Answer:
[tex] \int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}[/tex]
Because the [tex] \lim_{x\to -\infty} \frac{1}{x} =0[/tex]
The integral converges to [tex] \frac{1}{5}[/tex]
Step-by-step explanation:
For this case we want to find the following integral:
[tex] \int_{-\infty}^{-5} x^{-2}dx[/tex]
And we can solve the integral on this way:
[tex] \int_{-\infty}^{-5} x^{-2}dx= \frac{x^{-2+1}}{-2+1} \Big|_{-\infty}^{-5}[/tex]
[tex] \int_{-\infty}^{-5} x^{-2}dx= -\frac{1}{x} \Big|_{-\infty}^{-5} [/tex]
And if we evaluate the integral using the fundamental theorem of calculus we got:
[tex] \int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}[/tex]
Because the [tex] \lim_{x\to -\infty} \frac{1}{x} =0[/tex]
The integral converges to [tex] \frac{1}{5}[/tex]
