Final Answer

The average value of \(f(x)=\sqrt{9-x^{2}}\) on the interval \([-3,3]\) is \(\dfrac{3\pi}{4}\approx 2.36\).

Average Value Formula

For a continuous function \(f(x)\) on \([a,b]\), the average value is

\[
f_{\text{avg}}=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx.
\]

In this question, \(f(x)=\sqrt{9-x^{2}}\), \(a=-3\), \(b=3\), so \(b-a=6\). Therefore,

\[
f_{\text{avg}}=\frac{1}{6}\int_{-3}^{3}\sqrt{9-x^{2}}\,dx.
\]

Evaluating the Integral

The curve \(y=\sqrt{9-x^{2}}\) is the upper semicircle of radius 3 centered at the origin, because squaring both sides gives \(y^{2}=9-x^{2}\), or \(x^{2}+y^{2}=9\), the equation of a circle of radius 3.

The area of a full circle with radius 3 is \(\pi r^{2}=9\pi\), so the area of the upper semicircle (which equals the integral from −3 to 3) is

\[
\int_{-3}^{3}\sqrt{9-x^{2}}\,dx=\frac{1}{2}\cdot 9\pi=\frac{9\pi}{2}.
\]

Substituting this into the average-value expression gives

\[
f_{\text{avg}}=\frac{1}{6}\cdot\frac{9\pi}{2}=\frac{9\pi}{12}=\frac{3\pi}{4}.
\]

Numerical Approximation

To express the result numerically, compute

\[
\frac{3\pi}{4}\approx\frac{3\times3.14159265}{4}\approx2.35619\approx2.36,
\]

so the required average value of \(f(x)=\sqrt{9-x^{2}}\) on \([-3,3]\) rounded to two decimal places is 2.36.