Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 74

Find the area of the region above the hyperbola $\displaystyle y = \frac{2}{x-2}$.

By using vertical strips,


$\begin{equation} \begin{aligned} A &= \int^{-1}_{-4} \left( y_{\text{upper}} - y_{\text{lower}} \right) \\ \\ A &= \int^{-1}_{-4} \left(0 - \left( \frac{2}{x-2} \right) \right) dx \\ \\ A &= \int^{-1}_{-4} \frac{-2}{x-2} dx \\ \\ \text{Let } u =& x - 2 \\ \\ du =& dx \end{aligned} \end{equation}$


Make sure that the upper and lower units are in terms of $u$.


$\begin{equation} \begin{aligned} A &= -2 \int^{-1-2}_{-4-2} \left( \frac{1}{u} \right) du\\ \\ A &= - 2 \int^{-3}_{-6} \frac{du}{u}\\ \\ A &= -2 [\ln u]^{-3}_{-6}\\ \\ A &= -2 [\ln(-3)-\ln(-6)] \end{aligned} \end{equation}$



We can't evaluate the area since $\ln$ of negative number doesn't exist. However, since the function is reflected about $x = 2$ its area is equal to the region bounded by the curve, $x$-axis and the lines $x = 5$ and $x = 8$. $A = 1.3863$ square units.