Chapter 14, Problem 79RE

a.

To determine

Graph $f$ ,indicating the area $A$ under $a$ to $b$ .

Answer to Problem 79RE

  

Explanation of Solution

Given information:

A function $f$ is defined over an interval $\left[a,b\right]$ .

  $f\left(x\right)=\frac{1}{x^2}$ , $\left[1,4\right]$

Graph $f$ ,indicating the area $A$ under $a$ to $b$ .

Calculation:

The function $f\left(x\right)=\frac{1}{x^2}$ is defined on the interval $\left[1,4\right]$ .

Graph $f\left(x\right)=\frac{1}{x^2}$ and shade the region under the graph from $\left[1,4\right]$ .

  

Hence , $f\left(x\right)=\frac{1}{x^2}$ is graph.

b.

To determine

Approximate the area $A$ by partitioning $\left[a,b\right]$ into three subintervals of equal length and choosing $u$ as the left endpoint of each subinterval.

Answer to Problem 79RE

  $\boxed{A\approx \frac{49}{36}}$

Explanation of Solution

Given information:

A function $f$ is defined over an interval $\left[a,b\right]$ .

  $f\left(x\right)=\frac{1}{x^2}$ , $\left[1,4\right]$

Approximate the area $A$ by partitioning $\left[a,b\right]$ into three subintervals of equal length and choosing $u$ as the left endpoint of each subinterval.

Calculation:

Approximate the area $A$ by partitioning $\left[a,b\right]$ into three subintervals of equal length and choosing $u$ as the left endpoint of each subinterval.

The area under $f\left(x\right)=\frac{1}{x^2}$ can be found using the formula.

  $A\approx \left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)\right)$ where $\left[a,b\right]$ is the interval where the area is being found,

  $n$ is the number of subintervals, $u_1$ is the end point of the first subinterval, $f\left(u_1\right)$ is the height of endpoint, $u_2$ is the end point of the second subinterval, $f\left(u_2\right)$ is the height of second endpoint, $u_3$ is the end point of the third subinterval, $f\left(u_3\right)$ is the height of third endpoint.

Now calculate $\left(\frac{b-a}{n}\right)$ put $\left(a,b\right)=\left(1,4\right)$ and $n=3$

  $\begin{array}{l}\left(\frac{b-a}{n}\right)=\frac{4-1}{3}\\ =\frac{3}{3}\\ =1\end{array}$

Now partition the $\left(1,4\right)$ in to three subinterval each of length $1:\left[1,2\right],\left[2,3\right],\left[3,4\right]$

Because we are using the left endpoint of the subintervals, $u_1=1,u_2=2,u_3=3$ from the graph we see that $f\left(u_1\right)=1$ , $f\left(u_2\right)=\frac{1}{4}$ , $f\left(u_3\right)=\frac{1}{9}$

Now put all values in area function we get,

  $\begin{array}{l}A\approx \left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)\right)\\ A\approx \left(1\right)\left(1+\frac{1}{4}+\frac{1}{9}\right)\\ A\approx \left(1\right)\left(\frac{49}{36}\right)\\ A\approx \frac{49}{36}\end{array}$

Hence, the area of the graph of $f\left(x\right)$ on the interval $\left[1,4\right]$ broken into three equal subintervals is approximately equal to $\boxed{A\approx \frac{49}{36}}$

c.

To determine

Approximate the area $A$ by partitioning $\left[a,b\right]$ into six subintervals of equal length and choosing $u$ as the left endpoint of each subinterval.

Answer to Problem 79RE

  $\boxed{A\approx 1.02}$

Explanation of Solution

Given information:

A function $f$ is defined over an interval $\left[a,b\right]$ .

  $f\left(x\right)=\frac{1}{x^2}$ , $\left[1,4\right]$

Approximate the area $A$ by partitioning $\left[a,b\right]$ into six subintervals of equal length and choosing $u$ as the left endpoint of each subinterval.

Calculation:

Approximate the area $A$ by partitioning $\left[a,b\right]$ into three subintervals of equal length and choosing $u$ as the left endpoint of each subinterval.

The area under $f\left(x\right)=\frac{1}{x^2}$ can be found using the formula.

  $A\approx \left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)\right)$ where $\left[a,b\right]$ is the interval where the area is being found,

  $n$ is the number of subintervals, $u_1$ is the end point of the first subinterval, $f\left(u_1\right)$ is the height of endpoint, $u_2$ is the end point of the second subinterval, $f\left(u_2\right)$ is the height of second endpoint, $u_3$ is the end point of the third subinterval, $f\left(u_3\right)$ is the height of third endpoint.

Now calculate $\left(\frac{b-a}{n}\right)$ put $\left(a,b\right)=\left(1,4\right)$ and $n=6$

  $\begin{array}{l}\left(\frac{b-a}{n}\right)=\frac{4-1}{6}\\ =\frac{3}{6}\\ =\frac{1}{2}\end{array}$

Now partition the $\left(1,4\right)$ in to six subinterval each of length

  $\frac{1}{2}:\left[1,\frac{3}{2}\right],\left[\frac{3}{2},2,\right],\left[2,\frac{5}{2}\right],\left[\frac{5}{2},3\right],\left[3,\frac{7}{2},\right],\left[\frac{7}{2},4\right]$

Because we are using the left endpoint of the subintervals,

  $u_1=1,u_2=\frac{3}{2},u_3=2,u_4=\frac{5}{2},u_5=3,u_6=\frac{7}{2}$

from the graph we see that $f\left(u_1\right)=1,f\left(u_2\right)=\frac{4}{9},f\left(u_3\right)=\frac{1}{4},f\left(u_4\right)=\frac{4}{25},f\left(u_5\right)=\frac{1}{9},f\left(u_6\right)=\frac{4}{49}$ Now put all values in area function we get,

  $\begin{array}{l}A\approx \left(\frac{b-a}{n}\right)\left(f\left(u_1\right)=+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)+f\left(u_5\right)+f\left(u_6\right)\right)\\ A\approx \left(\frac{1}{2}\right)\left(1+\frac{4}{9}+\frac{1}{4}+\frac{4}{25}+\frac{1}{9}+\frac{4}{49}\right)\\ A\approx \left(\frac{1}{2}\right)\left(\frac{14}{9}+\frac{41}{100}+\frac{4}{49}\right)\\ \boxed{A\approx 1.02}\end{array}$

Hence, the area of the graph of $f\left(x\right)$ on the interval $\left[1,4\right]$ broken into six equal subintervals is approximately equal to $\boxed{A\approx 1.02}$

d.

To determine

Express the area $A$ as an integral.

Answer to Problem 79RE

  $\boxed{{\displaystyle \underset{1}{\overset{4}{\int }}\frac{1}{x^2}}dx}$

Explanation of Solution

Given information:

A function $f$ is defined over an interval $\left[a,b\right]$ .

  $f\left(x\right)=\frac{1}{x^2}$ , $\left[1,4\right]$

Express the area $A$ as an integral.

Calculation:

The area under $f\left(x\right)=\frac{1}{x^2}$ as an integral.

The area under the function $f\left(x\right)=\frac{1}{x^2}$ from $\left[1,4\right]$ can written as an integral ${\displaystyle \underset{1}{\overset{4}{\int }}\frac{1}{x^2}}dx$

Hence, the area $A$ as an integral is $\boxed{{\displaystyle \underset{1}{\overset{4}{\int }}\frac{1}{x^2}}dx}$ .

e.

To determine

Use graphing utility to approximate the integral.

Answer to Problem 79RE

  $\boxed{{\displaystyle \underset{1}{\overset{4}{\int }}\frac{1}{x^2}}dx=0.75}$

Explanation of Solution

Given information:

A function $f$ is defined over an interval $\left[a,b\right]$ .

  $f\left(x\right)=\frac{1}{x^2}$ , $\left[1,4\right]$

Use graphing utility to approximate the integral.

Calculation:

The area under $f\left(x\right)=\frac{1}{x^2}$ as an integral.

The area under the function $f\left(x\right)=\frac{1}{x^2}$ from $\left[1,4\right]$ can written as an integral ${\displaystyle \underset{1}{\overset{4}{\int }}\frac{1}{x^2}}dx$

Hence, the integral is $\boxed{{\displaystyle \underset{1}{\overset{4}{\int }}\frac{1}{x^2}}dx=0.75}$ .