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 $f$.