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

Jun 6, 2016

$\frac{4}{\pi}$

#### Explanation:

The average value of a function from point $x = a$ to point $x = b$ is given by:
$\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

For ${\sec}^{2} \left(x\right)$ on $\left[0 , \frac{\pi}{4}\right]$, this looks like:
$\frac{1}{\frac{\pi}{4} - 0} {\int}_{0}^{\frac{\pi}{4}} {\sec}^{2} x \mathrm{dx}$

The antiderivative of ${\sec}^{2} x$ is $\tan x$ because $\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$; that reduces our problem to:
$\frac{1}{\frac{\pi}{4}} {\left[\tan x\right]}_{0}^{\frac{\pi}{4}}$
$= \frac{4}{\pi} \left(\tan \left(\frac{\pi}{4}\right) - \tan \left(0\right)\right)$
$= \frac{4}{\pi} \left(1 - 0\right)$
$= \frac{4}{\pi} = 1.273 \ldots$