Chapter 12.5, Problem 29E

To determine

To complete: The table to show the approximate are of the region using the indicated numbers $n$ of rectangles of equal width.

Answer to Problem 29E

Completed table is,

n	4	8	20	50
Approximate area	12.516	11.848	11.481	11.341

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{1}{9}{x}^3+3$

  

Calculation:

From the given graph,

  $\begin{array}{l}a=0\\ b=3\end{array}$

So each rectangle has a width of $\frac{b-a}{n}=\frac{3}{n}$ ,

  $f\left(a+\frac{\left(b-a\right)}{n}i\right)=f\left(\frac{3i}{n}\right)\to \text{by using the right endpoints}$

  $\begin{array}{l}A\approx {\displaystyle \sum _{i=1}^{n}f\left(\frac{3i}{n}\right)\frac{3}{n}}\\ A\approx {\displaystyle \sum _{i=1}^n\left[\frac{1}{9}{\left(\frac{3i}{n}\right)}^3+3\right]}\frac{3}{n}\\ A\approx {\displaystyle \sum _{i=1}^n\left[\frac{9i^3}{n^4}+\frac{9}{n}\right]}\\ A\approx \frac{9}{n^4}{\displaystyle \sum _{i=1}^n{i}^3}+\frac{9}{n}{\displaystyle \sum _{i=1}^{n}1}\\ A\approx \frac{9}{n^4}\left(\frac{n^2{\left(n+1\right)}^2}{4}\right)+\frac{9}{n}\left(n\right)\to \text{By using}{\displaystyle \sum _{i=1}^n{i}^3}=\frac{n^2{\left(n+1\right)}^2}{4}\text{and}{\displaystyle \sum _{i=1}^{n}c}=cn\\ A\approx \frac{9{\left(n+1\right)}^2}{4n^2}+9\end{array}$

By using $A\approx \frac{9{\left(n+1\right)}^2}{4n^2}+9$ the completed table is,

n	4	8	20	50
Approximate area	12.516	11.848	11.481	11.341