Question

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

Answer 1

Answer:Convergent

Step-by-step explanation:

Given

Improper Integral I is given as

$I=\int^{-4}_{-\infty}\frac{3}{x^4}dx$

integration of $\frac{3}{x^4}$  is  -$\frac{1}{x^3}$

$I=-\left [ \frac{1}{x^3}\right ]^{-4}_{-\infty}$

$I=-\left [ \frac{1}{(-4)^3}-\frac{1}{(-\infty)^3}\right ]$

$I=-\left [ -\frac{1}{4^3}\right ]$

$I=\frac{1}{4^3}$

Integral is convergent at   $I=\frac{1}{64}$