Question

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

Answer 1

Answer:

$\int_{-\infty}^{-1} x^{-\frac{8}{3}} dx = -\frac{3}{5} [-1 - \infty]$

So then we see that the integral not converges since the limit $\lim_{x\to -\infty} \frac{1}{x^{5/3}}$ is not defined on this case.

So then this integral diverges.

Step-by-step explanation:

For this case we need to solve the following integral:

$\int_{-\infty}^{-1} x^{-\frac{8}{3}} dx$

So lets solve the integral in order to see if diverges or converges:

$= \frac{x^{-\frac{8}{3} +1}}{-\frac{8}{3}+1}$

$= \frac{x^{-\frac{5}{3}}}{-\frac{5}{3}} = -\frac{3}{5 x^{\frac{5}{3}}} \Big|_{-\infty}^{-1}$

And on this case we can evaluate like this:

$\int_{-\infty}^{-1} x^{-\frac{8}{3}} dx = -\frac{3}{5} [\frac{1}{(-1)^{5/3}} - \lim_{x\to -\infty} \frac{1}{x^{5/3}}]$

$\int_{-\infty}^{-1} x^{-\frac{8}{3}} dx = -\frac{3}{5} [-1 - \infty]$

So then we see that the integral not converges since the limit $\lim_{x\to -\infty} \frac{1}{x^{5/3}}$ is not defined on this case.

This integral diverges.