Chapter 5.10, Problem 4E

a.

To determine

To graph: the function $f(x)=\frac{1}{x^{1.1}}\text{and}g(x)=\frac{1}{x^{0.9}}$ in the viewing rectangles [0, 10] by [0, 1] and [0, 100] by [0, 1].

Answer to Problem 4E

Explanation of Solution

Given information: $f(x)=\frac{1}{x^{1.1}}\text{and}g(x)=\frac{1}{x^{0.9}}$ .

Calculation:

This is the graph of the function $f(x)=\frac{1}{x^{1.1}}$ on the viewing rectangle [0, 10] by [0, 1].

  

This is the graph of the function $f(x)=\frac{1}{x^{1.1}}$ on the viewing rectangle [0, 100] by [0, 1].

  

This is the graph of the function $g(x)=\frac{1}{x^{0.9}}$ on the viewing rectangle [0, 10] by [0, 1].

  

This is the graph of the function $g(x)=\frac{1}{x^{0.9}}$ on the viewing rectangle [0, 100] by [0, 1].

  

b.

To determine

To find: the areas under the graph of f and g from x =1 to x = t and evaluate for t =10, 100, ${10}^4,{10}^6,{10}^{10}\text{and}{10}^{20}.$

Answer to Problem 4E

The areas under the graph of f and g from x =1 to x = t is $10(t^{0.1}+t^{-0.1}-2)$ .

Explanation of Solution

Given information:

Calculation:

Graph of the function f is under the graph of a function g. Therefore:

  $\begin{array}{c}{\displaystyle \underset{1}{\overset{t}{\int }}(g(x)-f(x))dx}={\displaystyle \underset{1}{\overset{t}{\int }}(\frac{1}{x^{0.9}}}-\frac{1}{x^{1.1}})dx.\\ ={\displaystyle \underset{1}{\overset{t}{\int }}\frac{1}{x^{0.9}}}dx-{\displaystyle \underset{1}{\overset{t}{\int }}\frac{1}{x^{1.1}}}dx.\\ =\frac{x^{0.1}}{0.1}|\begin{array}{c}t\\ 1\end{array}-\frac{x^{-0.1}}{-0.1}|\begin{array}{c}t\\ 1\end{array}\\ =\frac{t^{0.1}}{0.1}-\frac{1^{0.1}}{0.1}-\frac{t^{-0.1}}{-0.1}+\frac{1^{-0.1}}{-0.1}\\ =\frac{t^{0.1}}{0.1}-\frac{2}{0.1}+\frac{t^{-0.1}}{0.1}\\ =10(t^{0.1}+t^{-0.1}-2).\end{array}$

Substituting value for t =10, 100, ${10}^4,{10}^6,{10}^{10}\text{and}{10}^{20}.$

  $\begin{array}{c}t=10\\ 10({10}^{0.1}+{10}^{-0.1}-2)=\mathrm{0.53.}\\ t=100\\ 10({100}^{0.1}+{100}^{-0.1}-2)=\mathrm{2.16.}\\ t={10}^4\\ 10({10}^{0.4}+{10}^{-0.4}-2)=\mathrm{9.1.}\\ t={10}^6\\ 10({10}^{0.6}+{10}^{-0.6}-2)=\mathrm{22.32.}\\ t={10}^{10}\\ 10({10}^1+{10}^{-1}-2)=81.\\ t={10}^{20}\\ 10({10}^2+{10}^{-2}-2)=\mathrm{980.1.}\end{array}$

c.

To determine

To find: the total area under each curve for $x\ge 1,$ if it exists.

Answer to Problem 4E

  $\underset{t\to \infty }{\lim }10(t^{0.1}+t^{-0.1}-2)=\infty .$

Explanation of Solution

Given information:

Calculation:

Graph of the function f is under the graph of a function g. Therefore:

  $\begin{array}{c}{\displaystyle \underset{1}{\overset{t}{\int }}(g(x)-f(x))dx}={\displaystyle \underset{1}{\overset{t}{\int }}(\frac{1}{x^{0.9}}}-\frac{1}{x^{1.1}})dx.\\ ={\displaystyle \underset{1}{\overset{t}{\int }}\frac{1}{x^{0.9}}}dx-{\displaystyle \underset{1}{\overset{t}{\int }}\frac{1}{x^{1.1}}}dx.\\ =\frac{x^{0.1}}{0.1}|\begin{array}{c}t\\ 1\end{array}-\frac{x^{-0.1}}{-0.1}|\begin{array}{c}t\\ 1\end{array}\\ =\frac{t^{0.1}}{0.1}-\frac{1^{0.1}}{0.1}-\frac{t^{-0.1}}{-0.1}+\frac{1^{-0.1}}{-0.1}\\ =\frac{t^{0.1}}{0.1}-\frac{2}{0.1}+\frac{t^{-0.1}}{0.1}\\ =10(t^{0.1}+t^{-0.1}-2).\end{array}$

Search limit of obtained result.

  $\underset{t\to \infty }{\lim }10(t^{0.1}+t^{-0.1}-2)=\infty .$