Chapter 5.10, Problem 41E

a.

To determine

To make: a table of approximate values of ${\displaystyle \underset{1}{\overset{t}{\int }}g(x)dx}$ for t =2, 5, 10, 100, 1000 and 10,000 using calculator and find ${\displaystyle \underset{1}{\overset{\infty }{\int }}g(x)dx}$ is convergent.

Answer to Problem 41E

Convergent.

Explanation of Solution

Given information: $g(x)=\frac{({\sin }^{2}x)}{x^2}$.

Calculation:

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

t	2	5	10	100	1000	10000
f (t)	0.447453	0.577101	0.621306	0.668479	0.672957	0.673407

From the table data, the integral ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{({\sin }^{2}x)}{x^2}}dx$ is convergent.

b.

To determine

To show: the ${\displaystyle \underset{1}{\overset{\infty }{\int }}g(x)dx}$ is convergent using Comparison Theorem.

Answer to Problem 41E

Explanation of Solution

Given information : $f(x)=\frac{1}{x^2}$.

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:

Recall that

  $\begin{array}{c}0\le {\sin }^{2}x\le 1.\\ \text{Therefore,}\\ 0.\frac{1}{x^2}\le \frac{1}{x^2}.{\sin }^{2}x\le 1.\frac{1}{x^2}.\\ 0\le \frac{{\sin }^{2}x}{x^2}\le \frac{1}{x^2}\\ \text{By}\text{Comparision}\text{Property}\text{of}\text{integrals},\\ {\displaystyle \underset{1}{\overset{\infty }{\int }}0dx\le {\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{{\sin }^{2}x}{x^2}}}\le {\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^2}dx}\\ 0\le {\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{{\sin }^{2}x}{x^2}}\le [-\frac{1}{x}]\begin{array}{c}\infty \\ 1\end{array}\\ 0\le {\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{{\sin }^{2}x}{x^2}}\le -\frac{1}{\infty }+1\\ 0\le {\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{{\sin }^{2}x}{x^2}}\le 1\end{array}$

Hence proved that the given integral converges.

c.

To determine

To graph: f and g on the same screen for $1\le x\le 10$ and explain intuitively why ${\displaystyle \underset{2}{\overset{\infty }{\int }}g(x)dx}$ is convergent.

Answer to Problem 41E

Explanation of Solution

Given information:

Calculation:

The graph of f and g on the same screen for $1\le x\le 10$ is shown below.

  

  $f(x)$ is the red curve and g(x) is the blue curve.

From the graph $0\le g(x)\le f(x)$ .

Recall that: m:math>∫1∞f(x)dx is the area under the graph for x >1

Since ${\displaystyle \underset{1}{\overset{\infty }{\int }}f(x)dx}$ is convergent, hence the area under the curve y = f(x) is finite.

Since 0< g(x) < f(x), so the area under g(x) is finite too.

Which in turns implies that ${\displaystyle \underset{1}{\overset{\infty }{\int }}g(x)dx}$ is convergent.