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 $f$.