# Question #da50f

Oct 2, 2014

${\int}_{- \infty}^{0} x {5}^{- {x}^{2}} \mathrm{dx}$

by the substitution $u = - {x}^{2}$.
$R i g h t a r r o w \frac{\mathrm{du}}{\mathrm{dx}} = - 2 x R i g h t a r r o w \mathrm{dx} = \frac{\mathrm{du}}{- 2 x}$
As $x$ goes from $- \infty$ to $0$, $u$ goes from $- \infty$ to $0$.

$= {\int}_{- \infty}^{0} x {5}^{u} \frac{\mathrm{du}}{- 2 x}$

by cancelling out $x$'s and pulling $- \frac{1}{2}$ out of the integral,

$= - \frac{1}{2} {\int}_{- \infty}^{0} {5}^{u} \mathrm{du}$

by the definition of improper integral,

$= - \frac{1}{2} {\lim}_{t \to - \infty} {\int}_{t}^{0} {5}^{u} \mathrm{du}$

by finding an anti-derivative,

$= - \frac{1}{2} {\lim}_{t \to - \infty} {\left[\frac{{5}^{u}}{\ln 5}\right]}_{t}^{0}$

by pulling $\frac{1}{\ln 5}$ out,

$= - \frac{1}{2 \ln 5} {\lim}_{t \to - \infty} \left(1 - {5}^{t}\right)$

since ${\lim}_{t \to - \infty} {5}^{t} = 0$,

$= - \frac{1}{2 \ln 5}$

Hence, the improper integral converges to $- \frac{1}{2 \ln 5}$.

I hope that this was helpful.