# What is the net area between f(x) = e^(2x)-xe^x  and the x-axis over x in [2, 4 ]?

Jul 7, 2016

$= \frac{1}{2} {e}^{8} - \frac{7}{2} {e}^{4} + {e}^{2}$

#### Explanation:

We can split the integral up:

${\int}_{2}^{4} \left({e}^{2 x} - x {e}^{x}\right) \mathrm{dx} = {\int}_{2}^{4} {e}^{2 x} \mathrm{dx} - {\int}_{2}^{4} x {e}^{x} \mathrm{dx}$

The first integral is easy

${\left[\frac{1}{2} {e}^{2 x}\right]}_{2}^{4} = \frac{1}{2} \left({e}^{8} - {e}^{4}\right)$

For the second, we need to use integration by parts:

$\int u \cdot v ' = u \cdot v - \int u ' \cdot v$

Let $u \left(x\right) = x$ and $v ' \left(x\right) = {e}^{x}$
then $u ' \left(x\right) = 1$ and $v \left(x\right) = {e}^{x}$

$\int x {e}^{x} \mathrm{dx} = x {e}^{x} - \int {e}^{x} \mathrm{dx} = {e}^{x} \left(x - 1\right)$

Therefore, we have

$\frac{1}{2} \left({e}^{8} - {e}^{4}\right) - {\left[{e}^{x} \left(x - 1\right)\right]}_{2}^{4}$

$\frac{1}{2} \left({e}^{8} - {e}^{4}\right) - \left[3 {e}^{4} - {e}^{2}\right]$

$= \frac{1}{2} {e}^{8} - \frac{7}{2} {e}^{4} + {e}^{2}$