Answer
$$
f(x)=\frac{3}{(x-1)^{3}} \text { for }(-\infty, 0]
$$
the area between the graph of the given function and the
x-axis over the given interval is given by:
$$
\begin{split}
\int_{-\infty}^{0} \frac{3}{(x-1)^{3}} d x &=-\frac{3}{2}.
\end{split}
$$
But, the area is positive, so the area between the graph of the given function and the x-axis over the given interval equal $\frac{3}{2}.$
Work Step by Step
$$
f(x)=\frac{3}{(x-1)^{3}} \text { for }(-\infty, 0]
$$
the area between the graph of the given function and the
x-axis over the given interval is given by:
$$
\begin{split}
\int_{-\infty}^{0} \frac{3}{(x-1)^{3}} d x &=\lim _{a \rightarrow-\infty} \int_{a}^{0} \frac{3}{(x-1)^{3}} d x \\
&=\left.\lim _{a \rightarrow-\infty}\left(\frac{3(x-1)^{-2}}{-2}\right)\right|_{a} ^{0} \\
&=\left.\lim _{a \rightarrow-\infty}\left(-\frac{3}{2(x-1)^{2}}\right)\right|_{a} ^{0}\\
&=\lim _{a \rightarrow-\infty}\left(-\frac{3}{2}+\frac{3}{2(a-1)^{2}}\right)\\
&=-\frac{3}{2}+\lim _{a \rightarrow-\infty}\left(\frac{3}{2(a-1)^{2}}\right)\\
&=-\frac{3}{2}+0\\
&=-\frac{3}{2}.
\end{split}
$$
Thus the given improper integral
$$
\int_{-\infty}^{0} \frac{3}{(x-1)^{3}} d x=-\frac{3}{2}.
$$
But, the area is positive, so the area between the graph of the given function and the x-axis over the given interval equal $\frac{3}{2}.$
