Thomas Calculus 13th [Solutions]

Section 5.4 The Fundamental Theorem of Calculus 373
44. (a)
(b)
t
4
5
t
3 1 1 5/2 1
x dx x 3sin x t 3sin t
5 5
0 2
1 x
0
d
dt
d
t 4 3 1 5/2 1 1 5 3/2 3 1 1 3/2
x dx d t 3sin t t t 3
dt 0 1 x
dt 5 5 2 1 t 2 t 2
2 t t
2 2
t 4 3 2 3 2 3 1 1 3/2
x dx t d t t t 3
0 1 x
1 t dt
1 t 2 t 2
2 t t
2 2
45.
x 2 dy 2
x
y 1 t dt 1 x 46. 1 dy 1
0
dx
1 t dx x
y dt , x 0
47.
0 sin 2 x sin 2 dy sin( ) 2 d ( ) (sin ) 1 1/2 sin x
x
0 dx
dx
2
2 x
y t dt t dt x x x x
48.
2 2 2 2
x 3 dy 3 3 2 3 2 3
sin
d
x x
sin 1 sin sin ( )
d
x
( ) sin
2 dx dx 2 2 dx 2
y x t dt x t dt t dt x x x t dt
2 6 x 3
2x sin x sin t dt
2
2
49.
x 2 x 2 dy 2 2
y t dt t dt
x x
1
2
3
2 2 2
t 4 t 4 dx x 4 x 4
0
50.
3 2 2
y x 3 10 3 10 3 10 3 10 3 10
(
0 t 1) dt dy
3 x ( 1) d
x ( 1) 3( 1) x ( 1)
dx 0 t dt dx 0 t dt x 0
t dt
51.
sin x
1
0 , dy
y dt x d
2 2
(sin x )
2
1 t
dx
1 sin x
dx
1
cos
2
(cos x)
x
cos x
cos x
cos x 1 since x
cos x
2
y tan x
dt dy 1 d
1 2
(tan ) sec 1
0 1 t dx 1 tan x dx
x x
sec x
52.
2 2 2
53.
x
2
e
2 2 1
1 dy 1 x 1 x x
y dt d e 2xe 2xe
2
0
dx 2
1 x
2
t
x
dx
e
e2
2
54.
x
1 3 2 1/3 dy x 1/3 x x
(2 ) (2 ) 2 /3 x
2 ln 2 2 4 x
y /3
x t dt t dt
d
ln 2
2 1
dx
dx
55.
1
sin t 1 1 2 1
0 cos dy
y t dt cos(sin x ) d (sin x
dx
dx
) 1 x 1
2
1 x
56.
1/ 1 1 1 1 1
y x 1 dy 1 1
sin sin d sin 1
1 t dt dx
x dx
x x x
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