# How do you test the improper integral int 2x^-3dx from [-1,1] and evaluate if possible?

Feb 21, 2018

The integral diverges

#### Explanation:

To calculate the improper integral, proceed as follows

${\int}_{- 1}^{1} 2 {x}^{-} 3 \mathrm{dx}$

$= {\lim}_{t \to {O}^{-}} \left({\int}_{-} {1}^{t} 2 {x}^{-} 3 \mathrm{dx}\right) + {\lim}_{t \to {O}^{+}} \left({\int}_{t}^{1} 2 {x}^{-} 3 \mathrm{dx}\right)$

The first part is

${\lim}_{t \to {O}^{-}} \left({\int}_{-} {1}^{t} 2 {x}^{-} 3 \mathrm{dx}\right)$

$= {\lim}_{t \to {O}^{-}} {\left[- \frac{1}{x} ^ 2\right]}_{1}^{t}$

$= {\lim}_{t \to {O}^{-}} \left(1 - \frac{1}{t} ^ 2\right)$

$= + \infty$

The integral diverges

The second part is

${\lim}_{t \to {O}^{+}} \left({\int}_{t}^{1} {x}^{-} 3 \mathrm{dx}\right)$

$= {\lim}_{t \to {O}^{+}} {\left[- \frac{1}{x} ^ 2\right]}_{t}^{1}$

$= {\lim}_{t \to {O}^{+}} \left(\frac{1}{t} ^ 2 - 1\right)$

$= + \infty$

The integral diverges