Chapter 9.4, Problem 6E

To determine

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

Answer to Problem 6E

The expression that represents the improper integral as a limit of definite integrals is $\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{2}{x^3}dx}$ and value of integral is $1$ .

Explanation of Solution

Given information:

The improper integral is ${\displaystyle {\int }_1^{\infty }\frac{2}{x^3}dx}$ .

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

  ${\displaystyle {\int }_1^{\infty }\frac{2}{x^3}dx}=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{2}{x^3}dx}$

To find the value of integral, first integrate and then simplify,

  $\begin{array}{l}{\displaystyle {\int }_1^{\infty }\frac{2}{x^3}dx}=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{2}{x^3}dx}\\ {\displaystyle {\int }_1^{\infty }\frac{2}{x^3}dx}=\underset{a\to \infty }{\lim }{\left[\frac{2x^{-3+1}}{-3+1}\right]}_1^a\text{ [}{\displaystyle \int x^n=\frac{x^{n+1}}{n+1}]}\\ {\displaystyle {\int }_1^{\infty }\frac{2}{x^3}dx}=\underset{a\to \infty }{\lim }{\left[\frac{2x^{-2}}{-2}\right]}_1^a\text{ [subtract]}\\ {\displaystyle {\int }_1^{\infty }\frac{2}{x^3}dx}=\underset{a\to \infty }{\lim }{\left[-\frac{1}{x^2}\right]}_1^a\text{ [divide]}\\ {\displaystyle {\int }_1^{\infty }\frac{2}{x^3}dx}=\underset{a\to \infty }{\lim }\left[-\frac{1}{a^2}+\frac{1}{1}\right]\text{ [simplify]}\\ {\displaystyle {\int }_1^{\infty }\frac{2}{x^3}dx}=0+1\text{ [simplify]}\\ {\displaystyle {\int }_1^{\infty }\frac{2}{x^3}dx}=1\end{array}$

Thus, the value is $1$ .