Chapter 11.4, Problem 38E

Solution

a.

To verify: the given function $f\left(x\right)=\sqrt{x-2}$ is nonnegative.

Yes, the function is nonnegative.

Given information:

The given function is $f\left(x\right)=\sqrt{x-2}$ .

Concept Used:

Nonnegative function: It has function values equal to or greater than zero $\left(f\left(x\right)\ge 0\right)$ . The domain of the function can be negative, but the outputs must be zero or greater in order for the function to be classified as non-negative. A positive function is similar, except that it does not include zero in its range.

Calculation:

First, draw the function $f\left(x\right)=\sqrt{x-2}$ .

The graph is shown below:

  

From the above graph, it is observed that the function is above the $x-$ axis at interval $\left[3,6\right]$ .

Therefore, the function is nonnegative at the given interval.

b.

To determine: the value of LRAM, RRAM and average approximations for the area under the graph of the function from $x=a$ to $x=b$ with $10,20,50,$ and $100$ approximating rectangles.

$N$	$LRAM$	$RRAM$	$Average$
10	4.51	4.81	4.66
20	4.59	4.74	4.66
50	4.64	4.70	4.67
100	4.65	4.68	4.66

Given information:

The given function is $f\left(x\right)=\sqrt{x-2}$ .

Concept Used:

  $\begin{array}{l}\text{PROGRAM:LRAM}\\ \text{:}\text{Input}\text{"A",A}\\ \text{:}\text{Input}\text{"B",B}\\ \text{:}\text{Input}\text{"N",N}\\ \text{:sum}\left(\text{seq}\left(\left(\left(\text{B-A}\right)\text{/N}\right){\text{Y}}_{\text{1}}\left(\text{A+K}\left(\left(\text{B-A}\right)\text{/N}\right)\right)\text{,K,0,N-1}\right)\right)\to \text{C}\\ \text{:Disp}\text{"AREA=",C}\end{array}$

  $\begin{array}{l}\text{PROGRAM:RRAM}\\ \text{:}\text{Input}\text{"A",A}\\ \text{:}\text{Input}\text{"B",B}\\ \text{:}\text{Input}\text{"N",N}\\ \text{:sum}\left(\text{seq}\left(\left(\left(\text{B-A}\right)\text{/N}\right){\text{Y}}_{\text{1}}\left(\text{A+K}\left(\left(\text{B-A}\right)\text{/N}\right)\right)\text{,K,0,N}\right)\right)\to \text{C}\\ \text{:Disp}\text{"AREA=",C}\end{array}$

Calculation:

The graph is shown below:

  

Here, $\text{B=6,A=3,}\text{and}\text{N=10}$ .

Now, use the LRAM program formula ${\displaystyle \sum _{K=0}^{N-1}f}\left(A+K\left(\frac{B-A}{N}\right)\right)\left(\frac{B-A}{N}\right)$ to evaluate the value of LRAM.

  $\begin{array}{c}{\displaystyle \sum _{K=0}^{10-1}f}\left(3+K\left(\frac{6-3}{10}\right)\right)\left(\frac{6-3}{10}\right)=\frac{3}{10}{\displaystyle \sum _{K=0}^{10-1}f}\left(3+K\left(\frac{3}{10}\right)\right)\\ =4.51\end{array}$

The graph is shown below:

  

Now, use the RRAM program formula ${\displaystyle \sum _{K=1}^{N}f}\left(A+K\left(\frac{B-A}{N}\right)\right)\left(\frac{B-A}{N}\right)$ to evaluate the value of RRAM.

  $\begin{array}{c}{\displaystyle \sum _{K=1}^{10}f}\left(3+K\left(\frac{6-3}{10}\right)\right)\left(\frac{6-3}{10}\right)=\frac{3}{10}{\displaystyle \sum _{K=1}^{10}f}\left(3+\frac{3K}{10}\right)\\ =4.81\end{array}$

Now, find the average approximation of the area by finding the average of LRAM and RRAM.

  $\begin{array}{c}\text{Average}=\frac{4.51+4.81}{2}\\ =4.66\end{array}$

Use the above procedure to evaluate the $20,50,$ and $100$ approximating rectangles.

$N$	$LRAM$	$RRAM$	$Average$
10	4.51	4.81	4.66
20	4.59	4.74	4.66
50	4.64	4.70	4.67
100	4.65	4.68	4.66

c.

To compare: the average area estimates in part (b) using 100 approximating rectangles with calculator NINT area estimate

The area from using 100 approximating rectangles is approximately equal to the NINT area.

Given information:

The given function is $f\left(x\right)=\sqrt{x+2}$ .

Concept Used:

The notation $\text{NINT}\left(f\left(x\right),x,a,b\right)$ to denote such a calculator approximation to ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ .

Calculation:

The graph is shown below:

  

The table is shown below:

$N$	$LRAM$	$RRAM$	$Average$
10	4.51	4.81	4.66
20	4.59	4.74	4.66
50	4.64	4.70	4.67
100	4.65	4.68	4.66

Here, $a=3,b=6\text{and}f\left(x\right)=\sqrt{x-2}$

  $\begin{array}{l}\underset{3}{\overset{6}{{\displaystyle \int }}}\sqrt{x-2}dx=4.66\hfill \\ \text{NINT}\left(f\left(x\right),x,3,6\right)=4.66\hfill \end{array}$

The area from using 100 approximating rectangles is approximately equal to the NINT area.