Chapter 12.5, Problem 35E

To determine

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

Answer to Problem 35E

n	Approximate Area
4	19
8	18.5
20	18.2
50	18.08
100	18.04
8	18

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

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

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

  $f\left(x\right)=9-x^2$

The $x$ axis b/w $x=-1$ and $x=3$ .

  

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

  $\begin{array}{c}W=\frac{b-a}{n}\\ =\frac{3-\left(-1\right)}{n}\\ =\frac{4}{n}\\ H=f\left[a+\frac{\left(b-a\right)i}{n}\right]\\ =f\left[-1+\frac{\left\{3-\left(-1\right)\right\}i}{n}\right]\\ =\left(\frac{1}{2}\left(-1+\frac{4i}{n}\right)+4\right)\end{array}$

rough area of the n rectangles can be calculated as:

  $\begin{array}{c}A\approx {\displaystyle \sum _{i=1}^{n}f}\left[a+\frac{\left(b-a\right)i}{n}\right]\left[\frac{b-a}{n}\right]\\ ={\displaystyle \sum _{i=1}^n\left(\frac{1}{2}\left(-1+\frac{4i}{n}\right)+4\right)\left[\frac{4}{n}\right]}\\ ={\displaystyle \sum _{i=1}^n\left[\frac{8i}{n^2}+\frac{14}{n}\right]}\\ \left\{{\displaystyle \sum _{i=1}^n\left(i\right)}=\frac{n\left(n+1\right)}{2}\right\}\\ A=\frac{8}{n^2}\frac{n\left(n+1\right)}{2}+14\\ =\frac{18n+4}{n}\end{array}$

Apply the following formula:

  $A\left(n\right)=\frac{18n+4}{n}$

Complete the given table using the above expression:

  $\begin{array}{c}A\left(n\right)=\frac{18n+4}{n}\\ \\ \text{for}n=4\\ A=\frac{18\left(4\right)+4}{\left(4\right)}\\ =19\\ \\ \text{for}n=8\\ A=\frac{18\left(8\right)+4}{\left(8\right)}\\ =18.5\\ \\ \text{for}n=20\\ A=\frac{18\left(20\right)+4}{20}\\ =18.2\\ \\ \text{for}n=50\\ A=\frac{18\left(50\right)+4}{50}\\ =18.08\\ \\ \text{for}n=100\\ A=\frac{18\left(100\right)+4}{100}\\ =18.04\\ \\ \text{for}n\to \infty \\ A=\underset{n\to \infty }{\lim }\frac{18n+4}{n}\\ =18\end{array}$

Therefore, by using above values the table as follows:

n	Approximate Area
4	19
8	18.5
20	18.2
50	18.08
100	18.04
8	18