Question

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

Answer 1

Answer:

$\int^{-1}_{-\infty} \frac{2}{x^{3}}dx=-1$

Step-by-step explanation:

We have:

$\int^{-1}_{-\infty} \frac{2}{x^{3}}dx$  

Now if integrate this using the power rule, we will have:

$\int^{-1}_{-\infty} \frac{2}{x^{3}}dx=-\frac{1}{x^{2}}|^{-1}_{-\infty}$  

Evaluating this integral we have:

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

Let's recall that 1/∞ is 0.

Therefore this integral converge at -1.

I hope it helps you!