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

Apr 13, 2017

${\int}_{2}^{\infty} {x}^{-} 2 \mathrm{dx} = \frac{1}{2}$

#### Explanation:

We will find the antiderivative of ${x}^{-} 2$ as normal. When we "evaluate" at infinity, we will take the limit of the antiderivative at infinity instead.

Note that:

$\int {x}^{-} 2 \mathrm{dx} = {x}^{-} \frac{1}{- 1} + C$

$\textcolor{w h i t e}{\int {x}^{-} 2 \mathrm{dx}} = - \frac{1}{x} + C$

So:

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

$\textcolor{w h i t e}{{\int}_{2}^{\infty} {x}^{-} 2 \mathrm{dx}} = \left[{\lim}_{x \rightarrow \infty} \left(- \frac{1}{x}\right)\right] - \left(- \frac{1}{2}\right)$

The limit approaches $0$:

$\textcolor{w h i t e}{{\int}_{2}^{\infty} {x}^{-} 2 \mathrm{dx}} = 0 - \left(- \frac{1}{2}\right)$

$\textcolor{w h i t e}{{\int}_{2}^{\infty} {x}^{-} 2 \mathrm{dx}} = \frac{1}{2}$

So, the area under $\frac{1}{x} ^ 2$ from $x = 2$ onwards infinitely is only $\frac{1}{2}$.