Chapter 11.5, Problem 23E

To determine

To show the approximate area of the region.

Answer to Problem 23E

The required table is shown in table 1.

Explanation of Solution

Given information:

The function is $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{9}{\mathrm{x}}^3$ and the graph of the area is shown below;

  

Approximate area of n rectangles can be calculated as,

  $\begin{array}{c}\mathrm{A}={\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{f}\left(\mathrm{a}+\frac{\left(\mathrm{b}-\mathrm{a}\right)\mathrm{i}}{\mathrm{n}}\right)}\left(\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\right)\\ ={\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\left(\frac{-4\mathrm{i}}{\mathrm{n}}+4\right)}\left(\frac{12}{\mathrm{n}}\right)\\ ={\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\left(\frac{-48\mathrm{i}}{{\mathrm{n}}^2}+\frac{48}{\mathrm{n}}\right)}\\ =\frac{-48}{{\mathrm{n}}^2}\cdot \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\frac{48}{\mathrm{n}}\mathrm{n}\end{array}$

Here,

  $\begin{array}{c}\mathrm{w}\mathrm{i}\mathrm{d}\mathrm{t}\mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ =\frac{3-0}{\mathrm{n}}\\ =\frac{3}{\mathrm{n}}\end{array}$

And,

  $\begin{array}{c}\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}=\mathrm{f}\left(\mathrm{a}+\frac{\left(\mathrm{b}-\mathrm{a}\right)\mathrm{i}}{\mathrm{n}}\right)\\ =\mathrm{f}\left(0+\frac{\left(3-0\right)\mathrm{i}}{\mathrm{n}}\right)\\ =\mathrm{f}\frac{3\mathrm{i}}{\mathrm{n}}\end{array}$

Now applying the formula,

  $\begin{array}{c}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{i}}^2}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ \mathrm{A}=\frac{-9}{{\mathrm{n}}^4}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{i}}^3}\\ =\frac{9}{{\mathrm{n}}^4}\cdot \frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{4}\\ =\frac{9}{{\mathrm{n}}^4}\frac{\left({\mathrm{n}}^2+2\mathrm{n}+1\right)}{{\mathrm{n}}^3}\end{array}$

As the sum of area of n rectangles are given as $\frac{9}{{\mathrm{n}}^4}\cdot \frac{{\mathrm{n}}^2+2\mathrm{n}+1}{{\mathrm{n}}^2}$ .

Substitute 4 for n in the expression $\frac{9}{{\mathrm{n}}^4}\cdot \frac{{\mathrm{n}}^2+2\mathrm{n}+1}{{\mathrm{n}}^2}$ .

  $\frac{9}{4}\cdot \frac{4^2+24+1}{4^2}=3.52$

Substitute 8 for n in the expression $\frac{9}{{\mathrm{n}}^4}\cdot \frac{{\mathrm{n}}^2+2\mathrm{n}+1}{{\mathrm{n}}^2}$ .

  $\frac{9}{4}\cdot \frac{8^2+2\cdot 8+1}{8^2}=2.85$

Substitute 20 for n in the expression $\frac{9}{{\mathrm{n}}^4}\cdot \frac{{\mathrm{n}}^2+2\mathrm{n}+1}{{\mathrm{n}}^2}$ .

  $\frac{9}{4}\cdot \frac{{20}^2+2\cdot 20+1}{{20}^2}=2.48$

Substitute 50 for n in the expression $\frac{9}{{\mathrm{n}}^4}\cdot \frac{{\mathrm{n}}^2+2\mathrm{n}+1}{{\mathrm{n}}^2}$ .

  $\frac{9}{4}\cdot \frac{{50}^2+2\cdot 50+1}{{50}^2}=2.34$

Complete the table for the area of given region.

$\mathrm{n}$	4	8	20	50
Approximate Area	$3.52$	$2.85$	$2.48$	$2.34$

Table 1

Therefore, the required table is shown in table 1.