# What is the average value of a function  f(t)= -2te^(-t^2) on the interval [0, 8]?

Oct 21, 2017

The average value is $\frac{1}{8 {e}^{64}} - \frac{1}{8} \approx - 0.125$

#### Explanation:

The average value of a function on the continuous interval $\left[a , b\right]$ is given by

$A = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

In this case we would have

$A = \frac{1}{8} {\int}_{0}^{8} - 2 t {e}^{- {t}^{2}} \mathrm{dt}$

This expression can be integrated using the substitution $u = - {t}^{2}$. Then $\mathrm{du} = - 2 t \mathrm{dt}$ and $\mathrm{dt} = \frac{\mathrm{du}}{- 2 t}$. We change the bounds of integration accordingly.

$A = \frac{1}{8} {\int}_{0}^{-} 64 {e}^{u} \mathrm{du}$

$A = \frac{1}{8} {\left[{e}^{u}\right]}_{0}^{-} 64$

A = 1/8[e^(-t^2)])_0^8

$A = \frac{1}{8} {e}^{- 64} - \frac{1}{8}$

If you want an approximation, use a calculator to get

$A = - 0.125$

Hopefully this helps!