Average Function Value — Question 7

Question 7

Find the average value of the function $$f(x)=\sqrt{1-x^2}$$ on the interval $[-1,1]$.

Question 7 - 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, $$f_{\text{avg}}=\frac{1}{2}\int_{-1}^{1}\sqrt{1-x^2}\,dx.$$

Method 1: Geometry

The graph of $$y=\sqrt{1-x^2}$$ is the upper half of a circle of radius $1$ centered at the origin.

The area under this curve from $x=-1$ to $x=1$ is the area of a semicircle: $$\frac{1}{2}\pi(1)^2=\frac{\pi}{2}.$$

Therefore, $$f_{\text{avg}}=\frac{1}{2}\cdot\frac{\pi}{2}=\frac{\pi}{4}.$$

Method 2: Trigonometric Substitution

Evaluate the integral directly.

Let $$x=\sin\theta, \qquad dx=\cos\theta\,d\theta.$$

Change the limits. When $x=-1$, $\theta=-\frac{\pi}{2}$. When $x=1$, $\theta=\frac{\pi}{2}$.

Substitute: $$\int_{-1}^{1}\sqrt{1-x^2}\,dx = \int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta.$$

Use the identity $$\cos^2\theta=\frac{1+\cos(2\theta)}{2}.$$

Then $$\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta = \frac12\int_{-\pi/2}^{\pi/2}(1+\cos 2\theta)\,d\theta.$$

Integrate: $$\frac12\left[\theta+\frac12\sin(2\theta)\right]_{-\pi/2}^{\pi/2}.$$

Since $\sin(\pi)=\sin(-\pi)=0$, this becomes $$\frac12\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right) = \frac{\pi}{2}.$$

Now compute the average value: $$f_{\text{avg}}=\frac{1}{2}\cdot\frac{\pi}{2}=\frac{\pi}{4}.$$

$$\boxed{\frac{\pi}{4}}$$