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

May 21, 2018

The answer is $= 6$

#### Explanation:

The improper integral is

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

There is an undefined point when $x = 0$

Therefore,

${\int}_{-} {1}^{1} {x}^{- \frac{2}{3} \mathrm{dx}} = {\lim}_{p \to 0} {\int}_{-} {1}^{p} {x}^{- \frac{2}{3}} \mathrm{dx} + {\lim}_{p \to 0} {\int}_{p}^{1} {x}^{- \frac{2}{3}} \mathrm{dx}$

${\lim}_{p \to 0} {\int}_{-} {1}^{p} {x}^{- \frac{2}{3}} \mathrm{dx} = {\lim}_{p \to 0} {\left[3 {x}^{\frac{1}{3}}\right]}_{-} {1}^{p}$

$= {\lim}_{p \to 0} \left(3 {p}^{\frac{1}{3}} + 3\right)$

$= 3$

${\lim}_{p \to 0} {\int}_{p}^{1} {x}^{- \frac{2}{3}} \mathrm{dx} = {\lim}_{p \to 0} {\left[3 {x}^{\frac{1}{3}}\right]}_{p}^{1}$

$= {\lim}_{p \to 0} \left(3 - 3 {p}^{\frac{1}{3}}\right)$

$= 3$

Finally,

${\int}_{-} {1}^{1} {x}^{- \frac{2}{3} \mathrm{dx}} = 3 + 3 = 6$