Chapter 14, Problem 78RE

a.

To determine

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

Answer to Problem 78RE

  

Explanation of Solution

Given information:

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

  $f(x)=x^2+3$ , $\left[0,6\right]$

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

Calculation:

The function $f(x)=x^2+3$ Is defined on the interval $\left[0,6\right]$

Graph $f(x)=x^2+3$ and shade the area under the graph from $\left[0,6\right]$

  

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 78RE

  $\boxed{58}$

Explanation of Solution

Given information:

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

  $f(x)=x^2+3$ , $\left[0,6\right]$

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 $\left[0,6\right]$ by partitioning the interval into $3$ subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

The area under $f(x)=x^2+3$ 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 $\left[a,b\right]=\left[0,6\right],andn=3$

  $\begin{array}{l}\frac{b-a}{n}=\frac{6-0}{3}\\ =\frac{6}{3}\\ =2\end{array}$

Partition the interval $\left[0,6\right]$ into $3$ subintervals each of length2: $\left[0,2\right],\left[2,4\right],and\left[4,6\right].$

Because we’re using the left endpoint of the subintervals, $u_1=0,u_2=2and{u}_3=4.$ from the graph we see that $f(u_1)=3,f\left(u_2\right)=7andf\left(u_3\right)=19$ plug these values and $\frac{b-a}{n}=2$ into the area formula

  $\begin{array}{l}A\approx \left(\frac{b-a}{n}\right)(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right))\\ \approx \left(2\right)\left(3+7+19\right)\\ \approx (2)(29)\\ \approx 58\end{array}$

The area under the graph of $f\left(x\right)$ on the interval $\left[0,6\right]$ when broken into equal subintervals is approximately equal to $58$ .

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 78RE

  $\boxed{73}$

Explanation of Solution

Given information:

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

  $f(x)=x^2+3$ , $\left[0,6\right]$

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 $\left[a,b\right]=\left[0,6\right],$ and $n=6$

  $\begin{array}{l}\frac{b-a}{n}=\frac{6-0}{6}\\ =\frac{6}{6}\\ =1\end{array}$

Partition the interval $\left[0,6\right]$ into $6$ subintervals each of length $1$ : $\left[0,1\right],\left[1,2\right],\left[2,3\right],\left[3,4\right],\left[4,5\right],and\left[5,6\right].$

Because we’re using the left endpoints of the subintervals, $u_1=0,u_2=1,u_3=2,u_4=3,u_5=4,and{u}_6=5$ from the graph of $f(x)=x^2+3$ we see that $f(u_1)=3,f(u_2)=4,f(u_3)=7,f(u_4)=12,f(u_5)=19,andf(u_6)=28.$ plug these values and $\frac{b-a}{n}=1$ into the area formula.

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

  $\begin{array}{l}\approx \left(1\right)\left(3+4+7+12+19+28\right)\\ \approx \left(1\right)\left(73\right)\\ \approx 73\end{array}$

Hence, the area under the graph of $f\left(x\right)$ on the interval $\left[0,6\right]$ when broken into 6 equal subintervals is approximately equal to $73$

d.

To determine

Express the area A as a integral.

Answer to Problem 78RE

  $\boxed{{\int }_0{}^6\left(x^2+3\right)dx}$

Explanation of Solution

Given information:

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

  $f(x)=x^2+3$ , $\left[0,6\right]$

Express the area A as a integral.

Calculation:

Write the area A under $f(x)=x^2+3$ as an integral.

Hence, the area under the function $f(x)=x^2+3$ from $\left[0,6\right]$ can be written as the integral ${\int }_0{}^6\left(x^2+3\right)dx$

e.

To determine

Use a graphing utility to approximate the integral.

Answer to Problem 78RE

  $\boxed{{\int }_0{}^6\left(x^2+3\right)dx=90}$

Explanation of Solution

Given information:

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

  $f(x)=x^2+3$ , $\left[0,6\right]$

Use a graphing utility to approximate the integral.

Calculation:

On a graphing utility enter the function $f(x)=x^2+3$ into $Y_1$ . Then hit MATH, 9:fnlnt(, VARS, Y-VARS, 1:Function, 1: $Y_1$ . Next enter a comma, X, comma, $0$ , comma, $6$ , and end parenthesis.

Hence, after hitting enter, the calculator finds that ${\int }_0{}^6\left(x^2+3\right)dx=90$ .