Chapter 12.5, Problem 24E

To determine

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

Answer to Problem 24E

Approximating the area of region is $4.125\text{square}\text{units}$ .

Explanation of Solution

Given information:

The given equation is,

  $\text{f}\left(\text{x}\right)\text{=}\frac{\text{1}}{\text{2}}{\left(\text{x-1}\right)}^{\text{3}}$

And, rectangles are $4$

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,

  $\text{f}\left(\text{x}\right)\text{=}\frac{\text{1}}{\text{2}}{\left(\text{x-1}\right)}^{\text{3}}$

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

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

The four intervals are as follows.

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

The right end point of each interval is $\frac{1}{2}i$

  $i=2,3,4,5,6,$

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

  $\begin{array}{c}{\displaystyle \sum _{i=2}^{4}f\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)}={\displaystyle \sum _{i=2}^{4}f\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)}\\ ={\displaystyle \sum _{i=2}^4\left(\frac{{\left(\frac{1}{2}i-1\right)}^3}{2}\right)\left(\frac{1}{2}\right)}\\ =\frac{1}{32}\left({\displaystyle \sum _{i=2}^4{i}^3-6i^2+12i+8}\right)\\ =\frac{1}{32}\left({\displaystyle \sum _{i=2}^4{i}^3-6i^2+12i+8-{\displaystyle \sum _{i=0}^1{i}^3-6i^2+12i+8}}\right)\end{array}$

Applying, summation formulas and properties,

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

Now,

  $\begin{array}{c}{\displaystyle \sum _{\text{i=2}}^{\text{4}}\text{f}\left(\frac{\text{1}}{\text{2}}\text{i}\right)\left(\frac{\text{1}}{\text{2}}\right)}\text{=}\frac{\text{1}}{\text{32}}\left[{\left(\frac{\text{6×7}}{\text{2}}\right)}^{\text{2}}\text{-}\left(\frac{\text{6×7×13}}{\text{1}}\right)\text{+12}\left(\frac{\text{6×7}}{\text{2}}\right)\text{-15}\right]\\ \text{=}\frac{\text{1}}{\text{32}}\left[\frac{\text{1764}}{\text{4}}\text{-546+252-15}\right]\\ \text{=}\frac{\text{132}}{\text{32}}\\ \text{=4}\text{.125}\end{array}$

Finally, Approximating the area of region is $4.125\text{square}\text{units}$ .