Answer:
[tex]A= -\dfrac{3}{2}[/tex]
area between curve and the x-axis, within the intervals [tex]x=[-\infty,0)[/tex]
Step-by-step explanation:
Given function is:
[tex]f(x) = \dfrac{x}{(x-1)^3}[/tex]
to find its area between the intervals [tex]x=[-\infty,0)[/tex], we'll need to integrate it.
[tex]A = \displaystyle{\int^0_{-\infty} {\dfrac{x}{(x-1)^3} \, dx}[/tex]
[tex]A = \left|- \dfrac{3}{2 \left(x - 1\right)^{2}}\right|^0_{-\infty}[/tex]
[tex]A = \left(- \dfrac{3}{2 \left(0 - 1\right)^{2}}\right)-\left(- \dfrac{3}{2 \left(-\infty - 1\right)^{2}}\right)[/tex]
[tex]A= \left(- \dfrac{3}{2}\right)-0[/tex]
[tex]A= -\dfrac{3}{2}\,\text{unit}^2[/tex]
