Chapter 11, Problem 106CR

To determine

To approximate: the area of the region using n = 4, n = 8, n = 20 and n =50 rectangular of equal width.

Answer to Problem 106CR

Explanation of Solution

Given information: Given function is $\mathrm{f}(\mathrm{x})=4\mathrm{x}-{\mathrm{x}}^2.$

  

Calculation:

  $\begin{array}{c}\mathrm{f}(\mathrm{x})=\frac{1}{4}{\mathrm{x}}^2,\mathrm{b}-\mathrm{a}=4-0=4\\ \mathrm{A}\approx {\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{f}\left(\frac{4\mathrm{i}}{\mathrm{n}}\right)}\left(\frac{4}{\mathrm{n}}\right)\\ \approx {\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\left[4\left(\frac{4\mathrm{i}}{\mathrm{n}}\right)-{\left(\frac{4\mathrm{i}}{\mathrm{n}}\right)}^2\right]}\left(\frac{4}{\mathrm{n}}\right)\\ ={\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\left[\frac{16\mathrm{i}}{\mathrm{n}}-\frac{16{\mathrm{i}}^2}{{\mathrm{n}}^2}\right]}\left(\frac{4}{\mathrm{n}}\right)\\ =\frac{64}{{\mathrm{n}}^2}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{i}}-\frac{64}{{\mathrm{n}}^3}{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{i}}^2}\\ =\frac{64}{{\mathrm{n}}^2}\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)-\frac{64}{{\mathrm{n}}^3}\left(\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\right)\\ \mathrm{A}\approx \frac{96{\mathrm{n}}^2+96\mathrm{n}-64{\mathrm{n}}^2-96\mathrm{n}-32}{3{\mathrm{n}}^2}\\ \mathrm{A}\approx \frac{32{\mathrm{n}}^2-32}{3{\mathrm{n}}^2}\\ \mathrm{A}\approx \frac{32}{3}-\frac{32}{3{\mathrm{n}}^2}.\end{array}$

The below table shows the approximate area of the region using n = 4, n = 8, n = 20 and n =50 rectangular of equal width.

n	4	8	20	50
Approximate area	10	10.5	10.64	10.662