Chapter 5.10, Problem 27E

To determine

To determine whether the integral ${\displaystyle \underset{-2}{\overset{14}{\int }}\frac{1}{\sqrt[4]{x+2}}dx}$ is convergent or divergent.

Answer to Problem 27E

This integral is convergent, and the value of the integral is $\frac{\text{32}}{\text{3}}$

Explanation of Solution

Given Information:

The integral is ${\displaystyle \underset{-2}{\overset{14}{\int }}\frac{1}{\sqrt[4]{x+2}}dx}$

Concept used: If the value of the integral is finite, it is convergent otherwise it is divergent.

Formula used:

  ${\displaystyle \int x^{n}dx}=\frac{x^{n+1}}{n+1}$

The given integral becomes as:

  $\begin{array}{l}={\left[\frac{{(x+2)}^{(-1/4)+1}}{(-1/4)+1}\right]}_{-2}^{14}=\frac{\text{4}}{\text{3}}{\left[{(x+2)}^{3/4}\right]}_{-2}^{14}\\ \text{=}\frac{\text{4}}{\text{3}}\left[{(16)}^{3/4}-0\right]=\frac{\text{4}}{\text{3}}\left[2^3\right]=\frac{\text{32}}{\text{3}},\text{finite}\text{number}\end{array}$

Since the given integral has finite value.

Hence, the given integral is convergent