Chapter 12.5, Problem 32E

To determine

To find: Approximating the area of region bounded by the graph of $f$ and the x-axis over the specified interval using the indicated numbers $n$ of rectangles of equal width and the exact area as $n\to \infty$ .

Answer to Problem 32E

Approximating the area of region with the number n of rectangles is

n	$4$	$8$	$20$	$50$	$100$	$\infty$
Approximate area	$44$	$46$	$47.2$	$47.68$	$47.84$	$48$

Explanation of Solution

Given information:

The given function is,

  $f\left(x\right)=20-2x$

Given intervals $\left[2,6\right]$

And, rectangles are $n$

Concept used:

The area of a region;

Let $f$ be continuous and non negative on the interval $\left[a,b\right]$ The area A of the region bounded by the graph of $f$ ,the $\text{x-axis}$ ,and the vertical lines $x=a$ and $x=b$ isgivenby

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

The given functionis,

  $f\left(x\right)=20-2x$

The area of the region bounded by the graph of $f\left(x\right)=20-2x$ and the x-axis between $x=2$ and $x=6$

The dimensions of the rectangles;

  $\begin{array}{c}width;\frac{b-a}{n}=\frac{6-2}{n}\\ =\frac{4}{n}\end{array}$

And,

  $\begin{array}{c}\text{Height}\text{;}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(2+\frac{\left(6-2\right)i}{n}\right)\\ =f\left(\frac{4i}{n}+2\right)\end{array}$

Given function; $f\left(x\right)=20-2x$

  $\begin{array}{c}f\left(x\right)=20-2\left(\frac{4i}{n}+2\right)\\ =16-\frac{8i}{n}\\ =\frac{16n-8i}{n}\end{array}$

Approximate the are as as the sum of the areas of $n$ rectangles.

  $\begin{array}{c}\text{A=}{\displaystyle \sum _{\text{i=1}}^{\text{n}}\text{f}\left(\text{a+}\frac{\left(\text{b-a}\right)\text{i}}{\text{n}}\right)\left(\frac{\text{b-a}}{\text{n}}\right)}\\ ={\displaystyle \sum _{i=1}^n\left(\frac{16n-8i}{n}\right)\left(\frac{4}{n}\right)}\\ ={\displaystyle \sum _{i=1}^n\frac{64}{n}-\frac{32i}{n^2}}\end{array}$

Applying summation formulas and properties.

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

Now,

  $\begin{array}{c}A=\frac{1}{n}{\displaystyle \sum _{i=1}^{n}64}-\frac{32}{n^2}{\displaystyle \sum _{i=1}^{n}i}\\ =\frac{64n}{n}-\frac{32}{n^2}\times \frac{n\left(n+1\right)}{2}\\ =64-\frac{16\left(n+1\right)}{n}\\ =\frac{48n-16}{n}\end{array}$

Obtain a more accurate approximation of the area of the region by increasing the number of rectangles.

  $A\left(n\right)=\frac{48n-16}{n}$

Take,

  $n=4$

  $\begin{array}{c}\text{A=}\frac{48\times 4-16}{4}\\ \text{=}\frac{176}{4}\\ \text{=44}\end{array}$

In this way,take $n=8,20,50,100\mathrm{....}\infty$

The exact area by taking the limit as $n$ approaches $\infty$

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left[\frac{48n-16}{n}\right]\\ =48\end{array}$

Therefore, the exact area is $\text{48}\text{square}\text{units}$

Now, values of $A=46,47.2,47.68,47.84\mathrm{....48}$

Finally, Approximating the area of region by using values in the following table.

n	$4$	$8$	$20$	$50$	$100$	$\infty$
Approximate area	$44$	$46$	$47.2$	$47.68$	$47.84$	$48$