Chapter 12.5, Problem 31E

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 31E

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

n	$4$	$8$	$20$	$50$	$100$	$\infty$
Approximate area	$40$	$38$	$36.8$	$36.32$	$36.16$	$36$

Explanation of Solution

Given information:

The givenfunction is,

  $f\left(x\right)=2x+5$

Given intervals $\left[0,4\right]$

And, rectangles are $n$

Concept used:

The area of a region;

Let $f$ becontinuousandnonnegativeontheinterval $\left[a,b\right]$ Thearea Aoftheregionboundedbythegraphof $f$ ,the $\text{x-axis}$ ,andtheverticallines $x=a$ 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 givenfunctionis,

  $\text{f}\left(\text{x}\right)\text{=2x+5}$

Thearea of the region bounded by the graph of $f\left(x\right)=2x+5$ and the x-axis between $x=0$ and $x=4$

Thedimensionsoftherectangles;

  $\begin{array}{c}width;\frac{b-a}{n}=\frac{4-0}{n}\\ =\frac{4}{n}\end{array}$

And,

  $\begin{array}{c}\text{Height}\text{;}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(0+\frac{\left(4-0\right)i}{n}\right)\\ =f\left(\frac{4i}{n}\right)\end{array}$

Given function; $f\left(x\right)=2x+5$

  $\begin{array}{c}f\left(x\right)=2\left(\frac{4i}{n}\right)+5\\ =\frac{8i}{n}+5\\ =\frac{8i+5n}{n}\end{array}$

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

  $\begin{array}{c}\text{A=}{\displaystyle \sum _{\text{i=1}}^{\text{n}}\text{f}\left(\text{a+}\frac{\left(\text{b-a}\right)\text{i}}{\text{n}}\right)\left(\frac{\text{b-a}}{\text{n}}\right)}\\ ={\displaystyle \sum _{i=1}^n\left(\frac{8i+5n}{n}\right)\left(\frac{4}{n}\right)}\\ ={\displaystyle \sum _{i=1}^n\frac{32i}{n^2}+\frac{20}{n}}\end{array}$

Applying Summationformulasandproperties.

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

Now,

  $\begin{array}{c}A=\frac{32}{n^2}{\displaystyle \sum _{i=1}^n\left(i\right)}+\frac{1}{n}{\displaystyle \sum _{i=1}^{n}20}\\ =\frac{32}{n^2}\times \frac{n\left(n+1\right)}{2}+\frac{20n}{n}\\ =\frac{16\left(n+1\right)}{n}+20\\ =\frac{36n+16}{n}\end{array}$

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

  $A\left(n\right)=\frac{36n+16}{n}$

Take,

  $n=4$

  $\begin{array}{c}\text{A=}\frac{36\times 4+16}{4}\\ \text{=}\frac{160}{4}\\ \text{=40}\end{array}$

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

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

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left[\frac{36n+16}{n}\right]\\ =36\end{array}$

Therefore, the exact area is $\text{36}\text{square}\text{units}$

Now, values of $A=38,36.8,36.32,36.16\mathrm{....36}$

Finally, toapproximating the area of region by using values inthe following table.

n	$4$	$8$	$20$	$50$	$100$	$\infty$
Approximate area	$40$	$38$	$36.8$	$36.32$	$36.16$	$36$