Chapter 8.4, Problem 4E

a.

To determine

To express: The improper integral, ${\displaystyle {\int }_1^{\infty }\frac{dx}{\sqrt{x}}}$ as a limit of definite integrals.

Answer to Problem 4E

The limit of definite integral will be, $\underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}$ .

Explanation of Solution

Given information: The given integral is ${\displaystyle {\int }_1^{\infty }\frac{dx}{\sqrt{x}}}$ .

Concept used: Consider that $f\left(x\right)$ is continuous on $\left[a,\infty \right)$ , then ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)dx=\underset{b\to \infty }{\lim }}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ .

Calculation:

Assume $I={\displaystyle {\int }_1^{\infty }\frac{dx}{\sqrt{x}}}$ . In order to express the improper integral as a limit definite integral, consider a real number $b$ such that,

  $I=\underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}$

So, the limit of definite integral will be, $\underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}$ .

b.

To determine

To evaluate: The value of the improper integral, ${\displaystyle {\int }_1^{\infty }\frac{dx}{\sqrt{x}}}$ .

Answer to Problem 4E

The given integral diverges.

Explanation of Solution

Given information: The given integral is ${\displaystyle {\int }_1^{\infty }\frac{dx}{\sqrt{x}}}$ .

Concept used: The formula for determining the value of the integral is ${\displaystyle \int x^{n}dx}=\frac{x^{n+1}}{n+1}+c$ .

Calculation:

From part a, the limit of integration is $\underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}$ .

Apply the rules of integration to determine the value of integral, ${\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}$ in terms of $b$ .

  $\begin{array}{l}{\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}={\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]}_1^b\\ \Rightarrow {\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}={\left[2\sqrt{x}\right]}_1^b\\ \Rightarrow {\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}=2\sqrt{b}-2\sqrt{1}\\ \Rightarrow {\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}=2\sqrt{b}-2\end{array}$

Determine the value of $\underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}$ .

  $\begin{array}{l}\underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}=\underset{b\to \infty }{\lim }\left[2\sqrt{b}-2\right]\\ \Rightarrow \underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}=2\sqrt{\infty }-2\\ \Rightarrow \underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{dx}{\sqrt{x}}}=\infty \end{array}$

Since, the value of the limit approaches to $\infty$ , the given integral diverges.