Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 82

Determine the area of the region bounded by the curve $\displaystyle y = \frac{2}{x-2}$, $x$-axis, $x = -4$ and $x = 1$.

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
\end{aligned}
\end{equation}
$

Let $u = x- 2$
$du = dx$

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.