Answer
$\frac{3}{4e^{2}}$
Work Step by Step
$u$ = $x$
$u'$ = $1$
$v'$ = $e^{-2x}$
$v$ = $-\frac{1}{2}e^{-2x}$
$\int{xe^{-2x}}dx$ = $-\frac{1}{2}xe^{-2x}+\frac{1}{2}\int{e^{-2x}}dx$ = $-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+C$
$\int_1^{\infty}{xe^{-2x}}dx$ = $\lim\limits_{R \to {\infty}}$$[-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}]|_1^R$ = $\frac{3}{4e^{2}}$
