Chapter 9.4, Problem 5E

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 5E

The expression that represents the improper integral as a limit of definite integrals is $\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{1}{x^4}dx}$ and value of integral is $\frac{1}{3}$ .

Explanation of Solution

Given information:

The improper integral is ${\displaystyle {\int }_1^{\infty }\frac{1}{x^4}dx}$ .

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

  ${\displaystyle {\int }_1^{\infty }\frac{1}{x^4}dx}=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{1}{x^4}dx}$

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

  $\begin{array}{l}{\displaystyle {\int }_1^{\infty }\frac{1}{x^4}dx}=\underset{a\to \infty }{\lim }{\displaystyle {\int }_1^a\frac{1}{x^4}dx}\\ {\displaystyle {\int }_1^{\infty }\frac{1}{x^4}dx}=\underset{a\to \infty }{\lim }{\left[\frac{x^{-4+1}}{-4+1}\right]}_1^a\text{ [}{\displaystyle \int x^n=\frac{x^{n+1}}{n+1}]}\\ {\displaystyle {\int }_1^{\infty }\frac{1}{x^4}dx}=\underset{a\to \infty }{\lim }{\left[\frac{x^{-3}}{-3}\right]}_1^a\text{ [subtract]}\\ {\displaystyle {\int }_1^{\infty }\frac{1}{x^4}dx}=\underset{a\to \infty }{\lim }\left[\frac{a^{-3}}{-3}-\frac{1^{-3}}{-3}\right]\text{ [simplify]}\\ {\displaystyle {\int }_1^{\infty }\frac{1}{x^4}dx}=0+\frac{1}{3}\text{ [simplify]}\\ {\displaystyle {\int }_1^{\infty }\frac{1}{x^4}dx}=\frac{1}{3}\end{array}$

Thus, the value is $\frac{1}{3}$ .