Single Variable Calculus, Chapter 6, 6.5, Section 6.5, Problem 10

Given that $f(x) = \sqrt{x}$ with interval $[0,4]$.

a.) Find the average value.


$\begin{equation} \begin{aligned} f_{ave} =& \frac{1}{b - a} \int^b_a f(x) dx \\ \\ f_{ave} =& \frac{1}{4 - 0} \int^4_0 \sqrt{x} dx \\ \\ f_{ave} =& \frac{1}{4} \left[ \frac{x^{\frac{3}{2}}}{\displaystyle \frac{3}{2}} \right]^4_0 \\ \\ f_{ave} =& \frac{4}{3} \end{aligned} \end{equation}$


b.) Find $C$ such that $f_{ave} = f(c)$.


$\begin{equation} \begin{aligned} f_{ave} =& f(c) \\ \\ \frac{4}{3} =& \sqrt{c} \\ \\ c =& \frac{4^2}{3^2} \\ \\ c =& \frac{16}{9} \end{aligned} \end{equation}$


c.) Sketch the graph of $f$ and a rectangle whose area is the same as the area under the graph of $f$.