Question

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Answer 1

Answer:

Since the $\lim_{x\to -\infty} \frac{1}{(x-2)^2}=0$ then we have this:

$\int_{-\infty}^{-1} (x-2)^{-3} dx =-\frac{1}{18}$

And we see that our integral on this case converged to -1/18.

Step-by-step explanation:

For this case we need to determine if the following integral converges or not:

$\int_{-\infty}^{-1} (x-2)^{-3} dx$

We can rewrite the integral like this:

$\int_{-\infty}^{-1} \frac{1}{(x-2)^3} dx$

Then we can use the substitution $u = x-2$ and then $du = dx$ and we have this:

$\int_{-\infty}^{-3} \frac{1}{u^3} du=\int_{-\infty}^{-3} u^{-3} du$

If we solve the integral we got:

$=\frac{u^{-3+1}}{-3+1} =-\frac{u^{-2}}{2}=-\frac{1}{2}\frac{1}{u^2}$

And then the integral would be equal to:

$\int_{-\infty}^{-1} (x-2)^{-3} dx = -\frac{1}{2(x-2)^2}\Big|_{-\infty}^{-1}$

And if we replace and using the fundamental theorem of calculus we got:

$= -\frac{1}{2} [\frac{1}{(-1-2)^2} -\lim_{x\to -\infty} \frac{1}{(x-2)^2}]$

Since the $\lim_{x\to -\infty} \frac{1}{(x-2)^2}=0$ then we have this:

$\int_{-\infty}^{-1} (x-2)^{-3} dx =-\frac{1}{18}$

And we see that our integral on this case converged to -1/18.