Thomas Calculus 13th [Solutions]

1 1
11. (a) Let u 4 5 t t ( u 4), dt du; t 0 u 4, t 1 u 9.
5 5
Section 5.6 Substitution and Area Between Curves 389
1 1 9 1 9 3/2 1/2
t 4 5 t dt ( u 4) u du u 4u du
0 25 4 25 4
9
1 2 5/2 8 3/2 1 2 8 2 8 506
u u
(243) (27) (32) 8)
25 5 3 4 25 5 3 5 3 375
(b) Use the same substitution as in (a); t 1 u 9, t 9 u 49.
9 1 49 3/2 1/2 1 2 5/2 8 3/2
t 4 5t dt u 4u du u u
1 25 9
25 5 3
1 2 8 2 8 86,744
(16,807) (343) (243) 27
25 5 3 5 3 375
12. (a) Let u 1 cos3t du 3sin 3t dt 1 du sin 3 t dt; t 0 u 0, t u 1 cos 1
3 6 2
/6 1
1 1 1 2 1 2
(1 cos3 t)sin 3 t dt u du u (1) (0) 1
0 0 3 3 2 6 6 6
0
2
1
(b) Use the same substitution as in part (a); t u 1, t u 1 cos 2
6 3
/3 2
1 1 1 2 1 2
(1 cos3 t)sin 3 t dt u du u (2) (1) 1
/6 1 3 3 2 6 6 2
1
2
2
13. (a) Let u 4 3sin z du 3cos z dz 1 du cos z dz; z 0 u 4, z 2 u 4
3
2
cos z
4
dz 1 1 du 0
0 4 3sin z 4 u 3
(b) Use the same substitution as in part (a); z u 4 3sin( ) 4, z u 4
cos z
4
dz 1 1 du 0
4 3sin z 4 u 3
2 2
14. (a) Let u 2 tan t du 1 sec t dt 2 du sec t dt; t u 2 tan 1, t 0 u 2
2 2 2 2 2 4
0 2 2 2 2 2 2
/2 2 tan t
2 sec t dt u
2 1
(2 du ) [ u ] 1 2 1 3
(b) Use the same substitution as in part (a); t u 1, t u 3
2 2
/2 2 3 2 3 2 2
/2 (2 tan t
2 )sec t dt
2
2 u du
1
[ u ] 1 3 1 8
5 4
15. Let u t 2 t du (5t 2) dt; t 0 u 0, t 1 u 3
1 5 4 3 1/2 3/2
3
2 2 3/2 2 3/2
t
0 2 t (5 t 2) dt u du u
0 3 0 3 (3) 3
(0) 2 3
dy
16. Let u 1 y du ; y 1 u 2, y 4 u 3
2 y
4 dy 3
1
3 2 1 3
du u du [ u ] 1 1 1
1
2
2
2
2
2
2 y (1 y ) u 3 2 6
49
9
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