Question

How do you evaluate the integral `int_0^oo x^2e^(-x^3) dx`?

Answer 1

The integral converges and gives a value of `1/3`

Explanation:

We have:

`I = lim_(c -> oo) int_0^c x^2e^(-x^3)dx`

Now integrate using substitution. If we let `u = -x^3`, then `du = -3x^2dx` and `dx= (du)/(-3x^2)`.

`I = lim_(c->oo) int_0^c x^2e^(u) -(du)/(3x^2)`

`I = lim_(c-> oo) int_0^c -1/3e^(u) du`

`I = lim_(c->oo) [-1/3e^(-x^3)]_0^c`

`I = lim_(c->oo) (-1/3e^(-c^3)) - (-1/3e^0))`

The limit to infinity of `-1/3e^(-c^3)` is clearly `0`, because the larger the value of `c`, the closer `-1/3e^(-c^3)` gets to `0`.

`I = 0 + 1/3`

`I = 1/3`

Hopefully this helps!