Question

If `intx^5e^(4x^2)dx=(1)/(48)e^(4x^2)(f(x))+c`, where c is contant of intergration then `f(x)` equals to (a) `-4x^3-1` (b) `-1-2x^3` (c) `4x^3+1` (d) `1-2x^3`
A. `-4x^(3)-1`
B. `4x^(3)+1`
C. `-2x^(3)-1`
D. `-2x^(3)+1`

Answer 1

Correct Answer - A
Given, `int x^(5) e^(-4x^(3))dx=(1)/(48)e^(-4x^(3))f(x)+C`
In LHS, put `x^(3) = t`
`rArr" "3x^(2)dx=dt`
So, `int x^(5)e^(-4x^(3))dx =(1)/(3)int te^(-4t)dt`
`=(1)/(3)[t(e^(-4t))/(-4)-int (e^(-4t))/(-4)dt]" "["using integration by parts"]`
`=(1)/(3)[(te^(4t))/(-4)+(e^(-4t))/(-16)]+C`
`=-(1)/(48)e^(-4t)[4t+1]+C`
`=-(e^(-4x^(3)))/(48)[4x^(3)+1]+C" "[therefore t=x^(3)]`
`therefore" "f(x) = -1 -4X^(3)` (comparing with given equation)