# How do you integrate int xe^(-x/2) by parts from [0,4]?

Mar 22, 2018

$- 4 \left(3 {e}^{-} 2 - 1\right)$

#### Explanation:

Let's select our values of $u , \mathrm{dv}$ and integrate and differentiate for $\mathrm{du} , v$:

$u = x$
$\mathrm{du} = \mathrm{dx}$

$\mathrm{dv} = {e}^{- \frac{x}{2}} \mathrm{dx}$
$v = - 2 {e}^{- \frac{x}{2}}$

Note that our bounds of integration will not change -- we're not performing any substitutions.

So, for definite integrals, we integrate by parts in the following way:

${\int}_{a}^{b} u \mathrm{dv} = u v {|}_{a}^{b} - {\int}_{a}^{b} v \mathrm{du}$

Thus,

${\int}_{0}^{4} x {e}^{- \frac{x}{2}} \mathrm{dx} = - 2 x {e}^{- \frac{x}{2}} {|}_{0}^{4} + 2 {\int}_{0}^{4} {e}^{-} \frac{x}{2} \mathrm{dx} = \left(- 8 {e}^{-} 2\right) - 4 {e}^{-} \left(\frac{x}{2}\right) | {0}^{4} = - 8 {e}^{-} 2 - 4 {e}^{-} 2 + 4 = - 12 {e}^{-} 2 + 4 = - 4 \left(3 {e}^{-} 2 - 1\right)$