Question

How do you evaluate `int (x^2(x^3 + 1)^3)` for [0, 1]?

Answer 1

`5/4`

Explanation:

`x^3+1=t => 3x^2dx=dt => x^2dx= dt/3`

`t_1 = 0^3+1 = 1`
`t_2 = 1^3+1 = 2`

`I = int_0^1 x^2(x^3+1)^3dx = int_1^2 t^3 dt/3 = 1/3 t^4/4= 1/12 t^4`

`I = 1/12 (2^4-1^4) = 1/12 (16-1) = 15/12 = 5/4`