## Calculus (3rd Edition)

$$\frac{3\pi}{2}$$
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 $$M=\frac{9\pi}{2}\frac{1}{3}=\frac{3\pi}{2}$$