Chapter 12, Problem 97RE

To determine

Calculate the table with Area of a Region.

Answer to Problem 97RE

  $\begin{array}{ccccc}n& 4& 8& 20& 50\\ \begin{array}{l}Approximate\\ Area\end{array}& 30& 25.5& 22.96& 21.98\end{array}$

Explanation of Solution

Given information:

Complete the table to show the approximate area of the region using the indicated numbers $n$ of rectangles of equal width.

  $\begin{array}{ccccc}n& 4& 8& 20& 50\\ \begin{array}{l}Approximate\\ Area\end{array}& & & & \end{array}$

  $f\left(x\right)=\frac{1}{4}{x}^2$

  

Calculation:

Consider the region.

  $f\left(x\right)=\frac{1}{4}{x}^2$

  

From the given region the value of $a$ is $0$ and $b$ is $4$ . Now the dimensions of rectangles are,

  $\begin{array}{l}width=\frac{b-a}{n}\\ =\frac{4-0}{n}\\ =\frac{4}{n}\\ height=f\left(a+\frac{\left(b-a\right)i}{n}\right)\end{array}$

On substituting the values of $a$ and $b$ ,

  $\begin{array}{l}height=f\left(0+\frac{\left(4-0\right)i}{n}\right)\\ =f\left(\frac{4i}{n}\right)\\ =\frac{1}{4}{\left(\frac{4i}{n}\right)}^2\\ =\frac{16i^2}{4n^2}\\ =\frac{4i^2}{n^2}\end{array}$

Now calculate the area of $n$ rectangles.

  $\begin{array}{l}A={\displaystyle \sum _{i=1}^{n}f\left(a+\frac{\left(b-a\right)i}{n}\right)}\left(\frac{\left(b-a\right)}{n}\right)\\ ={\displaystyle \sum _{i=1}^n\left(\frac{4i^2}{n^2}\right)}\left(\frac{4}{n}\right)\\ =\frac{64}{n^3}{\displaystyle \sum _{i=1}^n{i}^2}\end{array}$

Apply property of submission,

  ${\displaystyle \sum _{i=1}^{n}i=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}and{\displaystyle \sum _{i=1}^{n}c=nc}$

  $\begin{array}{l}A=\frac{64}{i^3}\times \frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ =\frac{32}{3}\left(\frac{2n^2+3n+1}{n^2}\right)\end{array}$

Now complete the table by substituting the $n=4$ according to the table in expression

  $\frac{32}{3}\left(\frac{2n^2+3n+1}{n^2}\right)$

  $\begin{array}{l}=\frac{32}{3}\left(\frac{2{\left(4\right)}^2+3\left(4\right)+1}{{\left(4\right)}^2}\right)\\ =30\end{array}$

Now put $n=8$ ,

  $\begin{array}{l}=\frac{32}{3}\left(\frac{2{\left(8\right)}^2+3\left(8\right)+1}{{\left(8\right)}^2}\right)\\ =25.5\end{array}$

Now put $n=20$ ,

  $\begin{array}{l}A=\frac{32}{3}\left(\frac{2{\left(20\right)}^2+3\left(20\right)+1}{{\left(20\right)}^2}\right)\\ =22.96\end{array}$

Now put $n=50$ ,

  $\begin{array}{l}A=\frac{32}{3}\left(\frac{2{\left(50\right)}^2+3\left(50\right)+1}{{\left(50\right)}^2}\right)\\ =21.98\end{array}$

Hence the table completion is,

  $\begin{array}{ccccc}n& 4& 8& 20& 50\\ \begin{array}{l}Approximate\\ Area\end{array}& 30& 25.5& 22.96& 21.98\end{array}$