Question

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

Answer 1

Answer:converge at $I=\frac{1}{3}$

Step-by-step explanation:

Given

Improper Integral I is given as

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

integration of $\frac{1}{x^2}$  is  -$\frac{1}{x}$

$I=\left [ -\frac{1}{x}\right ]^{\infty}_3$

substituting value

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

$I=-\left [ 0-\frac{1}{3}\right ]$

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

so the value of integral converges at $\frac{1}{3}$