# How do you integrate int2xe^x dx from 0 to 1?

Apr 5, 2015

By parts:

${\int}_{a}^{b} 2 x {e}^{x} \mathrm{dx} = u v - {\int}_{a}^{b} v \mathrm{du}$

Let u = $2 x$ and v = ${e}^{x}$.
$\mathrm{du}$ = $2 \mathrm{dx}$

${\int}_{0}^{1} 2 x {e}^{x} \mathrm{dx} = \left(2 x \cdot {e}^{x}\right) - {\int}_{0}^{1} {e}^{x} \cdot 2 \mathrm{dx}$
$= \left[\left(2 x \cdot {e}^{x}\right) - 2 {\int}_{0}^{1} {e}^{x} \mathrm{dx}\right] e v a l \left(0 \to 1\right)$
$= \left[2 \left(1\right) {e}^{1} - 2 \left(0\right) {e}^{0}\right] - 2 \left({e}^{1} - {e}^{0}\right)$
$= \left[2 e\right] - 2 e + 2$
$= 2$