## Calculus (3rd Edition)

$\frac{1}{60}.$
The average is given as follows $$\frac{1}{3-0} \int_{0}^{3}\frac{x}{(x^2+16)^{3/2}} d x=\frac{1}{6}\int_{0}^{3}2x(x^2+16)^{-3/2} d x\\=-\left.\frac{1}{3}(x^2+16)^{-1/2}\right|_{0 } ^{3}=-\frac{1}{3}\left( \frac{1}{5}-\frac{1}{4}\right)=\frac{1}{60}.$$