Average Function Value — Question 1

Question 1

Find the average value of the function $$f(x)=x\sqrt{1+x^2}$$ on the interval $[0,2]$.

Question 1 - Solution

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

Here, $a=0$ and $b=2$, so $$f_{\text{avg}}=\frac{1}{2}\int_{0}^{2} x\sqrt{1+x^2}\,dx.$$

Evaluate the integral using substitution. Let $$u=1+x^2.$$ Then $$du=2x\,dx \quad\Rightarrow\quad x\,dx=\frac12\,du.$$

Change the limits. When $x=0$, $u=1$. When $x=2$, $u=5$.

Substitute: $$\int_{0}^{2} x\sqrt{1+x^2}\,dx = \frac12\int_{1}^{5} u^{1/2}\,du.$$

Integrate: $$\frac12\int u^{1/2}\,du = \frac12\cdot\frac{2}{3}u^{3/2} = \frac{1}{3}u^{3/2}.$$

Apply the limits: $$\frac{1}{3}u^{3/2}\Big|_{1}^{5} = \frac{1}{3}\left(5^{3/2}-1\right).$$

Now divide by the interval length: $$f_{\text{avg}} = \frac{1}{2}\cdot\frac{1}{3}\left(5^{3/2}-1\right) = \frac{1}{6}\left(5^{3/2}-1\right).$$

$$\boxed{\frac{1}{6}\left(5^{3/2}-1\right)}$$