Answer
$c=\frac{1444}{225}$
Work Step by Step
The average value of $f(x)=\sqrt x$ on $[4,9]$ is:
$$\frac{1}{9-4}\int_{4}^{9} \sqrt x~dx$$
so:
$$f(c)=\frac{1}{9-4}\int_{4}^{9} \sqrt x~dx$$
$$f(c)=\frac{1}{5}\int_{4}^{9} \sqrt x~dx$$
$$\sqrt{c}=\frac{1}{5}\int_{4}^{9} \sqrt x~dx$$
$$\sqrt{c}=\frac{1}{5}\int_{4}^{9} x^{\frac{1}{2}}~dx$$
Using the $FTC I$ it follows:
$$\sqrt{c}=\frac{1}{5}[\frac{2}{3}x^{\frac{3}{2}}]_{4}^{9}$$
$$\sqrt{c}=\frac{1}{5}[\frac{2}{3}\cdot 9^{\frac{3}{2}}-(\frac{2}{3}\cdot 4^{\frac{3}{2}})]$$
$$\sqrt{c}=\frac{38}{15}$$
$$c=(\frac{38}{15})^{2}=\frac{1444}{225}$$
