Answer Key
839
3/2 . One should take C = − 1 3 .
The antiderivative is y = 1 3 ⎛
⎝ 2 x 3 +1 ⎞ ⎠
305 . No, because the integrand is discontinuous at x =1. 307 . u = sin ⎛ ⎝ t 2 ⎞ ⎠ ; the integral becomes 1 2 ∫ 0 0 udu . 309 . u = 2 ⎞ ⎠ ⎟ ; the integral becomes − ∫ 5/4 1 ⎛ ⎝ ⎜ 1+ ⎛ ⎝ t − 1 2 ⎞ ⎠
u du .
5/4
311 . u =1− t ; the integral becomes ∫ 1 −1 u cos( π (1− u )) du = ∫ 1 −1
u [cos π cos πu −sin π sin πu ] du
−1
=− ∫
u cos πudu
1
−1 1
= ∫
u cos πudu =0
since the integrand is odd. 313 . Setting u = cx and du = cdx gets you 1 b c − a c ∫ a / c b / c
⌠ ⌡ u = a u = b
− a ∫ a b
f ( u ) du
f ( cx ) dx = c
f ( u ) du .
c = 1 b
b − a
u 1− a |
1
⎛ ⎝ 1−
1− a ⎞
315 . ⌠ ⌡ 0 x
u =1− x 2 1
⌠ ⌡
⎛ ⎝ 1− x 2
⎞ ⎠
du u a = 1
⎠ . As x →1 the limit is
g ( t ) dt = 1 2
= 1
2(1− a )
2(1− a )
u =1− x 2
1 2(1− a ) if a <1, and the limit diverges to +∞ if a >1. 317 . ∫ t = π 0 b 1−cos 2 t × (− a sin t ) dt = ∫ t =0 π ab sin 2 tdt
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