Chapter 12.5, Problem 34E

To determine

To calculate: the given system of equation $f\left(x\right)=x^2+1$ .

Answer to Problem 34E

n	Approximate Area
4	57.75
8	55.188
20	53.67
50	53.067
100	52.867
8	158/3

Explanation of Solution

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

Calculation:

Let’s take the function f that is continuous and non negative on the interval (a,b). The area A of the region surrounded by the chart along x-axis is

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

Let’s take the subsequent figure of the given equation:

  $f\left(x\right)=x^2+1$

The $x$ axis b/w $x=4$ and $x=6$ .

  

Find the size of the rectangle made with x and y axis:

  $\begin{array}{c}W=\frac{b-a}{n}\\ =\frac{6-\left(4\right)}{n}\\ =\frac{2}{n}\\ H=f\left[a+\frac{\left(b-a\right)i}{n}\right]\\ =f\left[4+\frac{\left\{6-\left(4\right)\right\}i}{n}\right]\\ =\left({\left(4+\frac{2i}{n}\right)}^2+1\right)\end{array}$

rough area of the n rectangles can be calculated as:

  $\begin{array}{c}A\approx {\displaystyle \sum _{i=1}^{n}f}\left[a+\frac{\left(b-a\right)i}{n}\right]\left[\frac{b-a}{n}\right]\\ ={\displaystyle \sum _{i=1}^n\left({\left(4+\frac{2i}{n}\right)}^2+1\right)\left[\frac{2}{n}\right]}\\ ={\displaystyle \sum _{i=1}^n\left[\frac{8i^2}{n^3}+\frac{32i}{n^2}+\frac{34}{n}\right]}\\ \left\{\begin{array}{l}{\displaystyle \sum _{i=1}^n\left(i^2\right)}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ {\displaystyle \sum _{i=1}^n\left(i\right)}=\frac{n\left(n+1\right)}{2}\end{array}\right\}\\ A=\frac{8}{n^3}\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{32}{n^2}\frac{n\left(n+1\right)}{2}+34\\ =\frac{158n^2+60n+4}{3n^2}\end{array}$

Apply the following formula:

  $A\left(n\right)=\frac{158n^2+60n+4}{3n^2}$

Complete the given table using the above expression:

  $\begin{array}{c}A\left(n\right)=\frac{158n^2+60n+4}{3n^2}\\ \\ \text{for}n=4\\ A=\frac{158{\left(4\right)}^2+60\left(4\right)+4}{3{\left(4\right)}^2}\\ =57.75\\ \\ \text{for}n=8\\ A=\frac{158{\left(8\right)}^2+60\left(8\right)+4}{3{\left(8\right)}^2}\\ =55.188\\ \\ \text{for}n=20\\ A=\frac{158{\left(20\right)}^2+60\left(20\right)+4}{3{\left(20\right)}^2}\\ =53.67\\ \\ \text{for}n=50\\ A=\frac{158{\left(50\right)}^2+60\left(50\right)+4}{3{\left(50\right)}^2}\\ =53.067\\ \\ \text{for}n=100\\ A=\frac{158{\left(100\right)}^2+60\left(100\right)+4}{3{\left(100\right)}^2}\\ =52.867\\ \\ \text{for}n\to \infty \\ A=\underset{n\to \infty }{\lim }\frac{158n^2+60n+4}{3n^2}\\ =\frac{158}{3}\end{array}$

Therefore, by using above values the table as follows:

n	Approximate Area
4	57.75
8	55.188
20	53.67
50	53.067
100	52.867
8	158/3