Chapter 9.4, Problem 2E

a.

To determine

To find: the expression that represents the improper integral as a limit of a definite integrals.

Answer to Problem 2E

The expression that represents the improper integral as a limit of definite integrals is $\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{1}{x^{\frac{1}{3}}}dx}$ .

Explanation of Solution

Given information:

The improper integral is ${\displaystyle {\int }_1^{\infty }\frac{1}{x^{\frac{1}{3}}}dx}$ .

Calculation: by using definition of improper integrals with infinite integration limits, the improper integral can be written as:

  ${\displaystyle {\int }_1^{\infty }\frac{1}{x^{\frac{1}{3}}}dx}=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{1}{x^{\frac{1}{3}}}dx}$

b.

To determine

To find: the value of integral.

Answer to Problem 2E

The value of integral is $\infty$ .

Explanation of Solution

Given information:

The improper integral is ${\displaystyle {\int }_1^{\infty }\frac{1}{x^{\frac{1}{3}}}dx}$ .

Calculation: by using definition of improper integrals with infinite integration limits, the improper integral can be written as:

  ${\displaystyle {\int }_1^{\infty }\frac{1}{x^{\frac{1}{3}}}dx}=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{1}{x^{\frac{1}{3}}}dx}$

Now, evaluate the definite integrals first integrate then use limit,

  $\begin{array}{l}{\displaystyle {\int }_1^{\infty }\frac{1}{x^{\frac{1}{3}}}dx}=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{1}{x^{\frac{1}{3}}}dx}\\ {\displaystyle {\int }_1^{\infty }\frac{1}{x^{\frac{1}{3}}}dx}=\underset{a\to \infty }{\lim }{\left[\frac{3}{2}{x}^{\frac{2}{3}}\right]}_1^a\text{ [integrate]}\\ {\displaystyle {\int }_1^{\infty }\frac{1}{x^{\frac{1}{3}}}dx}=\underset{a\to \infty }{\lim }\left[\frac{3}{2}{a}^{\frac{2}{3}}-\frac{3}{2}{1}^{\frac{2}{3}}\right]\text{ [simplify]}\\ {\displaystyle {\int }_1^{\infty }\frac{1}{x^{\frac{1}{3}}}dx}=\underset{a\to \infty }{\lim }\left[\frac{3}{2}{a}^{\frac{2}{3}}-\frac{3}{2}\right]\text{ [subtract}]\\ {\displaystyle {\int }_1^{\infty }\frac{1}{x^{\frac{1}{3}}}dx}=\infty -\frac{3}{2}\text{ [simplify]}\\ {\displaystyle {\int }_1^{\infty }\frac{1}{x^{\frac{1}{3}}}dx}=\infty \text{ [add]}\end{array}$

Thus, the integral is $\infty$ .