Chapter 11.5, Problem 27E

To determine

the table for the area of the region.

Answer to Problem 27E

The required table is shown in table 1.

Explanation of Solution

Given information:

The function is $\mathrm{f}\left(\mathrm{x}\right)=9-{\mathrm{x}}^2$ and the interval is $\left[0,2\right]$ and the sum of areas of n rectangles is $\frac{46{\mathrm{n}}^3-12{\mathrm{n}}^2-4\mathrm{n}}{3{\mathrm{n}}^3}$ .

As the sum of area of n rectangles are given as $\frac{46{\mathrm{n}}^3-12{\mathrm{n}}^2-4\mathrm{n}}{3{\mathrm{n}}^3}$ .

Substitute 4 for n in the expression $\frac{46{\mathrm{n}}^3-12{\mathrm{n}}^2-4\mathrm{n}}{3{\mathrm{n}}^3}$ .

  $\begin{array}{l}\mathrm{A}=\frac{46{\left(4\right)}^3-12{\left(4\right)}^2-4\left(4\right)}{3{\left(4\right)}^3}\\ =14.25\end{array}$

Substitute 8 for n in the expression $\frac{46{\mathrm{n}}^3-12{\mathrm{n}}^2-4\mathrm{n}}{3{\mathrm{n}}^3}$ .

  $\begin{array}{l}\mathrm{A}=\frac{46{\left(8\right)}^3-12{\left(8\right)}^2-4\left(8\right)}{3{\left(8\right)}^3}\\ =14.8125\end{array}$

Substitute 20 for n in the expression $\frac{46{\mathrm{n}}^3-12{\mathrm{n}}^2-4\mathrm{n}}{3{\mathrm{n}}^3}$ .

  $\begin{array}{l}\mathrm{A}=\frac{46{\left(20\right)}^3-12{\left(20\right)}^2-4}{3{\left(20\right)}^3}\\ =15.13\end{array}$

Substitute 50 for n in the expression $\frac{46{\mathrm{n}}^3-12{\mathrm{n}}^2-4\mathrm{n}}{3{\mathrm{n}}^3}$ .

  $\begin{array}{l}\mathrm{A}=\frac{46{\left(50\right)}^3-12{\left(50\right)}^2-4}{3{\left(50\right)}^3}\\ =15.2528\end{array}$

Substitute 100 for n in the expression $\frac{46{\mathrm{n}}^3-12{\mathrm{n}}^2-4\mathrm{n}}{3{\mathrm{n}}^3}$ .

  $\begin{array}{l}\mathrm{A}=\frac{46{\left(100\right)}^3-12{\left(100\right)}^2-4\left(4\right)}{3{\left(100\right)}^3}\\ =15.2932\end{array}$

Apply the limits as n approaches to infinity to find the exact area of the region.

  $\begin{array}{c}\mathrm{A}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{46{\mathrm{n}}^3-12{\mathrm{n}}^2-4\mathrm{n}}{3{\mathrm{n}}^3}\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{{\mathrm{n}}^3\left(46-\frac{12}{\mathrm{n}}-\frac{4}{{\mathrm{n}}^3}\right)}{3{\mathrm{n}}^3}\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{46-\frac{12}{\mathrm{n}}-\frac{4}{{\mathrm{n}}^2}}{3}\\ =\frac{46}{3}\text{square}\text{units}\end{array}$

The table for the area can be constructed as,

$\mathrm{n}$	4	8	20	50	100	$\mathrm{\infty }$
Approximated Area	14.25	14.8125	15.13	15.2528	15.2932	$\frac{46}{3}$

Table 1

Therefore, the required table is shown in table 1.