Chapter 5.10, Problem 42E

a.

To determine

To make: a table of approximate values of ${\displaystyle \underset{2}{\overset{t}{\int }}g(x)dx}$ for t =5, 10, 100, 1000 and 10,000 using calculator and find ${\displaystyle \underset{2}{\overset{\infty }{\int }}g(x)dx}$ is convergent or divergent.

Answer to Problem 42E

Divergent.

Explanation of Solution

Given information: $g(x)=\frac{1}{\sqrt{x}-1}$

Calculation:

  $\text{Let}f(t)={\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}-1}}dx,$ Then

t	5	10	100	1000	10000
f (t)	3.830326718	6.801199648	23.32876922	69.02336139	208.1245598

The integral ${\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}-1}}dx$ looks divergent because the values of f(t) get larger and larger from table and are headed towards infinity.

b.

To determine

To show: the ${\displaystyle \underset{2}{\overset{\infty }{\int }}g(x)dx}$ is divergent using Comparison Theorem.

Answer to Problem 42E

Explanation of Solution

Given information: $f(x)=\frac{1}{\sqrt{x}}$.

Formula used:

Comparison property for integrals:

If

  $\begin{array}{l}h(x)\le f(x)\le g(x)\text{for}[a,b]\text{Then}\\ {\displaystyle \underset{a}{\overset{b}{\int }}h(x)dx\le {\displaystyle \underset{a}{\overset{b}{\int }}f(x)}}dx\le {\displaystyle \underset{a}{\overset{b}{\int }}g(x)dx}.\end{array}$

Calculation:

Given:

  $\begin{array}{l}{\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}-1}}dx\\ \text{Since,}\\ \sqrt{x}>\sqrt{x}-1\\ \text{then},\\ \frac{1}{\sqrt{x}}<\frac{1}{\sqrt{x}-1}\\ \text{Thus}\text{,}\\ {\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}}}dx\\ \text{is}\text{divergent}\text{,}\text{then}\text{by}\text{}\text{comparison}\text{theorem}.\\ {\displaystyle \underset{2}{\overset{\infty }{\int }}\frac{1}{\sqrt{x}-1}}dx\\ \text{also}\text{divergent}\text{.}\\ \end{array}$

c.

To determine

To graph : f and g on the same screen for $2\le x\le 20$ and explain intuitively why ${\displaystyle \underset{2}{\overset{\infty }{\int }}g(x)dx}$ is divergent.

Answer to Problem 42E

Explanation of Solution

Given information:

Calculation:

The graph of f and g on the same screen for $2\le x\le 20$ is shown below.

  

  $f(x)=\frac{1}{\sqrt{x}}$ is known to be divergent, and from the graph that it will always be less than g(x) for $x\ge 2$ . Since f stays under g and is divergent that means g(x) will also divergent.