Chapter 11, Problem 97RE

To determine

To calculate:

The approximate area of region:

  $f\left(x\right)=\frac{1}{4}{x}^2$

Answer to Problem 97RE

The table of approximate area of region:

$n$	$4$	$8$	$20$	$50$
Approximate area	$30$	$25.5$	$22.96$	$21.98$

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{1}{4}{x}^2$

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

  

Calculation:

Consider the function:

  $f\left(x\right)=\frac{1}{4}{x}^2$

From the given graph values are as follows:

  $\left(a,b\right)=\left(0,4\right)$

The dimensions of three rectangles:

  $\begin{array}{l}Width=W=\frac{b-a}{n}\\ W=\frac{4-0}{n}\\ W=\frac{4}{n}\\ Height=f\left(a+\frac{\left(b-a\right)i}{n}\right)\end{array}$

Put the value:

  $\begin{array}{l}Height=H=f\left(0+\frac{\left(4-0\right)i}{n}\right)\\ H=f\left(\frac{4i}{n}\right)\\ H=\frac{1}{4}{\left(\frac{4i}{n}\right)}^2\\ H=\frac{4i^2}{n^2}\end{array}$

The area of the rectangles:

  $\begin{array}{l}A={\displaystyle \sum _{i=1}^{n}f\left(a+\frac{\left(b-a\right)i}{n}\right)\left(\frac{\left(b-a\right)}{n}\right)}\\ A=\underset{i\to 1}{\lim }\left(\frac{4i^2}{n^2}\right)\left(\frac{4}{n}\right)\\ A=\frac{64}{n^2}\underset{i\to 1}{\lim }{i}^2\end{array}$

Know that:

  $\underset{i\to 1}{\lim }{i}^2=\frac{n\left(n+1\right)\left(n+2\right)}{6}$

Put the formula:

  $\begin{array}{l}A=\frac{64}{n^2}\frac{n\left(n+1\right)\left(n+2\right)}{6}\\ A=\frac{32}{3}\left(\frac{2n^2+3n+1}{n^2}\right)\end{array}$

Substitute the value in this expression:

  $\begin{array}{l}n=4\\ A=\frac{32}{3}\left(\frac{2{\left(4\right)}^2+3\left(4\right)+1}{{\left(4\right)}^2}\right)\\ A=30\end{array}$

Substitute the value in this expression:

  $\begin{array}{l}n=8\\ A=\frac{32}{3}\left(\frac{2{\left(8\right)}^2+3\left(8\right)+1}{{\left(8\right)}^2}\right)\\ A=25.5\end{array}$

Substitute the value in this expression:

  $\begin{array}{l}n=20\\ A=\frac{32}{3}\left(\frac{2{\left(20\right)}^2+3\left(20\right)+1}{{\left(20\right)}^2}\right)\\ A=22.96\end{array}$

Substitute the value in this expression:

  $\begin{array}{l}n=50\\ A=\frac{32}{3}\left(\frac{2{\left(50\right)}^2+3\left(50\right)+1}{{\left(50\right)}^2}\right)\\ A=21.98\end{array}$

Hence,

$n$	$4$	$8$	$20$	$50$
Approximate area	$30$	$25.5$	$22.96$	$21.98$