Chapter 11.5, Problem 31E

To determine

To calculate:

The exact area of given function.

Answer to Problem 31E

The exact area of region is $18$ 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	$19$	$38$	$18.5$	$18.2$	$18.08$	$18.04$

Explanation of Solution

Given Information:

Function: $f\left(x\right)=\frac{1}{2}x+4$

Interval: $\left[-1,3\right]$

Sum of Areas of $n$ Rectangles: $\frac{18n^2+4n}{n^2}$

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

Calculation:

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

Consider the following function over the interval $\left[-1,3\right]$

  $f\left(x\right)=\frac{1}{2}x+4$

Since the interval is $\left[-1,3\right]$

Then, the vertical lines are $x=-1$ and $x=3$

Then, the width of each rectangle

  $\begin{array}{l}\frac{b-a}{n}=\frac{3-\left(-1\right)}{n}\\ =\frac{4}{n}\end{array}$

And height of each rectangle is

  $\begin{array}{l}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(-1+\frac{4i}{n}\right)\\ =f\left(-1+\frac{4i}{n}\right)\\ =\frac{1}{2}\left(-1+\frac{4i}{n}\right)+4\\ =-\frac{1}{2}+\frac{2i}{n}+4\\ =\frac{7}{2}+\frac{2i}{n}\end{array}$

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

  $\begin{array}{l}A={\displaystyle \sum _{i=1}^{n}f\left(a+\frac{\left(b-a\right)i}{n}\right)}\left(\frac{b-a}{n}\right)\\ ={\displaystyle \sum _{i=1}^n\left(\frac{7}{2}+\frac{2i}{n}\right)}\left(\frac{4}{n}\right)\\ =\left(\frac{4}{n}\right)\left({\displaystyle \sum _{i=1}^n\frac{7}{2}+{\displaystyle \sum _{i=1}^n\frac{2i}{n}}}\right)\\ =\left(\frac{4}{n}\right)\left(\frac{7}{2}+\frac{2n\left(n+1\right)}{2n}\right)\\ =14+\frac{4\left(n+1\right)}{n}\end{array}$

That is,

  $A=14+\frac{4\left(n+1\right)}{n}$

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

  $\begin{array}{l}A=14+\frac{4\left(4+1\right)}{4}\\ =19\end{array}$

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

  $\begin{array}{l}A=14+\frac{4\left(8+1\right)}{8}\\ =18.5\end{array}$

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

  $\begin{array}{l}A=14+\frac{4\left(20+1\right)}{20}\\ =18.2\end{array}$

The value of the area for $n=50$

  $\begin{array}{l}A=14+\frac{4\left(50+1\right)}{50}\\ =18.08\end{array}$

The value of the area for $n=100$

  $\begin{array}{l}A=14+\frac{4\left(100+1\right)}{100}\\ =18.04\end{array}$

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,

  $\begin{array}{l}A=\underset{x\to \infty }{\lim }\left[14+\frac{4\left(n+1\right)}{n}\right]\\ =\underset{x\to \infty }{\lim }14+\underset{x\to \infty }{\lim }\frac{4\left(n+1\right)}{n}\\ =14+4\\ =18\end{array}$

Hence the exact area of region is $18$ .

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	$19$	$38$	$18.5$	$18.2$	$18.08$	$18.04$