Question

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

Answer 1

`3/4`

Explanation:

`int_0^1[x(x^2+1)]dx=int_0^1(x^3+x)dx`

`=[x^4/4+x^2/2]_0^1`

`=1/4+1/2=3/4`