Question

Find the value of the integral that converges. ∫^-5_-[infinity] x^-2 dx.

Answer 1

Answer:

$\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}$

Because the $\lim_{x\to -\infty} \frac{1}{x} =0$

The integral converges to $\frac{1}{5}$

Step-by-step explanation:

For this case we want to find the following integral:

$\int_{-\infty}^{-5} x^{-2}dx$

And we can solve the integral on this way:

$\int_{-\infty}^{-5} x^{-2}dx= \frac{x^{-2+1}}{-2+1} \Big|_{-\infty}^{-5}$

$\int_{-\infty}^{-5} x^{-2}dx= -\frac{1}{x} \Big|_{-\infty}^{-5}$

And if we evaluate the integral using the fundamental theorem of calculus we got:

$\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}$

Because the $\lim_{x\to -\infty} \frac{1}{x} =0$

The integral converges to $\frac{1}{5}$