If f(x) = 1x1/4 (1 + x1/4 )dx, f(0) = -6, then f(1) is equal to :
(1) 4 ( e 2 - 2 )
Put x = t4 dx = 4t3 dt
f(x) = 4t3t(1 + t)dt = 4 t21 + tdt
= 4 (t - 1 + 11 + t )dt
= 4 [ t22 - t + e(1 + t) ] + C, where t = x1/4
f(0) = -6 C = -6
f(1) = 4 [ 12 - 1 + e 2 ] - 6
= -2 + 4 e 2 - 6 = 4 e 2 - 8
= 4 ( e 2 - 2 )
Answer: (1)
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