Chapter 11.5, Problem 24E

To determine

To calculate: the given system of equation $f\left(x\right)=9-x^2$ .

Answer to Problem 24E

n	Approximate Area
4	21.66
8	19.75
20	18.69
50	18.27

Explanation of Solution

Given equations: $f\left(x\right)=9-x^2$

Calculation:

Let’s take the function f that is continuous and non negative on the interval (a,b). The area A of the region surrounded by the chart along x-axis is

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{a}g\left\{a+\frac{\left(b-a\right)i}{n}\right\}\left\{\frac{\left(b-a\right)}{n}\right\}}$

Let’s take the subsequent figure of the given equation:

  $f\left(x\right)=9-x^2$

The $x$ axis b/w $x=0$ and $x=3$ .

  

Find the size of the rectangle made with x and y axis:

  $\begin{array}{c}W=\frac{b-a}{n}\\ =\frac{3-0}{n}\\ =\frac{3}{n}\\ H=f\left[a+\frac{\left(b-a\right)i}{n}\right]\\ =f\left[0+\frac{\left(3-0\right)i}{n}\right]\\ =f{\left[\frac{3i}{n}\right]}^2\\ =9-{\left[\frac{3i}{n}\right]}^2\end{array}$

rough area of the n rectangles can be calculated as:

  $\begin{array}{c}A\approx {\displaystyle \sum _{i=1}^{n}f}\left[a+\frac{\left(b-a\right)i}{n}\right]\left[\frac{\left(b-a\right)}{n}\right]\\ ={\displaystyle \sum _{i=1}^n\left[9-{\left(\frac{3i}{n}\right)}^2\right]\left(\frac{3}{n}\right)}\\ ={\displaystyle \sum _{i=1}^n\left[\frac{27}{n}-\frac{27i^2}{n^3}\right]}\\ {\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\end{array}$

Apply the following formula:

  $\begin{array}{l}A=\frac{27}{n}n-\frac{27}{n^3}\times \frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ =\frac{9}{2}\left[\frac{4n^2+3n+1}{n^2}\right]\\ \end{array}$

Complete the given table using the above expression:

  $\begin{array}{c}A\approx \frac{9}{2}\left[\frac{4n^2+3n+1}{n^2}\right]\\ \\ \text{For} \text{n}=\text{4}\\ A=\frac{9}{2}\left[\frac{4{\left(4\right)}^2+3\left(4\right)+1}{{\left(4\right)}^2}\right]\\ =21.66\\ \\ \text{For} \text{n}=\text{8}\\ A=\frac{9}{2}\left[\frac{4{\left(8\right)}^2+3\left(8\right)+1}{{\left(8\right)}^{{}^2}}\right]\\ =\text{19}.\text{75}\\ \\ \text{For} \text{n}=\text{2}0\\ A=\frac{9}{2}\left[\frac{4{\left(20\right)}^2+3\left(20\right)+1}{{\left(20\right)}^{{}^2}}\right]\\ =18.69\\ \\ \text{For n }=\text{5}0\\ A=\frac{9}{2}\left[\frac{4{\left(50\right)}^2+3\left(50\right)+1}{{\left(50\right)}^{{}^2}}\right]\\ =18.27\end{array}$

Therefore, by using above values the table as follows:

n	Approximate Area
4	21.66
8	19.75
20	18.69
50	18.27