# What is the average value of a function  y=sec^2 x on the interval [0,pi/4]?

Oct 17, 2016

$\frac{4}{\pi}$

#### Explanation:

The average value of a function $y = f \left(x\right)$ over an interval $\left[a , b\right]$ is given by $\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

So for $y = {\sec}^{2} x$ we have:

$A v g = \frac{1}{\frac{\pi}{4} - 0} {\int}_{0}^{\frac{\pi}{4}} {\sec}^{2} x \mathrm{dx}$
$= \frac{1}{\frac{\pi}{4}} {\int}_{0}^{\frac{\pi}{4}} {\sec}^{2} x \mathrm{dx}$
$= \frac{4}{\pi} {\left[\tan x\right]}_{0}^{\frac{\pi}{4}}$
$= \frac{4}{\pi} \left\{\tan \left(\frac{\pi}{4}\right) - \tan 0\right\}$
$= \frac{4}{\pi} \left\{\tan \left(\frac{\pi}{4}\right) - \tan 0\right\}$
$= \frac{4}{\pi} \left\{1 - 0\right\}$
$= \frac{4}{\pi}$