Answer
$\frac{(ln(5))^2}{8}$
Work Step by Step
$f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$
Substitute $x=ln(u)$
$dx = \frac{1}{u}du$
Change the lower and upper bounds by plugging them into $u$.
$f_{avg}=\frac{1}{4}\int_{ln(0)}^{ln(5)}xdx$
$=\frac{1}{4}\left[\frac{1}{2}x^2\right]_{0}^{ln5}$
$=\frac{(ln(5))^2}{8}$
