# How do you find the integral from 0 to 2 of xe^(2x) dx?

Jun 6, 2015

$\setminus {\int}_{0}^{2} x {e}^{2 x} \mathrm{dx} = \frac{3}{4} {e}^{4} + \frac{1}{4}$

$\setminus \int u \mathrm{dv} = u v - \int v \mathrm{du}$

Let $u = x , \setminus \implies \mathrm{du} = \mathrm{dx}$

Let $\mathrm{dv} = {e}^{2 x} \mathrm{dx} , \setminus \implies v = \frac{1}{2} {e}^{2 x}$

Substitute $v$ and $u$ into the top expression

$\setminus {\int}_{0}^{2} x {e}^{2 x} \mathrm{dx} = {\left[\frac{x}{2} {e}^{2 x}\right]}_{0}^{2} - {\int}_{0}^{2} \frac{1}{2} {e}^{2 x} \mathrm{dx}$

$\setminus {\int}_{0}^{2} x {e}^{2 x} \mathrm{dx} = \left({e}^{4} - 0\right) - {\left[\frac{1}{4} {e}^{2 x}\right]}_{0}^{2}$

$\setminus {\int}_{0}^{2} x {e}^{2 x} \mathrm{dx} = \frac{3}{4} {e}^{4} + \frac{1}{4}$