Chapter 10.5, Problem 1QR

To determine

To explain:

Given improper integral converges or diverges.

Answer to Problem 1QR

Given improper integral converges.

Explanation of Solution

Given information:

  ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{4/3}}dx}$

Given integral is ,

  ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{4/3}}dx}$

By comparing $f\left(x\right)$ and $g\left(x\right)$ within interval $\left[1,\infty \right)$

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{4/3}}dx}\\ ={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^p}dx}\to {\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{4/3}}dx}\end{array}$

It converges.

Integral converges when $p>1$ ,

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{\infty }{\int }}g\left(x\right)dx}\\ ={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}\left(x+1\right)}dx}\end{array}$

It converges.

Since, $f\left(x\right)\ge g\left(x\right)$ and ${\displaystyle \underset{1}{\overset{\infty }{\int }}f\left(x\right)dx}$ converges, use comparison test.