# How do you find the average value of the function for f(x)=e^x, -1<=x<=1?

Apr 28, 2018

$\overline{f} \left(x\right) = \frac{1}{2} \setminus \left(e - \frac{1}{e}\right) \approx 1.1752$

#### Explanation:

By definition, the average value of a function $f \left(x\right)$ over a domain $\left[a , b\right]$ is given by:

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

So, for the given function, $f \left(x\right) = {e}^{x}$ with $- 1 \le x \le 1$, we have:

$\overline{f} \left(x\right) = \frac{1}{1 - \left(- 1\right)} \setminus {\int}_{- 1}^{1} \setminus {e}^{x} \setminus \mathrm{dx}$

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

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} \setminus \left(e - {e}^{- 1}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} \setminus \left(e - \frac{1}{e}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \approx 1.1752$