Pieszkowo3gc4

2023-03-17

How to determine if the improper integral converges or diverges $\int \frac{1}{\sqrt{x}}$ from 0 to infinity?

Haylee French

Beginner2023-03-18Added 7 answers

Although the integral in this instance is so straightforward to evaluate that we can just compute it and check if the value is bounded, we could use the comparison test for improper integrals.

${\int}_{0}^{\infty}\frac{1}{\sqrt{x}}dx={\int}_{0}^{\infty}{x}^{-\frac{1}{2}}={\left[2\sqrt{x}\right]}_{0}^{\infty}=\underset{x\to \infty}{lim}\left(2\sqrt{x}\right)-2\sqrt{0}=\underset{x\to \infty}{lim}\left(2\sqrt{x}\right)=\infty$

This means that the integral diverges.

${\int}_{0}^{\infty}\frac{1}{\sqrt{x}}dx={\int}_{0}^{\infty}{x}^{-\frac{1}{2}}={\left[2\sqrt{x}\right]}_{0}^{\infty}=\underset{x\to \infty}{lim}\left(2\sqrt{x}\right)-2\sqrt{0}=\underset{x\to \infty}{lim}\left(2\sqrt{x}\right)=\infty$

This means that the integral diverges.

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