Answer
$5$
Work Step by Step
\[\begin{align}
& f\left( x \right)=\frac{5}{{{\left( x-2 \right)}^{2}}},\text{ }\left( -\infty ,1 \right] \\
& \text{The area is given by} \\
& A=\int_{-\infty }^{1}{\frac{5}{{{\left( x-2 \right)}^{2}}}}dx \\
& \text{Using the definition of improper integrals} \\
& A=\underset{a\to -\infty }{\mathop{\lim }}\,\int_{a}^{1}{\frac{5}{{{\left( x-2 \right)}^{2}}}}dx \\
& A=\underset{a\to -\infty }{\mathop{\lim }}\,\left[ -\frac{5}{x-2} \right]_{a}^{1} \\
& A=\underset{a\to -\infty }{\mathop{\lim }}\,\left[ -\frac{5}{1-2}+\frac{5}{a-2} \right] \\
& A=\underset{a\to -\infty }{\mathop{\lim }}\,\left[ 5+\frac{5}{a-2} \right] \\
& \text{Evaluate when }a\to -\infty \\
& A=\underset{a\to -\infty }{\mathop{\lim }}\,\left[ 5+\frac{5}{a-2} \right] \\
& A=5+\frac{5}{-\infty -2} \\
& A=5 \\
\end{align}\]
