Thomas Calculus 13th [Solutions]

Chapter 5 Additional and Advanced Exercises 423
3. 1
x
1
x
1
x
y f ( t) sin a( x t) dt f ( t) sin ax cos at dt f ( t)cos ax sin at dt
a 0 a 0 a 0
sin ax x
cos ax x
dy
f ( t) cos at dt f ( t)sin at dt
a 0 a 0
dx
x sin ax x x cos ax x
cos ax f ( t) cos at dt d f ( t) cos at dt sin ax f ( t)sin at dt d f ( t)sin at dt
0 a dx 0 0 a dx 0
cos ax x
( )cos at sin ax
x
( ( )cos ) sin ( )sin at cos ax ( ( )sin )
0 f t dt a
f x ax ax 0
f t dt a
f x ax
x
x
2
dy
d y
cos ax f ( t)cos at dt sin ax f ( t)sin at dt . Next,
dx
0 0
2
dx
a sin ax x ( )cos at (cos ) ( )cos at cos ( )sin at (sin ) ( )sin at
0 f t dt ax d
x x d
x
dx 0 f t dt a ax 0 f t dt ax dx 0
f t dt
x
x
a sin ax f ( t) cos at dt (cos ax) f ( x)cos ax a cos ax f ( t) sin at dt (sin ax) f ( x) sin ax
0 0
x
x
2
a sin ax f ( t) cos at dt a cos ax f ( t) sin at dt f ( x ). Therefore, y a y
0 0
a 2 sin
cos
cos ax x ( ) sin at sin ( ) cos at ( ) ( ) cos at ( ) sin at
0 f t dt a ax x ax
x ax x
0 f t dt f x a a 0 f t dt a 0
f t dt
f ( x).
Note also that y (0) y(0) 0.
4.
x y 1 dt d ( x)
d y 1 dt d
y 1 dt dy from the chain rule
0 2 0 2 0 2
1 4t dx dx
1 4t dy
1 4t
dx
1 dy dy
2 d y
1 1 4 y . Then
2
1 4 y
dx dx
2
dx
dy
2
4 y 1 4 y
dx
2 2
2
d 2 2 dy
1 4y
d 1 4y
dx dy dx
1/2
4 y
2
1 2
dy
d y
1 4 y (8 y) 4 y . Thus 4 y , and the constant of proportionality
2
dx
2
1 4 y 1 4 y
dx
is 4.
5. (a)
(b)
2 2
x
x
2
f ( t) dt x cos x d f ( t) dt cos x x sin x f ( x )(2 x) 0 dx 0
cos x xsin
x
2
f ( x ) cos x xsin
x
2x
cos 2 2 sin 2
Thus, x 2 f (4)
1
4 4
f ( x) 3 f ( x)
2 1 3 1 3 3
t dt t f ( x) f ( x) x cos x f ( x) 0 3
0
3 3
3x cos x f ( x) 3 3x cos x
f (4) 3 3(4) cos 4
3
12
6.
a
2
f ( x) dx a a sin a cos a . Let
0
2 2 2
a
2
F( a) f ( t) dt f ( a) F ( a ). Now F( a) a a sin a cos a
0
2 2 2
1 1 2
f ( a) F ( a) a sin a a cos a sin a f
sin cos sin 1 1
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
7.
b
( ) 2 1 2 ( ) b
( ) 1 2 1/2
f x dx b f b d f x dx
2
( b 1) (2 b ) b f ( x ) x
db
b
x
1 1 2 2
1 1
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