Single Variable Calculus, Chapter 6, 6.5, Section 6.5, Problem 8

Determine the average value of the function $\displaystyle f(x) = \frac{3}{(1 + x)^2}$ on the interval $[1,6]$.


$
\begin{equation}
\begin{aligned}

f_{ave} =& \frac{1}{b - a} \int^b_a f(x) dx
\\
\\
f_{ave} =& \frac{1}{6 - 1} \int^6_1 \frac{3}{(1 + x)^2} dx
\\
\\
\text{Let } u =& 1 + x
\\
\\
du =& dx

\end{aligned}
\end{equation}
$


Also, make sure that your upper and lower limits are also in terms of $u$.


$
\begin{equation}
\begin{aligned}

f_{ave} =& \frac{3}{5} \int^{1 + (6)}_{1 + (1)} \frac{1}{u^2} du
\\
\\
f_{ave} =& \frac{3}{5} \int^7_2 u^{-2} du
\\
\\
f_{ave} =& \frac{3}{5} \left[ \frac{u^{-1}}{-1} \right]^7_2
\\
\\
f_{ave} =& \frac{3}{5} \left[ \frac{-1}{7} - \left( \frac{-1}{2} \right) \right]
\\
\\
f_{ave} =& \frac{3}{14}

\end{aligned}
\end{equation}
$