# How do you test the improper integral int (2x-1)^-3dx from [0,1/2] and evaluate if possible?

Feb 19, 2018

The integral is divergent.

#### Explanation:

To calculate the improper integral, proceed as follows :

First, calculate the indefinite integral

${\int}_{0}^{t} {\left(2 x - 1\right)}^{-} 3 \mathrm{dx}$

$= {\left[- \frac{1}{4} {\left(2 x - 1\right)}^{-} 2\right]}_{0}^{t}$

$= {\left[\frac{1}{4} {\left(2 x - 1\right)}^{-} 2\right]}_{t}^{0}$

$= \frac{1}{4} - \frac{1}{4} {\left(2 t - 1\right)}^{-} 2$

Second, determine the limits as $t \to \frac{1}{2}$

${\lim}_{t \to \frac{1}{2}} {\int}_{0}^{t} {\left(2 x - 1\right)}^{-} 3 \mathrm{dx}$

$= {\lim}_{t \to \frac{1}{2}} \left(\frac{1}{4} - \frac{1}{4} {\left(2 t - 1\right)}^{-} 2\right)$

$= \frac{1}{4} - \frac{1}{4} {\left(\frac{1}{0}\right)}^{2}$

$= - \infty$