# How do you integrate x^2 e^-5x dx?

Feb 19, 2015

Hello,

I think you want integrate ${x}^{2} {e}^{- 5 x}$, not ${x}^{2} {e}^{- 5} x$...

Use two integrations by parts :
$\setminus {\int}_{0}^{x} u \left(t\right) v ' \left(t\right) \mathrm{dt} = {\left[u \left(t\right) v \left(t\right)\right]}_{0}^{x} - \setminus {\int}_{0}^{x} u ' \left(t\right) v \left(t\right) \mathrm{dt}$.
You can prove that if you write $\left(u v\right) ' = u ' v + u v '$ and if you integrate between $0$ and $x$.

Let $F$ be the function define by $F \left(x\right) = \setminus {\int}_{0}^{x} {t}^{2} {e}^{- 5 t} \mathrm{dt}$.

Take $u \left(t\right) = {t}^{2}$ and $v \left(t\right) = - \setminus \frac{1}{5} {e}^{- 5 t}$. So you have
$u ' \left(t\right) = 2 t$ and $v ' \left(t\right) = {e}^{- 5 t}$

Apply the integration by parts formula :

$F \left(x\right) = {\left[- \setminus \frac{1}{5} {t}^{2} {e}^{- 5 t}\right]}_{0}^{x} + \setminus \frac{2}{5} \setminus {\int}_{0}^{x} t {e}^{- 5 t} \mathrm{dt}$

Use integration by parts again to calculate $\setminus {\int}_{0}^{x} t {e}^{- 5 t} \mathrm{dt}$. Take
$u \left(t\right) = t$ and $v \left(t\right) = - \setminus \frac{1}{5} {e}^{- 5 t}$. So you have
$u ' \left(t\right) = 1$ and $v ' \left(t\right) = {e}^{- 5 t}$, and

$\setminus {\int}_{0}^{x} t {e}^{- 5 t} \mathrm{dt} = {\left[- \setminus \frac{1}{5} t {e}^{- 5 t}\right]}_{0}^{x} + \setminus \frac{1}{5} \setminus {\int}_{0}^{x} {e}^{- 5 t} \mathrm{dt} = {\left[- \setminus \frac{1}{5} t\right]}_{0}^{x} + \setminus \frac{1}{5} {\left[- \setminus \frac{1}{5} {e}^{- 5 t}\right]}_{0}^{x}$

You can simplify :

$\setminus {\int}_{0}^{x} t {e}^{- 5 t} \mathrm{dt} = - \setminus \frac{1}{5} x {e}^{- 5 x} - \setminus \frac{1}{25} \left({e}^{- 5 x} - 1\right)$

Finally, plug that in the above expression :

$F \left(x\right) = - \setminus \frac{1}{5} {x}^{2} {e}^{- 5 x} - \setminus \frac{2}{5} \left(\setminus \frac{1}{5} x {e}^{- 5 x} + \setminus \frac{1}{25} \left({e}^{- 5 x} - 1\right)\right)$

Primitives (or antiderivatives) of ${x}^{2} {e}^{- 5 x}$ are

$- \left(\setminus \frac{1}{5} {x}^{2} + \setminus \frac{2}{25} x + \frac{2}{125}\right) {e}^{- 5 x} + c$

where $c \setminus \in \mathbb{R}$.