Question

How do you determine if the improper integral converges or diverges `int 5x^(2)e^(-x^(3))` from 1 to infinity?

Answer 1

Hence, the integral converges,
and the value of the integral is `5/(3e)`

Explanation:

As this is an improper integral (we have an infinite upper bound), we must examine the behaviour of the integral at the upper limit:

First note that `d/dx e^(-x^3) = -3x^2e^(-x^3)`

And so:

`int \ 5x^2 e^(-x^3) \ dx = -5/3 e^(-x^3) + c`

So for our improper integral we have:

`int_1^oo \ 5x^2 e^(-x^3) \ dx =lim_(a rarr oo) int_1^a \ 5x^2 e^(-x^3) \ dx`
`" " = lim_(a rarr oo) [-5/3 e^(-x^3)]_1^a`

`" " = -5/3 \ lim_(a rarr oo) [ 1/e^(x^3) ]_1^a`

`" " = -5/3 \ lim_(a rarr oo) (1/e^(a^2) - 1/e)`
`" " = -5/3 (0 - 1/e)`
`" " = 5/(3e)`