Chapter 12.5, Problem 29E

To determine

To find: Approximating the area of region using the indicated numbers $n$ of rectangles of equal width.

Answer to Problem 29E

Approximating the area of region with the number n of rectangles is

n	4	8	20	50
Approximate area	5.15625	5.5078125	5.78625	5.9118

Explanation of Solution

Given information:

The given function is,

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

And, rectangles are $n$

Concept used:

The area of a region;

Let $f$ be continuous and nonnegative on the interval $\left[a,b\right]$ The area A of the region bounded by the graph of $f$ , the $\text{x-axis}$ ,and the vertical lines $x=a$ and $x=b$ is given by

  $A=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\underset{\text{height}}{\underbrace{\left(a+\frac{\left(b-a\right)i}{n}\right)}}}\underset{\text{width}}{\underbrace{\left(\frac{b-a}{n}\right)}}$

The given function is,

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

The area of the region bounded by the graph of $f\left(x\right)=\frac{1}{9}{x}^3$ and the x-axis between $x=0$ and $x=3$

The dimensions of the rectangles;

  $\begin{array}{c}width;\frac{b-a}{n}=\frac{3-0}{n}\\ =\frac{3}{n}\end{array}$

And,

  $\begin{array}{c}\text{Height}\text{;}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(0+\frac{\left(3-0\right)i}{n}\right)\\ =f\left(\frac{3i}{n}\right)\end{array}$

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

  $\begin{array}{c}f\left(x\right)=\frac{1}{9}{\left(\frac{3i}{n}\right)}^3\\ =\frac{1}{9}\times \frac{27i^3}{n^3}\\ =\frac{3i^3}{n^3}\end{array}$

Approximate the areas as the sum of the areas of $n$ rectangles.

  $\begin{array}{c}\text{A=}{\displaystyle \sum _{\text{i=1}}^{\text{n}}\text{f}\left(\text{a+}\frac{\left(\text{b-a}\right)\text{i}}{\text{n}}\right)\left(\frac{\text{b-a}}{\text{n}}\right)}\\ ={\displaystyle \sum _{i=1}^n\left(\frac{3i^3}{n^3}\right)\left(\frac{3}{n}\right)}\\ ={\displaystyle \sum _{i=1}^n\frac{9i^3}{n^4}}\end{array}$

Applying Summation formulas and properties.

  ${\displaystyle \sum _{i=1}^n{i}^3=\frac{n^2{\left(n+1\right)}^2}{4}}$

Now,

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

We can obtain a more accurate approximation of the area of the region by increasing the number of rectangles.

  $A\left(n\right)=\frac{9}{4}\left(\frac{n^2+2n+1}{n^2}\right)$

Take,

  $n=4$

  $\begin{array}{c}\text{A=}\frac{9}{4}\left(\frac{{\left(4\right)}^2+2\times 4+1}{4^2}\right)\\ \text{=}\frac{\text{9}}{4}\left(\frac{16+8+1}{16}\right)\\ \text{=3}\text{.52}\end{array}$

In this way, we take $n=8,20,50\mathrm{....}$

Now, values of $A=2.85,2.48,2.34\mathrm{....}$

Finally,to approximating the area of region by using values in the following table.

n	4	8	20	50
Approximate area	3.52	2.85	2.48	2.34