Chapter 11, Problem 17CT

To determine

To Calculate:

The area of region bounded.

Answer to Problem 17CT

The required area of the region by the limit process is $A=\frac{34}{3}$ square unit.

Explanation of Solution

Given Information:

Function: $f\left(x\right)=7-x^2$

Interval: $\left[0,2\right]$

Calculation:

To get the area of the region by the limit process

Consider the following function over the interval $\left[0,2\right]$

  $f\left(x\right)=7-x^2$

First approximate the area as the sum of the area of $n$ rectangles

Since, the approximate area as the sum of the area of $n$ rectangles are

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

Then,

  $\begin{array}{l}A={\displaystyle \sum _{i=1}^{n}f\left(a+\frac{\left(b-a\right)i}{n}\right)}\left(\frac{b-a}{n}\right)\\ ={\displaystyle \sum _{i=1}^{n}f\left(0+\frac{\left(2-0\right)i}{n}\right)}\left(\frac{2-0}{n}\right)\\ =\left(\frac{2}{n}\right){\displaystyle \sum _{i=1}^{n}f\left(\frac{2i}{n}\right)}\\ =\left(\frac{2}{n}\right){\displaystyle \sum _{i=1}^n\left[7-{\left(\frac{2i}{n}\right)}^2\right]}\\ =\left(\frac{2}{n}\right)\left[{\displaystyle \sum _{i=1}^{n}7-\frac{4}{n^2}{\displaystyle \sum _{i=1}^n{i}^2}}\right]\\ =\left(\frac{2}{n}\right)\left[7n-\frac{2}{n}-\frac{\left(n+1\right)\left(2n+1\right)}{3}\right]\\ =\left[14-\frac{4\left(2n^2+3n+1\right)}{3n^2}\right]\end{array}$

That is

  $A=\left[14-\frac{4\left(2n^2+3n+1\right)}{3n^2}\right]$

Since, the exact area of the region is

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

Then,

  $\begin{array}{l}A=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f}\left(a+\frac{\left(b-a\right)i}{n}\right)\left(\frac{b-a}{n}\right)\\ =\underset{x\to \infty }{\lim }\left(14-\frac{4\left(2n^2+3n+1\right)}{3n^2}\right)\\ =\left[\underset{x\to \infty }{\lim }14-\underset{x\to \infty }{\lim }\frac{4\left(2n^2\right)\left(1+\frac{3n}{2n^2}+\frac{1}{2n^2}\right)}{3n^2}\right]\\ =14-\frac{8}{3}\\ =\frac{42-8}{3}\\ =\frac{34}{3}\end{array}$

Hence, the required area of the region by the limit process is $A=\frac{34}{3}$ square unit.