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

Mar 2, 2017

Hence, the integral converges,
and the value of the integral is $\frac{5}{3 e}$

#### 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 $\frac{d}{\mathrm{dx}} {e}^{- {x}^{3}} = - 3 {x}^{2} {e}^{- {x}^{3}}$

And so:

$\int \setminus 5 {x}^{2} {e}^{- {x}^{3}} \setminus \mathrm{dx} = - \frac{5}{3} {e}^{- {x}^{3}} + c$

So for our improper integral we have:

${\int}_{1}^{\infty} \setminus 5 {x}^{2} {e}^{- {x}^{3}} \setminus \mathrm{dx} = {\lim}_{a \rightarrow \infty} {\int}_{1}^{a} \setminus 5 {x}^{2} {e}^{- {x}^{3}} \setminus \mathrm{dx}$
$\text{ } = {\lim}_{a \rightarrow \infty} {\left[- \frac{5}{3} {e}^{- {x}^{3}}\right]}_{1}^{a}$

$\text{ } = - \frac{5}{3} \setminus {\lim}_{a \rightarrow \infty} {\left[\frac{1}{e} ^ \left({x}^{3}\right)\right]}_{1}^{a}$

$\text{ } = - \frac{5}{3} \setminus {\lim}_{a \rightarrow \infty} \left(\frac{1}{e} ^ \left({a}^{2}\right) - \frac{1}{e}\right)$
$\text{ } = - \frac{5}{3} \left(0 - \frac{1}{e}\right)$
$\text{ } = \frac{5}{3 e}$