Question

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

Answer 1

Answer:Divergent

Step-by-step explanation:

Given

Improper Integral I is given as

$I=\int^{\infty}_{4}\frac{2}{\sqrt{x}}dx$

Integration of $\frac{2}{\sqrt{x}}$  is  $4\sqrt{2}$

$I=\left [ 4\sqrt{x}\right ]^{\infty}_{4}$

$I=4\left [ \sqrt{x}\right ]^{\infty}_{4}$

$I=4\left [ \sqrt{\infty}-\sqrt{4}\right ]$

I tends to $\infty$ so Integral is divergent in nature