Chapter 12.5, Problem 30E

To determine

To calculate: the given system of equation $f\left(x\right)=3-\frac{x^3}{4}$ .

Answer to Problem 30E

n	Approximate Area
4	5.61
8	6.11
20	6.39
50	6.49

Explanation of Solution

Given equations: $f\left(x\right)=3-\frac{x^3}{4}$

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:

  $f\left(x\right)=3-\frac{x^3}{4}$

The $x$ axis b/w $x=-1$ and $x=2$ .

  

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

  $\begin{array}{c}W=\frac{b-a}{n}\\ =\frac{2-\left(-1\right)}{n}\\ =\frac{3}{n}\\ H=f\left[a+\frac{\left(b-a\right)i}{n}\right]\\ =f\left[-1+\frac{\left\{2-\left(-1\right)\right\}i}{n}\right]\\ =3-\frac{{\left[\frac{3i^3}{n^3}\right]}^3}{4}\\ =\left[-\frac{27i^3}{4n^3}+\frac{27i^2}{4n^2}-\frac{9i}{4n}+\frac{11}{4}\right]\end{array}$

rough area of the n rectangles can be calculated as:

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

  $={\displaystyle \sum _{i=1}^n\left[-\frac{27i^3}{4n^3}+\frac{27i^2}{4n^2}-\frac{9i}{4n}+\frac{11}{4}\right]\left[\frac{3}{n}\right]}$

  $={\displaystyle \sum _{i=1}^n\left[-\frac{81i^3}{4n^4}+\frac{81i^2}{4n^3}-\frac{27i}{4n^2}+\frac{33}{4n}\right]}$

  $=-\frac{81}{4n^4}{\displaystyle \sum _{i=1}^n\left(i^3\right)}+\frac{81}{4n^3}{\displaystyle \sum _{i=1}^n\left(i^2\right)-\frac{27}{4n^2}{\displaystyle \sum _{i=1}^n\left(i\right)}+{\displaystyle \sum _{i=1}^n\left(\frac{33}{4n}\right)}}$

  $\left\{\begin{array}{l}{\displaystyle \sum _{i=1}^n\left(i^3\right)}=\frac{n^2{\left(n+1\right)}^2}{4}\\ {\displaystyle \sum _{i=1}^n\left(i^2\right)}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ {\displaystyle \sum _{i=1}^n\left(i\right)}=\frac{n\left(n+1\right)}{2}\\ {\displaystyle \sum _{i=1}^n\left(c\right)}=nc\end{array}\right\}$

  $A=-\frac{81}{4n^4}\frac{n^2{\left(n+1\right)}^2}{4}+\frac{81}{4n^3}\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{27}{4n^2}\left(\frac{n\left(n+1\right)}{2}\right)+\frac{33}{4}$

  $=\frac{33}{4}-\frac{27}{16}\left(\frac{n^2+2n+1}{n^2}\right)$

Apply the following formula:

  $A\left(n\right)=\frac{33}{4}-\frac{27}{16}\left(\frac{n^2+2n+1}{n^2}\right)$

Complete the given table using the above expression:

  $\begin{array}{c}A\left(n\right)=\frac{33}{4}-\frac{27}{16}\left(\frac{n^2+2n+1}{n^2}\right)\\ \\ \text{for}n=4\\ A=\frac{33}{4}-\frac{27}{16}\left(\frac{{\left(4\right)}^2+2\left(4\right)+1}{{\left(4\right)}^2}\right)\\ =5.61\\ \\ \text{for}n=8\\ A=\frac{33}{4}-\frac{27}{16}\left(\frac{{\left(8\right)}^2+2\left(8\right)+1}{{\left(8\right)}^2}\right)\\ =6.11\\ \\ \text{for}n=20\\ A=\frac{33}{4}-\frac{27}{16}\left(\frac{{\left(20\right)}^2+2\left(20\right)+1}{{\left(20\right)}^2}\right)\\ =6.39\\ \\ \text{for}n=50\\ A=\frac{33}{4}-\frac{27}{16}\left(\frac{{\left(50\right)}^2+2\left(50\right)+1}{{\left(50\right)}^2}\right)\\ =6.49\end{array}$

Therefore, by using above values the table as follows:

n	Approximate Area
4	5.61
8	6.11
20	6.39
50	6.49