Chapter 12.5, Problem 23E

To determine

To find: Approximating the area of region using the indicated number of rectangles of equal width.

Answer to Problem 23E

Approximating the area of region is $1.2656\text{square}\text{units}$ .

Explanation of Solution

Given information:

The given equation is,

  $f\left(x\right)=\frac{1}{4}{x}^3$

And, rectangles are $8$

Concept used:

The area of a region;

Let $f$ be continuous and non negative 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,

  $f\left(x\right)=\frac{1}{4}{x}^3$

The length of the interval along the $\text{x-axis}$ is $2$ and the reare eight rectangles,the width of each rectangle is $\frac{2}{8}=\frac{1}{4}$

To obtain the height of each rectangle, evaluate $f$ at the right end point of each interval.

The eight intervals are as follows.

  $\left[0,\frac{1}{4}\right],\left[\frac{1}{4},\frac{1}{2}\right],\left[\frac{1}{2},\frac{3}{4}\right],\left[\frac{3}{4},1\right],\left[1,\frac{5}{4}\right],\left[\frac{5}{4},\frac{3}{2}\right],\left[\frac{3}{2},\frac{7}{4}\right]$ and $\left[\frac{\text{7}}{\text{4}}\text{,2}\right]$

The right end point of each intervalis $\frac{1}{4}i$

  $i=1,2,3,4,5,6,7,8$

Hence,the sum of the areas of the eight rectangles is

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{8}f\left(\frac{1}{4}i\right)\left(\frac{1}{4}\right)}={\displaystyle \sum _{i=1}^{8}f\left(\frac{1}{4}i\right)\left(\frac{1}{4}\right)}\\ ={\displaystyle \sum _{i=1}^8\left(\frac{{\left(\frac{1}{4}i\right)}^3}{4}\right)\left(\frac{1}{4}\right)}\\ =\frac{1}{1024}{\displaystyle \sum _{i=1}^8{i}^3}\end{array}$

Applying, sum mation formulas and properties,

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

Now,

  $\begin{array}{c}{\displaystyle \sum _{\text{i=1}}^{\text{8}}\text{f}\left(\frac{\text{1}}{\text{4}}\text{i}\right)\left(\frac{\text{1}}{\text{4}}\right)}\text{=}\frac{{\text{8}}^{\text{2}}{\left(\text{8+1}\right)}^{\text{2}}}{\text{1024×4}}\\ \text{=}\frac{\text{64×81}}{\text{1024×4}}\\ \text{=}\frac{\text{5184}}{\text{4096}}\\ \text{=1}\text{.2656}\end{array}$

Thus,the approximating the area of region is $1.2656\text{square}\text{units}$ .