# What is the average value of a function f(t)=−4sec(t)tan(t) on the interval [0, pi/4]?

Oct 8, 2016

$\frac{16 - 16 \sqrt{2}}{\pi}$

#### Explanation:

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

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

Thus, where $f \left(t\right) = - 4 \sec \left(t\right) \tan \left(t\right)$ and the interval is $\left[0 , \frac{\pi}{4}\right]$, we see its average value on that interval is:

$\frac{1}{\frac{\pi}{4} - 0} {\int}_{0}^{\frac{\pi}{4}} - 4 \sec \left(t\right) \tan \left(t\right) \mathrm{dt}$

$= \frac{- 4}{\frac{\pi}{4}} {\int}_{0}^{\frac{\pi}{4}} \sec \left(t\right) \tan \left(t\right) \mathrm{dt}$

Note that since $\frac{d}{\mathrm{dt}} \sec \left(t\right) = \sec \left(t\right) \tan \left(t\right)$, so the integral of $\sec \left(t\right) \tan \left(t\right)$ is $\sec \left(t\right) + C$.

$= - \frac{16}{\pi} {\left[\sec \left(t\right)\right]}_{0}^{\frac{\pi}{4}}$

$= - \frac{16}{\pi} \left[\sec \left(\frac{\pi}{4}\right) - \sec \left(0\right)\right]$

Note that $\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, so $\sec \left(\frac{\pi}{4}\right) = \frac{2}{\sqrt{2}} = \sqrt{2}$.

$= - \frac{16}{\pi} \left(\sqrt{2} - 1\right)$

$= \frac{16 \left(1 - \sqrt{2}\right)}{\pi}$