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

May 16, 2017

Integrate by substitution and evalute.

#### Explanation:

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

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

Let $u = 2 x - 1$ so that $\mathrm{du} = 2 u$ and $\int {u}^{-} 3 \mathrm{du} = - \frac{1}{2} {u}^{-} 2 + C$

So,

${\lim}_{b \rightarrow - \infty} \frac{1}{2} {\int}_{b}^{0} {\left(2 x - 1\right)}^{-} 3 2 \mathrm{dx} = {\lim}_{b \rightarrow - \infty} {\left[- \frac{1}{4} {\left(2 x - 1\right)}^{-} 2\right]}_{b}^{0}$

$= {\lim}_{b \rightarrow - \infty} \left(- \frac{1}{4} + \frac{1}{4 {\left(2 b - 1\right)}^{2}}\right)$

$= - \frac{1}{4}$