Chapter 11.5, Problem 41E

To determine

To calculate:

The area of the region.

Answer to Problem 41E

The required area of the region by limit process is $A=\frac{3}{4}$ square unit.

Explanation of Solution

Given Information:

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

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

Calculation:

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

  $f\left(x\right)=2x-x^3$

The graph of function $f\left(x\right)=2x-x^3$ is follows,

  

The above graph shows the $f\left(x\right)=2x-x^3$

Now with the help of $A\approx {\displaystyle \sum _{i=1}^{n}f}\left(a+\frac{\left(b-a\right)i}{n}\right)\left(\frac{b-a}{n}\right)$ find the approximate area.

Therefore,

  $\begin{array}{l}A=\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^{n}f}\left(0+\frac{\left(1-0\right)i}{n}\right)\left(\frac{1-0}{n}\right)\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\left(2\left(\frac{i}{n}\right)-{\left(\frac{i}{n}\right)}^3\right)}\left(\frac{1}{n}\right)\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\left(\frac{2i}{n^2}-\frac{i^3}{n^4}\right)}\right)\end{array}$

Now use the summation formula as

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{i}^3=\frac{n^2{\left(n+1\right)}^2}{4}}\\ {\displaystyle \sum _{i=1}^{n}i=}\frac{n\left(n+1\right)}{2}\end{array}$

Hence,

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

Hence, the area is $\frac{3}{4}$ square units.