Chapter 11.5, Problem 22E

To determine

To show the approximate area of the region.

Answer to Problem 22E

The required areas are shown below.

Explanation of Solution

Given information:

The function is $\mathrm{f}\left(\mathrm{x}\right)=-\frac{1}{3}\mathrm{x}+4$ .

  

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,

  $\begin{array}{c}\mathrm{w}\mathrm{i}\mathrm{d}\mathrm{t}\mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\\ =\frac{12-0}{\mathrm{n}}\\ =\frac{12}{\mathrm{n}}\end{array}$

And,

  $\begin{array}{c}\mathrm{H}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}=\mathrm{f}\left(\mathrm{a}+\frac{\left(\mathrm{b}-\mathrm{a}\right)\mathrm{i}}{\mathrm{n}}\right)\\ =\mathrm{f}\left(0+\frac{\left(12-0\right)\mathrm{i}}{\mathrm{n}}\right)\\ =\mathrm{f}\frac{12\mathrm{i}}{\mathrm{n}}\\ =\frac{-4\mathrm{i}}{\mathrm{n}}+4\end{array}$

Now applying the formula,

  $\begin{array}{c}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{i}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ \mathrm{A}=\frac{-24}{\mathrm{n}}\left(\mathrm{n}+1\right)+48\\ =\frac{24\left(\mathrm{n}-1\right)}{\mathrm{n}}\end{array}$

As the sum of area of n rectangles are given as $\frac{24\left(\mathrm{n}-1\right)}{\mathrm{n}}$ .

Substitute 4 for n in the expression $\frac{24\left(\mathrm{n}-1\right)}{\mathrm{n}}$ .

  $\frac{24\left(4-1\right)}{4}=18$ .

Substitute 8 for n in the expression $\frac{24\left(\mathrm{n}-1\right)}{\mathrm{n}}$ .

  $\frac{24\left(8-1\right)}{8}=21$ .

Substitute 20 for n in the expression $\frac{24\left(\mathrm{n}-1\right)}{\mathrm{n}}$ .

  $\frac{24\left(20-1\right)}{20}=22.8$ .

Substitute 50 for n in the expression $\frac{24\left(\mathrm{n}-1\right)}{\mathrm{n}}$ .

  $\frac{24\left(50-1\right)}{50}=23.52$ .

The table for the area can be constructed as,

Therefore, the required areas are shown above.