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

Aug 5, 2017

#### Explanation:

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

Integrate by substitution $u = x - 1$ to get

$= {\lim}_{b \rightarrow {1}^{-}} 3 {\left[\sqrt[3]{x - 1}\right]}_{0}^{b}$

$= {\lim}_{b \rightarrow {1}^{-}} 3 \left(\sqrt[3]{b - 1} - \sqrt[3]{- 1}\right)$

$= 3 \left(0 - \left(- 1\right)\right) = 3$