Average Function Value — Question 9

Question 9

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

Question 9 - Solution

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

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

Notice that $\sqrt{x^2+1}$ is an even function, so $$\int_{-1}^{1}\sqrt{x^2+1}\,dx = 2\int_{0}^{1}\sqrt{x^2+1}\,dx.$$

Thus, $$f_{\text{avg}}=\int_{0}^{1}\sqrt{x^2+1}\,dx.$$

Evaluate the integral using trigonometric substitution. Let $$x=\tan\theta, \qquad dx=\sec^2\theta\,d\theta.$$

When $x=0$, $\theta=0$. When $x=1$, $\theta=\frac{\pi}{4}$.

Substitute: $$\int_{0}^{1}\sqrt{x^2+1}\,dx = \int_{0}^{\pi/4}\sec\theta\cdot\sec^2\theta\,d\theta = \int_{0}^{\pi/4}\sec^3\theta\,d\theta.$$

Use the standard result $$\int \sec^3\theta\,d\theta = \frac12\left(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\right).$$

Apply the limits: $$\frac12\left(\sec\theta\tan\theta+\ln(\sec\theta+\tan\theta)\right) \Big|_{0}^{\pi/4}.$$

Evaluate: $$\sec\!\left(\frac{\pi}{4}\right)=\sqrt{2}, \quad \tan\!\left(\frac{\pi}{4}\right)=1,$$ $$\sec(0)=1, \quad \tan(0)=0.$$

Thus, $$f_{\text{avg}} = \frac12\left(\sqrt{2}+\ln(1+\sqrt{2})\right).$$

$$\boxed{\frac12\left(\sqrt{2}+\ln(1+\sqrt{2})\right)}$$