Chapter 11.5, Problem 25E

To determine

To calculate:

The table of approximate area value of the following region:

  $f\left(x\right)=\frac{1}{9}{x}^3$

Answer to Problem 25E

The table of the approximate area of the region is

$n$	$4$	$8$	$20$	$50$
Approximate area	$3.52$	$2.85$	$2.48$	$2.34$

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{1}{9}{x}^3$

  

$n$	$4$	$8$	$20$	$50$
Approximate area	-	-	-	-

No of rectangles: $3$ rectangles

Width of the rectangle:

  $\begin{array}{l}W=\frac{b-a}{n}\\ W=\frac{3-0}{n}\\ W=\frac{3}{n}\end{array}$

Height of the rectangle:

  $\begin{array}{l}H=f\left(a+\frac{\left(b-a\right)i}{n}\right)\\ H=f\left(a+\frac{\left(3-0\right)i}{n}\right)\\ H=f\left(a+\frac{\left(b-a\right)i}{n}\right)\\ H=f\left(\frac{3i}{n}\right)\\ H=\frac{3i^3}{n}\end{array}$

Formula Used:

By using this formula:

  ${\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(n+2\right)}{6}$

Calculation:

Consider this equation:

  $f\left(x\right)=\frac{1}{9}{x}^3$

The area A of the region bounded by the graph along x axis is

  $\begin{array}{l}A=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(a+\frac{\left(b-a\right)i}{n}\right)\left(\frac{b-a}{n}\right)}\\ A={\displaystyle \sum _{i=1}^n\left(\frac{3i^3}{n^3}\right)\left(\frac{3}{n}\right)}\end{array}$

By using this formula:

  ${\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(n+2\right)}{6}$

Comparing with this equation:

  $\begin{array}{l}A=\frac{9}{n^4}{\displaystyle \sum _{i=1}^{n}f\left(i^2\right)}=\frac{9}{n^4}{\displaystyle \sum _{i=1}^n\frac{n^2{\left(n+1\right)}^2}{4}}\\ A=\frac{9}{4}\left(\frac{n^2+2n+1}{n^2}\right)\end{array}$

Put the value of $n=4$ ,

  $\begin{array}{l}n=4\\ A=\frac{9}{4}\left(\frac{{\left(4\right)}^2+2\times 4+1}{{\left(4\right)}^2}\right)\\ A=3.52\end{array}$

Put the value of $n=8$ ,

  $\begin{array}{l}n=8\\ A=\frac{9}{4}\left(\frac{{\left(8\right)}^2+2\times 8+1}{{\left(8\right)}^2}\right)\\ A=2.85\end{array}$

Put the value of $n=20$ ,

  $\begin{array}{l}n=20\\ A=\frac{9}{4}\left(\frac{{\left(20\right)}^2+2\times 20+1}{{20}^2}\right)\\ A=2.48\end{array}$

Put the value of $n=50$ ,

  $\begin{array}{l}n=50\\ A=\frac{9}{4}\left(\frac{{\left(50\right)}^2+2\times 50+1}{{50}^2}\right)\\ A=2.34\end{array}$

Using all this value the table as follows:

$n$	$4$	$8$	$20$	$50$
Approximate area	$3.52$	$2.85$	$2.48$	$2.34$