# How do you test the improper integral intx^2 dx from (-oo, oo) and evaluate if possible?

Apr 11, 2017

It is divergent.

#### Explanation:

We deal with the indefinite integral as normal: So:

$\int \setminus {x}^{2} \setminus \mathrm{dx} = \frac{1}{3} {x}^{3} \setminus \setminus \left(+ c\right)$

Then as we are dealing with an infinite integration limit we use the limit definition to get:

${\int}_{- \infty}^{\infty} \setminus {x}^{2} \setminus \mathrm{dx} = {\left[\setminus \frac{1}{3} {x}^{3} \setminus\right]}_{- \infty}^{\infty}$

$\text{ } = {\lim}_{n \rightarrow \infty} {\left[\setminus \frac{1}{3} {x}^{3} \setminus\right]}_{- n}^{n}$

$\text{ } = {\lim}_{n \rightarrow \infty} \frac{1}{3} \left({n}^{3} - {\left(- n\right)}^{3}\right)$

$\text{ } = {\lim}_{n \rightarrow \infty} \frac{1}{3} \left({n}^{3} + {n}^{3}\right)$

$\text{ } = {\lim}_{n \rightarrow \infty} \frac{2}{3} {n}^{3}$

Which is clearly divergent (and therefore undefined)

Apr 11, 2017

The integral:

${\int}_{- \infty}^{\infty} {x}^{2} \mathrm{dx}$

is divergent.

#### Explanation:

We have for $t > 0$:

${\int}_{- t}^{t} {x}^{2} \mathrm{dx} = {\left[{x}^{3} / 3\right]}_{- t}^{t} = {t}^{3} / 3 + {t}^{3} / 3 = \frac{2}{3} {t}^{3}$

So:

${\int}_{- \infty}^{\infty} {x}^{2} \mathrm{dx} = {\lim}_{t \to \infty} {\int}_{- t}^{t} {x}^{2} \mathrm{dx} = {\lim}_{t \to \infty} \frac{2}{3} {t}^{3} = + \infty$

so the improper integral is divergent.