Chapter 14, Problem 77RE

a.

To determine

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

Answer to Problem 77RE

  

Explanation of Solution

Given information:

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

  $f(x)=4-x^2$ , $\left[-1,2\right]$

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

Calculation:

The function $f(x)=4-x^2$ Is defined on the interval $\left[-1,2\right]$

Graph $f(x)=4-x^2$ and shade the area under the graph from $\left[-1,2\right]$

  

b.

To determine

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

Answer to Problem 77RE

  $\boxed{10}$

Explanation of Solution

Given information:

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

  $f(x)=4-x^2$ , $\left[-1,2\right]$

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

Calculation:

Approximate the area A under $f(x)$ on the interval $\left[-1,2\right]$ by partitioning the interval into $3$ subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

The area under $f(x)=4-x^2$ can be found using the formula $A\approx \left(\frac{b-a}{n}\right)(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right))$

Where $[a,b]$ is the interval where the area is being found, $n$ is the number of subintervals, $u_1$ is the endpoint of the first subinterval, $f\left(u_1\right)$ is the height of that endpoint, $u_2$ is the endpoint of the second subinterval $f\left(u_2\right)$ is the height of that second endpoint, and so on. There are $3$ endpoint used because there are $3$ subintervals within the given interval. First calculation the factor $\left(\frac{b-a}{n}\right)$ by plugging in $\left[a,b\right]=\left[-1,2\right],andn=3$

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

Partition the interval $\left[-1,2\right]$ into $3$ subintervals each of length1: $\left[-1,0\right],\left[0,1\right],and\left[1,2\right].$

Because we’re using the left endpoint of the subintervals, $u_1=-1,u_2=0and{u}_3=1.$ from the graph we see that $f(u_1)=3,f\left(u_2\right)=4andf\left(u_3\right)=3$ plug these values and $\frac{b-a}{n}=1$ into the area formula

  $\begin{array}{l}A\approx \left(\frac{b-a}{n}\right)(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right))\\ \approx \left(1\right)\left(3+4+3\right)\\ \approx (1)(10)\\ \approx 10\end{array}$

The area under the graph of $f\left(x\right)$ on the interval $\left[-1,2\right]$ when broken into equal subintervals is approximately equal to $10$ .

c.

To determine

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

Answer to Problem 77RE

  $\boxed{\frac{77}{8}}$

Explanation of Solution

Given information:

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

  $f(x)=4-x^2$ , $\left[-1,2\right]$

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

Calculation:

First calculate by plugging in and $\left(\frac{b-a}{n}\right)$ by plugging in $\left[a,b\right]=\left[-1,2\right],$ and $n=6$

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

Partition the interval $\left[-1,2\right]$ into $6$ subintervals each of length $\frac{1}{2}$ : $\left[-1,-\frac{1}{2}\right],\left[-\frac{1}{2},0\right],\left[0,\frac{1}{2}\right],\left[\frac{1}{2},1\right],\left[1,\frac{3}{2}\right],and\left[\frac{3}{2},2\right].$

Because we’re using the left endpoints of the subintervals, $u_1=-1,u_2=-\frac{1}{2},u_3=0,u_4=\frac{1}{2},u_5=1,and{u}_6=\frac{3}{2}.$ from the graph of $f(x)=4-x^2$ we see that $f(u_1)=3,f(u_2)=\frac{15}{4},f(u_3)=4,f(u_4)=\frac{15}{4},f(u_5)=3,andf(u_6)=\frac{7}{4}.$ plug these values and $\frac{b-a}{n}=\frac{1}{2}$ into the area formula.

  $A\approx \left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)+f\left(u_5\right)+f\left(u_6\right)\right)$

  $\begin{array}{l}\approx \left(\frac{1}{2}\right)\left(3+\frac{15}{4}+4+\frac{15}{4}+3+\frac{7}{4}\right)\\ \approx \left(\frac{1}{2}\right)\left(10+\frac{37}{4}\right)\\ \approx \left(\frac{1}{2}\right)\left(\frac{77}{4}\right)\end{array}$

Hence, the area under the graph of $f\left(x\right)$ on the interval $\left[-1,2\right]$ when broken into 6 equal subintervals is approximately equal to $\frac{77}{8}$

d.

To determine

Express the area A as a integral.

Answer to Problem 77RE

  $\boxed{{\int }_{-1}{}^6\left(4-x^2\right)dx}$

Explanation of Solution

Given information:

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

  $f(x)=4-x^2$ , $\left[-1,2\right]$

Express the area A as a integral.

Calculation:

Write the area A under $f(x)=4-x^2$ as an integral.

Hence, the area under the function $f(x)=4-x^2$ from $\left[-1,2\right]$ can be written as the integral ${\int }_{-1}{}^6\left(4-x^2\right)dx$

e.

To determine

Use a graphing utility to approximate the integral.

Answer to Problem 77RE

  $\boxed{{\int }_{-1}{}^2\left(4-x^2\right)dx=9}$

Explanation of Solution

Given information:

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

  $f(x)=4-x^2$ , $\left[-1,2\right]$

Use a graphing utility to approximate the integral.

Calculation:

On a graphing utility enter the function $f(x)=4-x^2$ into $Y_1$ . Then hit MATH, 9:fnlnt(, VARS, Y-VARS, 1:Function, 1: $Y_1$ . Next enter a comma, X, comma, $-1$ , comma, $2$ , and end parenthesis.

Hence, after hitting enter, the calculator finds that ${\int }_{-1}{}^2\left(4-x^2\right)dx=9$ .