Answer
$1$
Work Step by Step
\[\begin{align}
& f\left( x \right)=\frac{1}{{{\left( x-1 \right)}^{2}}},\text{ }\left( -\infty ,0 \right] \\
& \text{The area under the curve is given by } \\
& A=\int_{-\infty }^{0}{\frac{1}{{{\left( x-1 \right)}^{2}}}}dx \\
& \text{By the definition of improper integrals} \\
& A=\underset{a\to -\infty }{\mathop{\lim }}\,\int_{a}^{0}{\frac{1}{{{\left( x-1 \right)}^{2}}}}dx \\
& A=\underset{a\to -\infty }{\mathop{\lim }}\,\left[ -\frac{1}{x-1} \right]_{a}^{0} \\
& A=\underset{a\to -\infty }{\mathop{\lim }}\,\left[ -\frac{1}{0-1}+\frac{1}{a-1} \right] \\
& A=\underset{a\to -\infty }{\mathop{\lim }}\,\left[ 1+\frac{1}{a-1} \right] \\
& \text{Evaluate when }a\to -\infty . \\
& A=1+\frac{1}{-\infty -1} \\
& A=1+0 \\
& A=1 \\
\end{align}\]
