Chapter 14, Problem 76RE

a.

To determine

Plot a graph.

Answer to Problem 76RE

  

Explanation of Solution

Given information:

The function $f\left(x\right)=2x+8$ is defined on the interval $\left[0,4\right]$ .

Graph f.

In (b)-(e), approximate the area $A$ under $f$ from $x=0$ to $x=4$ as follows:

Calculation:

Graph

  $f\left(x\right)=2x+3$ on the interval $\left[0,4\right]$ .

  

b.

To determine

Partition $\left[0,4\right]$ into four subintervals of equal length

Answer to Problem 76RE

  $\boxed{\text{Approx}.\text{ 2}0 }$

Explanation of Solution

Given information:

The function $f\left(x\right)=2x+8$ is defined on the interval $\left[0,4\right]$ .

Partition $\left[0,4\right]$ into four subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Calculation:

Approximate the area $A$ under $f\left(x\right)$ on the interval $\left[0,4\right]$ by partitioning the interval into 4 subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

The area under $f\left(x\right)=2x+8$ can be found using the formula

  $A\approx \left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f(u_2)+f(u_3)+f(u_4)\right)$ where $\left[a,b\right]$ is the interval where is being found, n is the subintervals, $u_1$ is the endpoint of the first subinterval, $f\left(u_1\right)$ is the height of the endpoint, $u_2$ is the endpoint of the second subinterval, $f\left(u_2\right)$ is the height of that second end point, and so on.

There are four endpoints used because there are four subintervals within the given interval. First calculate the factor $\left(\frac{b-a}{n}\right)$ by plugging in $\left[a,b\right]=\left[0,4\right]$ and $n=4$

  $\begin{array}{l}\frac{b-a}{n}=\frac{4-0}{4}\\ =\frac{4}{4}\\ =1\end{array}$

Hence, partition the interval $\left[0,4\right]$ into $4$ subintervals each of length 1: $\left[0,1\right],\left[1,2\right],\left[2,3\right]and\left[3,4\right]$

Because we’re using the left endpoints of the subintervals. $u_1=0,u_2=1,u_3=2and{u}_4=3$ .

From the graph we see that $f\left(u_1\right)=8,f\left(u_2\right)=6,f\left(u_3\right)=4$ and $f(u_4)=2$ . Plug these values and $\frac{b-a}{n}=1$ into the area formula

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

  $\begin{array}{l}\approx (1)(8+6+4+2)\\ \approx (1)(20)\\ \approx 20\end{array}$

The area under the graph of $f(x)$ on the interval $\left[0,4\right]$ when broken into 4 equal subintervals is approximately equal to $20$ .

c.

To determine

Partition $\left[0,4\right]$ into four subintervals of equal length

Answer to Problem 76RE

  $\boxed{\text{Approx}.\text{ 12}}$

Explanation of Solution

Given information:

The function $f\left(x\right)=2x+8$ is defined on the interval $\left[0,4\right]$ .

Partition $\left[0,4\right]$ into four subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

Calculation:

Approximate the area $A$ under $f\left(x\right)$ on the interval $\left[0,4\right]$ by partitioning the interval into 4 subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

Since the interval in part (b) was also partitioned into 4 subintervals of equal length we, can again use the fact that $\frac{b-a}{n}=1$ and the subintervals are $\left[0,1\right],\left[1,2\right],\left[2,3\right]and\left[3,4\right]$ .

Because we’re using the right endpoints of the subintervals, $u_1=1,u_2=2,u_3=3$ and $u_4=4$

From the graph of $f\left(x\right)=2x+8$ , we see that $f\left(u_1\right)=6,f\left(u_2\right)=4,f\left(u_3\right)=2$ and $f(u_4)=0$ . Plug these values and $\frac{b-a}{n}=1$ into the area formula

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

  $\begin{array}{l}\approx \left(1\right)\left(6+4+2+0\right)\\ \approx \left(1\right)\left(12\right)\\ \approx 12\end{array}$

The area under the graph of $f(x)$ on the interval $\left[0,4\right]$ when broken into 4 equal subintervals is approximately equal to $12$ .

d.

To determine

Partition $\left[0,4\right]$ into eight subintervals of equal length.

Answer to Problem 76RE

  $\boxed{\text{Approx 18}}$

Explanation of Solution

Given information:

The function $f\left(x\right)=2x+8$ is defined on the interval $\left[0,4\right]$ .

Partition $\left[0,4\right]$ into eight subintervals of equal length and choose u as the left endpoint of each subinterval.

Calculation:

First calculate $\left(\frac{b-a}{n}\right)$ by plugging in $\left[a,b\right]=\left[0,4\right]$ , and $n=8$

  $\frac{b-a}{n}=\frac{4-0}{8}$

  $\begin{array}{l}=\frac{4}{8}\\ =\frac{1}{2}\end{array}$

Partition the interval $\left[0,4\right]$ into $8$ subintervals each of length $\frac{1}{2}:\left[0,\frac{1}{2}\right],\left[\frac{1}{2},1\right],\left[1,\frac{3}{2}\right]$ , $\left[\frac{3}{2},2\right],\left[2,\frac{5}{2}\right],\left[\frac{5}{2},3\right],\left[3,\frac{7}{2}\right]and\left[\frac{7}{2},4\right]$ .

Because we’re using the left endpoints of the subintervals, $u_1=0,u_2=\frac{1}{2},u_3=1,u_4=\frac{3}{2},u_5=2,u_6=\frac{5}{2},u_7=3$ and $u_8=\frac{7}{2}$ . From the graph of $f\left(x\right)=2x+8$ we see that $f(u_1)=8,f(u_2)=7,f(u_3)=6,f\left(u_4\right)=5,f(u_5)=4,f(u_6)=3,f(u_7)=2$ and $f(u_8)=1$ . Plug these values and $\frac{b-a}{n}=\frac{1}{2}$ into the area formula

  $A\approx \left(\frac{b-a}{n}\right)(f(u_1)+f(u_2)+f(u_3)+f(u_4)+f(u_5)+f(u_6)+f(u_7)+f(u_8))$

  $\begin{array}{l}\approx \left(\frac{1}{2}\right)\left(8+7+6+5+4+3+2+1\right)\\ \approx \left(\frac{1}{2}\right)\left(36\right)\\ \approx 18\end{array}$

Hence, the area under the graph of $f(x)$ on the interval $\left[0,4\right]$ when broke into 8 equal subintervals is approximately equal to $18$ .

e.

To determine

Partition $\left[0,4\right]$ into eight subintervals of equal length.

Answer to Problem 76RE

  $\boxed{\text{Approx}.\text{ 28} }$

Explanation of Solution

Given information:

The function $f\left(x\right)=2x+8$ is defined on the interval $\left[0,4\right]$ .

Partition $\left[0,4\right]$ into eight subintervals of equal length and choose u as the right endpoint of each subinterval.

Calculation:

Since the interval in part (d) was also partitioned into 8 subintervals of equal length, we can again use the fact that $\frac{b-a}{n}=\frac{1}{2}$ and the subintervals are $\left[0,\frac{1}{2}\right],\left[\frac{1}{2},1\right],\left[1,\frac{3}{2}\right]$ , $\left[\frac{3}{2},2\right],\left[2,\frac{5}{2}\right],\left[\frac{5}{2},3\right],\left[3,\frac{7}{2}\right]and\left[\frac{7}{2},4\right]$ because we’re using the right endpoints of the subintervals, $u_1=\frac{1}{2},u_2=1,u_3=\frac{3}{2},u_4=2,u_5=\frac{5}{2},u_6=3,u_7=\frac{7}{2}$ and $u_8=4$ .

From the graph of $f\left(x\right)=2x+8$ we see that

  $f(u_1)=7,f(u_2)=6,f(u_3)=5,f\left(u_4\right)=4,f(u_5)=3,f(u_6)=2,f(u_7)=1$ and $f(u_8)=0$ . Plug these values and $\frac{b-a}{n}=\frac{1}{2}$ into the area formula

  $A\approx \left(\frac{b-a}{n}\right)(f(u_1)+f(u_2)+f(u_3)+f(u_4)+f(u_5)+f(u_6)+f(u_7)+f(u_8))$

  $\begin{array}{l}\approx \left(\frac{1}{2}\right)\left(7+6+5+4+3+2+1+0\right)\\ \approx \left(\frac{1}{2}\right)\left(28\right)\\ \approx 28\end{array}$

Hence, the area under the graph of $f(x)$ on the interval $\left[0,4\right]$ when broke into 8 equal subintervals is approximately equal to $28$

f.

To determine

What is the actual area $A$ ?

Answer to Problem 76RE

  $\boxed{16}$

Explanation of Solution

Given information:

The function $f\left(x\right)=2x+8$ is defined on the interval $\left[0,4\right]$ .

What is the actual area $A$ ?

Calculation:

The actual area under $f(x)$ is the area of a right triangle formed by $f(x)$ and the x-axis. The base has a length of $4$ because the interval has a length of $4$ . The height of the triangles is $8$ because at the end of interval, $x=0$ so $f(0)=8$ . Plug these values into the area of the triangle

  $\begin{array}{l}A=\frac{1}{2}basegheight\\ =\frac{1}{2}(4)(8)\\ =\frac{1}{2}(32)\\ =16\end{array}$

Hence, the actual area under $f\left(x\right)=2x+8$ is $16$