Single Variable Calculus, Chapter 6, 6.5, Section 6.5, Problem 6

Determine the average value of the function $\displaystyle f(x) =\sec^2 \left( \frac{x}{2} \right)$ on the interval $\displaystyle \left[0, \frac{\pi}{2}\right]$.


$\begin{equation} \begin{aligned} f_{ave} =& \frac{1}{b - a} \int^b_a f(x) dx \\ \\ f_{ave} =& \frac{1}{\displaystyle \frac{\pi}{2} - 0} \int^{\frac{\pi}{2}}_0 \sec^2 \left( \frac{x}{2} \right) dx \\ \\ \text{Let } u =& \frac{x}{2} \\ \\ du =& \frac{dx}{2} \end{aligned} \end{equation}$


Also, make sure that your upper and lower limits are also in terms of $u$.


$\begin{equation} \begin{aligned} f_{ave} =& \frac{2}{\displaystyle \frac{\pi}{2}} \int^{\frac{\frac{\pi}{2}}{2}}_{\frac{0}{2}} \sec^2 u du \\ \\ f_{ave} =& \frac{4}{\pi} [\tan u]^{\frac{\pi}{4}}_0 \\ \\ f_{ave} =& \frac{4}{\pi} \left[ \tan \left( \frac{\pi}{4} \right) - \tan 0 \right] \\ \\ f_{ave} =& \frac{4}{\pi} \end{aligned} \end{equation}$