Answer
$f_{avg}=\frac{4}{3}$
Work Step by Step
Given: Average value of the function $f(x)=\sqrt x$ over the interval $[0,4]$ is given by
$f_{avg}=\frac{1}{4-0}\int_{0}^{4}\sqrt xdx$
$=\frac{1}{4}[\frac{x^{3/2}}{3/2}]_{0}^{4}$
$=\frac{1}{4}[\frac{2}{3}x\sqrt x]_{0}^{4}$
$=\frac{1}{4}[\frac{2}{3}4\sqrt 4]$
Hence, $f_{avg}=\frac{4}{3}$
