Integrals · Question 50

If f(x) = 1x1/4 (1 + x1/4 )dx, f(0) = -6, then f(1) is equal to :

1. 4 ( e 2 - 2 )
2. e2 2 + 2
3. 2 - e2
4. 4 ( e 2 + 2 )

Answer

(1) 4 ( e 2 - 2 )

Solution

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)