Single Variable Calculus, Chapter 6, 6.5, Section 6.5, Problem 12

Given that $\displaystyle f(x) = \frac{2x}{(1 + x^2)^2}$ with interval $[0,2]$.

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}{2 - 0} \int^2_0 \frac{2x}{(1 + x^2)^2} dx
\\
\\
\text{Let } u =& 1 + x^2
\\
\\
du =& 2x dx

\end{aligned}
\end{equation}
$


Make sure that your upper and lower limits are also in terms of $u$.


$
\begin{equation}
\begin{aligned}

f_{ave} =& \frac{1}{2} \left( \frac{1}{2} \right) \int^{1 + (2)^2}_{1 + (0)^2} \frac{2}{u^2} du
\\
\\
f_{ave} =& \frac{1}{2} \int^5_1 u^{-2} du
\\
\\
f_{ave} =& \frac{1}{2} \left[ \frac{u^{-1}}{-1} \right]^5_1
\\
\\
f_{ave} =& \frac{1}{2} \left[ - \frac{1}{5} \left( \frac{-1}{1} \right) \right]
\\
\\
f_{ave} =& \frac{2}{5}

\end{aligned}
\end{equation}
$


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


$
\begin{equation}
\begin{aligned}

f_{ave} =& f(c)
\\
\\
\frac{2}{5} =& \frac{2c}{(1 + c^2)^2}
\\
\\
(1 + c^2)^2 =& 5c
\\
\\
1 + 2c^2 + c^4 =& 5c
\\
\\
c^4 + 2c^2 - 5c + 1 =&0

\end{aligned}
\end{equation}
$


We got 4 values of $c$. However, we omit the complex roots. So, we have it supposed to have 4 values of $c$. So that,


$
\begin{equation}
\begin{aligned}

& c \approx 0.21979
\\
& c \approx 1.20684

\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 $f$.