Average Function Value — Question 6

Question 6

Find the average value of the function $$f(x)=\sin^2 x$$ on the interval $[0,\pi]$.

Question 6 - Solution

The average value of a function on $[a,b]$ is $$f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.$$

Here, $$f_{\text{avg}}=\frac{1}{\pi}\int_{0}^{\pi}\sin^2 x\,dx.$$

Use the identity $$\sin^2 x=\frac{1-\cos(2x)}{2}.$$

Then $$\int_{0}^{\pi}\sin^2 x\,dx = \frac12\int_{0}^{\pi}(1-\cos 2x)\,dx.$$

Integrate: $$\frac12\left[x-\frac12\sin(2x)\right]_{0}^{\pi} = \frac12(\pi-0) = \frac{\pi}{2}.$$

Now divide by the interval length: $$f_{\text{avg}}=\frac{1}{\pi}\cdot\frac{\pi}{2}=\frac12.$$

$$\boxed{\frac12}$$