Chapter 11.5, Problem 44E

To determine

To calculate:

The area of the region.

Answer to Problem 44E

The required area of the region by limit process is $A=\frac{1}{12}$ square unit.

Explanation of Solution

Given Information:

Function: $f\left(x\right)=x^2-x^3$

Interval: $\left[0,1\right]$

Calculation:

To find the area of the region by the limit process

Consider the following function over the interval $\left[0,1\right]$

  $f\left(x\right)=x^2-x^3$

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

Since, the approximate area as the sum 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,

  $\begin{array}{l}A={\displaystyle \sum _{i=1}^{n}f}\left(a+\frac{\left(b-a\right)i}{n}\right)\left(\frac{b-a}{n}\right)\\ ={\displaystyle \sum _{i=1}^{n}f}\left(0+\frac{\left(1-0\right)i}{n}\right)\left(\frac{1-0}{n}\right)\\ =\left(\frac{1}{n}\right){\displaystyle \sum _{i=1}^{n}f}\left(\frac{i}{n}\right)\\ =\left(\frac{1}{n}\right){\displaystyle \sum _{i=1}^n\left[{\left(\frac{i}{n}\right)}^2-{\left(\frac{i}{n}\right)}^3\right]}\\ =\left(\frac{1}{n}\right){\displaystyle \sum _{i=1}^n\left[{\left(\frac{i}{n}\right)}^2-4{\displaystyle \sum _{i=1}^n{\left(\frac{i}{n}\right)}^3}\right]}\\ =\left(\frac{1}{n}\right)\left[\frac{1}{n^2}\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{1}{n^3}\frac{n^2{\left(n+1\right)}^2}{4}\right]\\ =\left(\frac{1}{n}\right)\left[\frac{\left(n+1\right)\left(2n+1\right)}{6n}-\frac{{\left(n+1\right)}^2}{4n}\right]\\ =\left[\frac{\left(2n^2+3n+1\right)}{6n^2}-\frac{\left(n^2+2n+1\right)}{4n^2}\right]\end{array}$

That is the approximate area as the sum of the area of $n$ rectangle is

  $A=\left[\frac{\left(2n^2+3n+1\right)}{6n^2}-\frac{\left(n^2+2n+1\right)}{4n^2}\right]$

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,

  $\begin{array}{l}A=\underset{x\to \infty }{\lim }\left[\frac{\left(2n^2+3n+1\right)}{6n^2}-\frac{\left(n^2+2n+1\right)}{4n^2}\right]\\ =\underset{x\to \infty }{\lim }\frac{\left(2n^2+3n+1\right)}{6n^2}-\underset{x\to \infty }{\lim }\frac{\left(n^2+2n+1\right)}{4n^2}\\ =\frac{2}{6}-\frac{1}{4}\\ =\frac{1}{12}\end{array}$

Hence, the required area of the region by the limit process is $A=\frac{1}{12}$ square unit.