Find the average value $\bar{y}$ of the function $f( x) =\cos ^{4}x$ over the closed interval $[0,\pi]$.

Solution The average value $\bar{y}$ of a function $f$ over $[a,b]$ is $\bar{y}=\dfrac{1}{b-a}\int_{a}^{b}f(x)\,dx.$

For $f( x) =\cos ^{4}x$ on $[0,\pi]$, we have

Now

To find $\int_{0}^{\pi }\cos ^{2}(2x)\, dx,$ we use the identity $\cos ^{2}\theta =\dfrac{1+\cos (2\theta )}{2}$ again to write $\cos ^{2}(2x) =\dfrac{1+\cos (4x) }{2}$. Then $$\begin{eqnarray*} \int_{0}^{\pi }\cos ^{2}(2x) \,dx &=&\int_{0}^{\pi }\dfrac{ 1+\cos (4x) }{2}\,dx=\dfrac{1}{2}\left[ \int_{0}^{\pi }dx+\int_{0}^{\pi }\cos ( 4x) \,dx\right]\\[7pt] &=& \dfrac{1}{2}\left[ \pi +\int_{0}^{4\pi }\cos u\,\dfrac{du}{4}\right] \qquad {\color{#0066A7}{\hbox{\(u=4x\); \(du=4\,dx\)}}} \end{eqnarray*}$$

So, from (1), $$\bar{y}=\dfrac{1}{4\pi }\left[ \pi +0+\dfrac{\pi }{2}\right] =\dfrac{3}{8}$$