Answer
$$\frac{3\pi}{2} $$
Work Step by Step
Given $$f(x)=\sqrt{9-x^{2}}, \quad[0,3]$$
The average value is given by
\begin{align*}
M&=\frac{1}{b-a}\int_{a}^{b}f(x)dx\\
&=\frac{1}{3}\int_{0}^{3}\sqrt{9-x^2}dx
\end{align*}
Since
$$ \int_{0}^{3}\sqrt{9-x^2}dx$$ represents the area of one-quarter of a circle of radius $3$, then$$ \int_{0}^{3}\sqrt{9-x^2}dx=\frac{9\pi}{4} $$
Hence
$$M=\frac{9\pi}{2}\frac{1}{3}=\frac{3\pi}{2} $$
