4 4
V
x x x x
b
b b
2 xe dx 2 xe dx 2 lim xe e 2 lim be e
0 0 b
0 b
1 2
V
x
2
2x 1 2x b
1 2b
e dx e dx lim e lim e 1
0 0 b
2 0 b
2 2 2
71.
/2
b
A (sec x tan x) dx lim ln |sec x tan x| ln |sec x| lim ln 1 tan b ln 1 0
0
0 secb
b
b
lim ln 1 sin b ln 2
b
2
2 2
72. (a)
(b)
/2 2 /2 2 /2 2 2 /2 2 2
V sec x dx tan x dx sec x tan x dx sec x sec x 1 dx
0 0 0 0
/2 2
dx
0 2
/2 2 2 /2
2
b
Souter
2 sec x 1 sec x tan x dx 2 sec x(sec x tan x) dx lim tan x
0 0
b
0
2 2
lim tan b 0 lim tan b S outer diverges;
b
b
2 2
2 2 2
2
/2 4
Sinner
2 tan x 1 sec x dx
0
/2 2 2
b
2 2
2 tan x sec x dx lim tan x lim tan b 0 lim tan b S
0
inner
b
0
b b
diverges
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