# How do you find the integral of t^3 e^(-t^2)?

May 1, 2018

$\int {t}^{3} {e}^{- {t}^{2}} \mathrm{dt} = - \frac{1}{2} {e}^{- {t}^{2}} \left({t}^{2} + 1\right) + c$, where $c$ is a constant

#### Explanation:

Let ${t}^{2} = x$, then $2 t \mathrm{dt} = \mathrm{dx}$ and

$\int {t}^{3} {e}^{- {t}^{2}} \mathrm{dt} = \frac{1}{2} \int x {e}^{- x} \mathrm{dx}$

Now let us use integration by parts, which states that

$\int u \left(x\right) v ' \left(x\right) \mathrm{dx} = u \left(x\right) v \left(x\right) - \int v \left(x\right) u ' \left(x\right) \mathrm{dx}$

Let $u = x$ and $\mathrm{dv} = {e}^{-} x$, then $\mathrm{du} = 1$ and $v = - {e}^{- x}$

and hence $\int x {e}^{- x} \mathrm{dx} = - x {e}^{- x} + \int {e}^{- x} \mathrm{dx}$

= $- x {e}^{- x} - {e}^{- x} + {c}_{1}$ where ${c}_{1}$ is a constant.

Hence $\int {t}^{3} {e}^{- {t}^{2}} \mathrm{dt} = \frac{1}{2} \int x {e}^{- x} \mathrm{dx}$

= $\frac{1}{2} \left[- x {e}^{- x} - {e}^{- x} + {c}_{1}\right]$

= $- \frac{1}{2} x {e}^{- x} - \frac{1}{2} {e}^{- x} + {c}_{1} / 2$

= $- \frac{1}{2} {t}^{2} {e}^{- {t}^{2}} - \frac{1}{2} {e}^{- {t}^{2}} + c$

= $- \frac{1}{2} {e}^{- {t}^{2}} \left({t}^{2} + 1\right) + c$, where $c$ is another constant