# What is the average value of a function sec^2x on the interval [pi/6, pi/4]?

##### 1 Answer
Jun 6, 2016

$\frac{12 - 4 \sqrt{3}}{\pi}$

#### Explanation:

The average value of the function $f \left(x\right)$ on the interval $\left[a , b\right]$ is equivalent to

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

Thus, the average value of $f \left(x\right) = {\sec}^{2} x$ on $\left[\frac{\pi}{6} , \frac{\pi}{4}\right]$ is

$\frac{1}{\frac{\pi}{4} - \frac{\pi}{6}} {\int}_{\frac{\pi}{6}}^{\frac{\pi}{4}} {\sec}^{2} x \mathrm{dx}$

Note that $\int {\sec}^{2} x \mathrm{dx} = \tan x + C$, since the derivative of $\tan x$ is ${\sec}^{2} x$.

Also, note that $\frac{1}{\frac{\pi}{4} - \frac{\pi}{6}} = \frac{1}{\frac{\pi}{12}} = \frac{12}{\pi}$.

The average value then equals:

$\frac{1}{\frac{\pi}{4} - \frac{\pi}{6}} {\int}_{\frac{\pi}{6}}^{\frac{\pi}{4}} {\sec}^{2} x \mathrm{dx} = \frac{12}{\pi} {\left[\tan x\right]}_{\frac{\pi}{6}}^{\frac{\pi}{4}}$

$= \frac{12}{\pi} \left[\tan \left(\frac{\pi}{4}\right) - \tan \left(\frac{\pi}{6}\right)\right] = \frac{12}{\pi} \left[1 - \frac{\sqrt{3}}{3}\right] = \frac{12}{\pi} \left[\frac{3 - \sqrt{3}}{3}\right]$

$= \frac{4}{\pi} \left[3 - \sqrt{3}\right] = \frac{12 - 4 \sqrt{3}}{\pi}$