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 $u$.


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