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

Jan 16, 2018

The integral converges to $= 3$

#### Explanation:

The critical point is when $x = \frac{1}{2}$

Therefore,

The integral is

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

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

$= {\lim}_{t \to \left(\frac{1}{2}\right)} {\left[\frac{3}{2} {\left(2 x - 1\right)}^{\frac{1}{3}}\right]}_{0}^{t} + {\lim}_{t \to \left(\frac{1}{2}\right)} {\left[\frac{3}{2} {\left(2 x - 1\right)}^{\frac{1}{3}}\right]}_{t}^{1}$

$= {\lim}_{t \to \left(\frac{1}{2}\right)} \left(\frac{3}{2} \cdot {\left(2 t - 1\right)}^{\frac{1}{3}} - \frac{3}{2} \cdot {\left(- 1\right)}^{\frac{1}{3}}\right) + {\lim}_{t \to \left(\frac{1}{2}\right)} \left(\frac{3}{2} \cdot {\left(1\right)}^{\frac{1}{3}} - \frac{3}{2} \cdot {\left(2 t - 1\right)}^{\frac{1}{3}}\right)$

$= \frac{3}{2} + \frac{3}{2}$

$= 3$