Chapter 11.5, Problem 30E

To determine

To calculate:

The exact area of the given function:

Answer to Problem 30E

The table of the approximate area of the region is

$n$	$4$	$8$	$20$	$50$	$100$	$\infty$
area	$57.75$	$55.1875$	$53.67$	$53.067$	$52.8668$	$52.66$

Explanation of Solution

Given information:

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

Function: $f\left(x\right)=x^2+1$

Integral: $\left[4,6\right]$

Sum of Areas of n rectangles: $\frac{158n^3+60n^2+4n}{3n^3}$

Formula Used:

By using this formula:

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

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ {\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{2}\end{array}$

Calculation:

Consider this equation:

  $f\left(x\right)=x^2+1$

By using this formula:

  $\begin{array}{l}A=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(a+\frac{\left(b-a\right)i}{n}\right)\left(\frac{b-a}{n}\right)}\\ A=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(4+\frac{\left(6-4\right)i}{n}\right)\left(\frac{6-4}{n}\right)}\\ A=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left({\left(4+\frac{2i}{n}\right)}^2+1\right)\left(\frac{2}{n}\right)}\\ A=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{34}{n}+\frac{8i^2}{n^3}+\frac{32}{n^2}\right)}\end{array}$

By using this formula:

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ {\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{2}\end{array}$

Comparing with this equation:

  $\begin{array}{l}A=34+\frac{8}{n^2}.\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{32}{n^2}.\frac{n\left(n+1\right)}{2}\\ A=34+\frac{4}{3}\left(\frac{2n^2+3n+1}{n^2}\right)+\frac{16\left(n+1\right)}{n}\\ A=\frac{158n^2+60n+4}{3n^2}\end{array}$

Put the value of $n=4$ :

  $\begin{array}{l}n=4\\ A=\frac{158{\left(4\right)}^2+60\left(4\right)+4}{3{\left(4\right)}^2}\\ A=57.75\end{array}$

Put the value of $n=8$ :

  $\begin{array}{l}n=8\\ A=\frac{158{\left(8\right)}^2+60\left(8\right)+4}{3{\left(8\right)}^2}\\ A=55.1875\end{array}$

Put the value of $n=20$ :

  $\begin{array}{l}n=20\\ A=\frac{158{\left(20\right)}^2+60\left(20\right)+4}{3{\left(20\right)}^2}\\ A=53.67\end{array}$

Put the value of $n=50$ :

  $\begin{array}{l}n=50\\ A=\frac{158{\left(50\right)}^2+60\left(50\right)+4}{3{\left(50\right)}^2}\\ A=53.0672\end{array}$

Put the value of $n=100$ :

  $\begin{array}{l}n=100\\ A=\frac{158{\left(100\right)}^2+60\left(100\right)+4}{3{\left(100\right)}^2}\\ A=52.8668\end{array}$

Put the value of $n=\infty$ :

  $\begin{array}{l}n=\infty \\ A=\underset{x\to \infty }{\lim }\frac{158n^2+60n+4}{3n^2}\\ A=\frac{158}{3}\end{array}$

The exact area is $52.66$ Sq units.

$n$	$4$	$8$	$20$	$50$	$100$	$\infty$
area	$57.75$	$55.1875$	$53.67$	$53.067$	$52.8668$	$52.66$