How do you integrate $x e^{x}$ $xe^x$ from $- \infty$ $-oo$ to 0?

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Answer 1 · 2016-07-01T08:12:46

$= - 1$ $= - 1$

$int _(-oo)^0 dx qquad x e^x$

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

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

So we have

$[x e^x]_(-oo)^0 - int dx qquad e^x$

$= [ e^x (x - 1)]_(-oo)^0$

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

$= - 1$

NB for the bit in blue, $lim_{x to - oo} x e^x = lim_{x to - oo} x/e^(-x)$ which is indeterminate $oo/oo$ so we use L'Hopital

$= lim_{x to - oo} 1/(-e^(-x))$

$=- lim_{x to - oo} e^(x)$

$=- exp( lim_{x to - oo} x)$

$= -e^(- oo) = 0$

How do you integrate xe^xxexxe^x from -oo−∞-oo to 0?