# What is the average value of the function f(t)=te^(-t^2 ) on the interval [0,5]?

Oct 22, 2016

It is $\frac{1}{10} \left(1 - {e}^{-} 25\right)$

#### Explanation:

$\frac{1}{5 - 0} {\int}_{0}^{5} t {e}^{- {t}^{2}} \mathrm{dt} = - \frac{1}{10} {\int}_{0}^{5} {e}^{- {t}^{2}} \left(- 2 t\right) \mathrm{dt}$

$= - \frac{1}{10} {\left[{e}^{- {t}^{2}}\right]}_{0}^{5}$

$= - \frac{1}{10} \left({e}^{-} 25 - {e}^{0}\right)$

$= \frac{1}{10} \left(1 - {e}^{-} 25\right)$