Chapter 11, Problem 98RE

To determine

To complete: The given table to show the approximate area of the given region.

Answer to Problem 98RE

  $\begin{array}{lllll}n\hfill & 4\hfill & 8\hfill & 20\hfill & 50\hfill \\ \mathrm{Approximate}Area\hfill & 10\hfill & 10.5\hfill & 10.64\hfill & 10.6624\hfill \end{array}$

Explanation of Solution

Given information:

Given function is $f(x)=4x-x^2$ and given incomplete table is:

  $\begin{array}{lllll}n\hfill & 4\hfill & 8\hfill & 20\hfill & 50\hfill \\ \mathrm{Approximate}Area\hfill & \hfill & \hfill & \hfill & \hfill \end{array}$

Calculation:

Let us find the dimensions of the rectangles:

Width $=\frac{b-a}{n}=\frac{4-0}{n}=\frac{4}{n}$

Height $=f(a+\frac{(b-a)i}{n})=f(0+\frac{4i}{n})=f(\frac{4i}{n})=4(\frac{4i}{n})-{(\frac{4i}{n})}^2=\frac{16i}{n}-\frac{16i^2}{n^2}=\frac{16}{n}(i-\frac{i^2}{n})$

Approximating the area as the sum of the areas of $n$ rectangles,

  $A={\displaystyle \sum _{i=1}^{n}f(a+\frac{(b-a)i}{n})(\frac{b-a}{n})}$

  $\begin{array}{l}={\displaystyle \sum _{i=1}^n(\frac{16}{n}(i-\frac{i^2}{n}))(\frac{4}{n})}\\ =\frac{64}{n^2}{\displaystyle \sum _{i=1}^n(i-\frac{i^2}{n}})\\ =\frac{64}{n^2}{\displaystyle \sum _{i=1}^{n}i-\frac{64}{n^3}{\displaystyle \sum _{i=1}^n{i}^2}}\\ =\frac{64}{n^2}(\frac{n(n+1)}{2})-\frac{64}{n^3}(\frac{n(n+1)(2n+1)}{6})\\ =\frac{32(n+1)}{n}-\frac{32(n+1)(2n+1)}{3n^2}\end{array}$

  $n=4,8,20,50;$ the complete table is:

  $\begin{array}{lllll}n\hfill & 4\hfill & 8\hfill & 20\hfill & 50\hfill \\ \mathrm{Approximate}Area\hfill & 10\hfill & 10.5\hfill & 10.64\hfill & 10.6624\hfill \end{array}$