Average Function Value — Question 2

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

Notice that $\ln(1+x^2)$ is an even function, since $$\ln(1+(-x)^2)=\ln(1+x^2).$$ Therefore, $$\int_{-1}^{1}\ln(1+x^2)\,dx = 2\int_{0}^{1}\ln(1+x^2)\,dx.$$

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

Evaluate the integral using integration by parts. Let $$u=\ln(1+x^2), \qquad dv=dx.$$ Then $$du=\frac{2x}{1+x^2}\,dx, \qquad v=x.$$

Apply integration by parts: $$\int \ln(1+x^2)\,dx = x\ln(1+x^2)-\int \frac{2x^2}{1+x^2}\,dx.$$

Rewrite the integrand: $$\frac{2x^2}{1+x^2} = 2-\frac{2}{1+x^2}.$$

Thus, $$\int \ln(1+x^2)\,dx = x\ln(1+x^2)-\int\left(2-\frac{2}{1+x^2}\right)dx.$$

Integrate: $$x\ln(1+x^2)-2x+2\arctan x.$$

Evaluate from $0$ to $1$: $$\left[x\ln(1+x^2)-2x+2\arctan x\right]_{0}^{1} = \ln 2-2+\frac{\pi}{2}.$$

Therefore, $$\boxed{\ln 2-2+\frac{\pi}{2}}$$