Chapter 11.5, Problem 32E

To determine

To calculate:

The exact area of given function.

Answer to Problem 32E

The exact area of region is $4$ and the approximate area of the region bounded by the graph $f$ and $x-$ axis over the interval is,

$n$	$4$	$8$	$20$	$50$	$100$	$\infty$
Area	$5$	$4.5$	$4.2$	$4.08$	$4.04$	$4$

Explanation of Solution

Given Information:

Function: $f\left(x\right)=\frac{1}{2}x+1$

Interval: $\left[-2,2\right]$

Sum of Areas of $n$ Rectangles: $\frac{4n^2+4n}{n^2}$

$n$	$4$	$8$	$20$	$50$	$100$	$\infty$
Area

Calculation:

To approximate the area of the region for each finite value of $n$

Consider the following function over the interval $\left[-2,2\right]$

  $f\left(x\right)=\frac{1}{2}x+1$

Since the interval is $\left[-2,2\right]$

Then, the vertical lines are $x=-2$ and $x=2$

Then, the width of each rectangle

  $\begin{array}{l}\frac{b-a}{n}=\frac{2-\left(-2\right)}{n}\\ =\frac{4}{n}\end{array}$

And height of each rectangle is

  $\begin{array}{l}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(-2+\frac{\left(2+2\right)i}{n}\right)\\ =f\left(-2+\frac{4i}{n}\right)\\ =\frac{1}{2}\left(-2+\frac{4i}{n}\right)+1\\ =-1+\frac{2i}{n}+1\\ =\frac{2i}{n}\end{array}$

Now, the approximate the area as the sum of the area of $n$ rectangles.

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

That is $\frac{4\left(n+1\right)}{n}$

Then, value of the area for $n=4$

  $\begin{array}{l}A=\frac{4\left(4+1\right)}{4}\\ =5\end{array}$

And the value of the area for $n=8$

  $\begin{array}{l}A=\frac{4\left(8+1\right)}{8}\\ =\frac{9}{2}\\ =4.5\end{array}$

Then, the value of the area for $n=20$

  $\begin{array}{l}A=\frac{4\left(20+1\right)}{20}\\ =4.2\end{array}$

The value of the area for $n=50$

  $\begin{array}{l}A=\frac{4\left(50+1\right)}{50}\\ =4.08\end{array}$

The value of the area for $n=100$

  $\begin{array}{l}A=\frac{4\left(100+1\right)}{100}\\ =4.04\end{array}$

Since,

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

Then,

  $\begin{array}{l}A=\underset{x\to \infty }{\lim }\left[\frac{4\left(n+1\right)}{n}\right]\\ =\underset{x\to \infty }{\lim }\left[\frac{4\left(1+\frac{1}{n}\right)}{n}\right]\\ =\underset{x\to \infty }{\lim }\left[\frac{4\left(1+\frac{1}{n}\right)}{1}\right]\\ =4\end{array}$

Hence the exact area of region is $4$ .

Therefore, the approximate area of the region bounded by the graph $f$ and $x-$ axis over the interval,

$n$	$4$	$8$	$20$	$50$	$100$	$\infty$
Area	$5$	$4.5$	$4.2$	$4.08$	$4.04$	$4$