Average Function Value — Question 3

Question 3 - Solution

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

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

Evaluate the integral using integration by parts.

Let $$u=x^2, \qquad dv=\cos x\,dx.$$ Then $$du=2x\,dx, \qquad v=\sin x.$$

Apply integration by parts: $$\int x^2\cos x\,dx = x^2\sin x-\int 2x\sin x\,dx.$$

Apply integration by parts again to the remaining integral. Let $$u=2x, \qquad dv=\sin x\,dx.$$ Then $$du=2\,dx, \qquad v=-\cos x.$$

Thus, $$\int 2x\sin x\,dx = -2x\cos x+2\int \cos x\,dx = -2x\cos x+2\sin x.$$

Substitute back: $$\int x^2\cos x\,dx = x^2\sin x+2x\cos x-2\sin x.$$

Evaluate from $0$ to $\pi$: $$\left[x^2\sin x+2x\cos x-2\sin x\right]_{0}^{\pi}.$$

At $x=\pi$: $$\pi^2\sin\pi+2\pi\cos\pi-2\sin\pi=-2\pi.$$

At $x=0$: $$0+0-0=0.$$

Thus, $$\int_{0}^{\pi} x^2\cos x\,dx=-2\pi.$$

Divide by the interval length: $$f_{\text{avg}}=\frac{1}{\pi}(-2\pi)=-2.$$

$$\boxed{-2}$$