# How do you integrate xe^x from -oo to 0?

Jul 1, 2016

$= - 1$

#### Explanation:

${\int}_{- \infty}^{0} \mathrm{dx} q \quad x {e}^{x}$

using IBP ie $\int u v ' = u v - \int u ' v$ where, here:

$u = x , u ' = 1$
$v ' = {e}^{x} , v = {e}^{x}$

So we have

${\left[x {e}^{x}\right]}_{- \infty}^{0} - \int \mathrm{dx} q \quad {e}^{x}$

$= {\left[{e}^{x} \left(x - 1\right)\right]}_{- \infty}^{0}$

= [ e^0 (0 - 1)] - [ \color{blue}{e^(-oo) (-oo) } - 1)]

$= - 1$

NB for the bit in blue, ${\lim}_{x \to - \infty} x {e}^{x} = {\lim}_{x \to - \infty} \frac{x}{e} ^ \left(- x\right)$ which is indeterminate $\frac{\infty}{\infty}$so we use L'Hopital

$= {\lim}_{x \to - \infty} \frac{1}{- {e}^{- x}}$

$= - {\lim}_{x \to - \infty} {e}^{x}$

$= - \exp \left({\lim}_{x \to - \infty} x\right)$

$= - {e}^{- \infty} = 0$