Average Function Value — Question 5

Question 5

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

Question 5 - 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 $a=0$ and $b=1$, so $$f_{\text{avg}}=\int_{0}^{1}\frac{x}{1+x^2}\,dx.$$

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 $x=1$, $u=2$.

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

Integrate: $$\frac12\int \frac{1}{u}\,du = \frac12\ln u.$$

Apply the limits: $$\frac12\ln u\Big|_{1}^{2} = \frac12\ln 2.$$

$$\boxed{\frac{1}{2}\ln 2}$$