Thomas Calculus 13th [Solutions]

Chapter 5 Practice Exercises 413
33.
e
2ln
1 2 1
A x dx
1 0 2 u du
x
[ u ] 0 1, where u = ln x and du 1 dx;
x = 1 u = 0, x = e u = 1
x
20 20
34. (a) A 1
20
1 dx ln x ln 20 ln10 ln ln 2, and
10 x
10
10
2
1
2
A2 dx ln x ln 2 ln1 ln 2
1 x
1
(b) A kb
1
kb
1 ln ln ln ln kb ln b ln ln ,
ka x dx x ka
kb ka ka a
b a and A b
2 1
b
ln ln ln
a x
dx x a
b a
35.
2 x
1 dy
y x dt 2x
1
1 t dx x
2
d y
dx
2 1 ; 1
y (1) 1 1
1 t
1 and y (1) 2 1 3
x
2 2
36.
x dy
2
d y
2
y 1 1/2
(1 2 sec ) 1 2 sec 2 (sec ) (sec tan )
0 t dt dx
x dx 2
x x x sec x (tan x );
0
dy
x 0 y (1 2 sec t ) dt 0 and x 0 1 2 sec0 3
0
dx
x
37. sin t
dy
y dt 3 sin x ; x 5
5 t
dx x
y
5 sin t dt
5 t
3 3
38.
x 2
y
1 2 sin t dt 2 so that
dy
dx
2 1 2
2 sin x; x 1 y 2 sin t dt 2 2
1
39.
dy
dx
1
1
dy dx y sin x C;
x = 0 and y = 0
2 2
1 x
1 x
1
0 sin 0 C C = 0
y
sin
1
x
dy
40. 1 1
1 dy 1 1 dx y tan ( x) x C;
dx 2 2
x 1 1 x
1
1
x = 0 and y = 1 1 tan 0 0 C C = 1 y tan ( x) x 1
41.
dy 1
1
dy dx y sec x C;
dx 2 2
x x 1 x x 1
x = 2 and y =
1 1 2 1 2
3 3 3
sec 2 C C sec 2 y sec ( x) , x > 1
dy 1 2 1 2
dx 1 x 1 x
1 x 1 x
42. dy dx y tan x 2sin x C;
2 2
x = 0 and y = 2
2 2
1 1
2 tan 0 2sin 0 C C = 2
1 1
1 1
y tan x 2sin x 2
43. Let u cos x du sin x dx du sin x dx
1/2 1/2
2(cos x) sin x dx 2 u ( du) 2
1/2
1
2
1/2 1/2 1/2
u du 2 u C 4u C 4(cos x)
C
2
44. Let u tan x du sec x dx
3/2 2
1/2
(tan x)
sec x dx u 3/2 du u C 2u 1/2 C 2 C
1/2
(tan x)
1
2
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