Thomas Calculus 13th [Solutions]

Section 5.6 Substitution and Area Between Curves 393
45.
2/2 dy
2
du
2
, where u 2 y and du 2 dy; y 1 u 2, y
1 2 2 2
y 4 y 1 u u 1
2
u 2,
1
2
1 1
sec u sec 2 sec 2
2
4 3 12
3 y dy 1
16 u 1
du, where u 5y 1 and du 5 dy; y 0 u 1, y 3 u 16,
0 5y
1 25 1 u
46.
1/2
1/2 1/2 3/2 1/2
16
1 u u du 1 2 u 2u
1 2 64 8 2 2 36
25 25 3 1 25 3 3 25
2
47. Let u 4 x du 2 x dx 1 du x dx; x 2 u 0, x 0 u 4, x 2 u 0
2
0 2 2 2 4
1 1/2 0
1 1/2 4
1 1/2 4 1/2
A x
2 4 x dx x
0 4 x dx u du u du
0 2 4 2 2 u du u du
0 2
0
3/2
4
2 3/2 3/2
u 2 (4) 2 (0) 16
3 0 3 3 3
48. Let u 1 cos x du sin x dx; x 0 u 0, x u 2
2
2 2
2 2
2 0
0 (1 cos x ) sin x dx u du u
0 2
0
2 2
2
49. Let u 1 cos x du sin x dx du sin x dx; x u 1 cos ( ) 0, x 0 u 1 cos 0 2
0 2 1/2 2 1/2 3/2
2
3/2 3/2 5/2
A 3(sin x) 1 cos x dx 3 u ( du) 3 u du 2u
2(2) 2(0) 2
0 0 0
50. Let u sin x du cos x dx 1
Because of symmetry about
sin
0 u du [ cos u ] 0 ( cos ) ( cos 0) 2
du cos x dx; x u sin 0, x 0 u
2 2
0
x , A 2 (cos x)(sin( sin x)) dx 2 (sin u)
1 du
2 /2 2 0 2
2 2 1 cos 2x
51. For the sketch given, a 0, b ; f ( x) g( x) 1 cos x sin x ;
2
(1 cos 2 x) 1 1 sin 2x
A dx (1 cos 2 x) dx x
1 [( 0) (0 0)]
0 2 2 0 2 2
0
2 2
2 2 2 2
52. For the sketch given, a , b ; f ( t) g( t) 1 sec t ( 4 sin t) 1 sec t 4sin t;
3 3 2 2
/3
1 2 2 1
/3 2 2 1
/3 2
/3 (1 cos 2 t)
A sec t 4sin t dt sec t dt 4 sin t dt sec t dt 4
dt
/3 2 2 /3 2 /3 /3 2
/3 /3 sin 2
/3
1 2 1 /3 t
sec t dt 2 (1 cos 2 t) dt [tan t] 4
2 /3 /3 2 /3 2 t
3 4 3
2
/3
3 3
2 4 2 2 4
53. For the sketch given, a 2, b 2; f ( x) g( x) 2 x ( x 2 x ) 4 x x ;
2 2 4
A (4 x x ) dx 4x x 32 32 32 32 64 64 320 192 128
2 3 5 3 5 3 5 3 5 15 15
2
2 5 2
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