# How do you find the average value of f(x)=1/(x^2+1) as x varies between [-1,1]?

Nov 7, 2016

Evaluate $\frac{1}{1 - \left(- 1\right)} {\int}_{-} {1}^{1} \frac{1}{{x}^{2} + 1} \mathrm{dx} = \frac{\pi}{4}$

#### Explanation:

The average value of $f$ on the interval $\left[a , b\right]$ is

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

We also need:

$\int \frac{1}{{x}^{2} + 1} \mathrm{dx} = {\tan}^{-} 1 x + C$

or, you may prefer the notation

$\int \frac{1}{{x}^{2} + 1} \mathrm{dx} = \arctan x + C$.

The average value is

$\frac{1}{1 - \left(- 1\right)} {\int}_{-} {1}^{1} \frac{1}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} {\left[{\tan}^{-} 1 x\right]}_{-} {1}^{1}$

$= \frac{1}{2} \left[\frac{\pi}{4} - \left(- \frac{\pi}{4}\right)\right]$

$= \frac{\pi}{4}$