Answer
$\dfrac{\ln (2)}{2}$
or, $\approx 0.3465$
Work Step by Step
We are given that $f(x)=\dfrac{x}{x^2+1}$, $x=0$ to $x=1$
Apply formula: $\overline{f}=\dfrac{1}{b-a}\int_a^b f(x) \ dx$
So, we have: $\overline{f}=\dfrac{1}{1-0}\int_{0}^1 \dfrac{x}{x^2+1} \ dx=\dfrac{1}{2}\int_0^1 \dfrac{2x}{x^2+1} \ dx$
or, $=\dfrac{1}{2} [\ln (x^2+1)]_0^1$
or, $=\dfrac{1}{2} [\ln (1^2+1)-\ln [(0)^2+1]]$
or, $=\dfrac{\ln (2)}{2}$
or, $\approx 0.3465$
