Chapter 12.5, Problem 44E

To determine

To find: The area of region bounded by the graph of $f$ and the x-axis over the specified interval using the limit process.

Answer to Problem 44E

The area of region is $4\text{square}\text{units}$

Explanation of Solution

Given information:

The given function is,

  $g\left(x\right)=4x-x^3$

Given intervals $\left[0,2\right]$

Concept used:

The area of a region;

Let $f$ be continuous and nonnegative on the interval $\left[a,b\right]$ The area A of the region bounded by the graph of $f$ , the $\text{x-axis}$ ,and the vertical lines $x=a$ and $x=b$ is given by

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

The given function is,

  $g\left(x\right)=4x-x^3$

The area of the region bounded by the graph of $g\left(x\right)=4x-x^3$ and the x-axis between $x=0$ and $x=2$

The dimensions of the rectangles;

$\begin{array}{c}width;\frac{b-a}{n}=\frac{2-0}{n}\\ =\frac{2}{n}\end{array}$

And,

  $\begin{array}{c}\text{Height}\text{;}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(0+\frac{\left(2-0\right)i}{n}\right)\\ =f\left(\frac{2i}{n}\right)\end{array}$

Given function; $g\left(x\right)=4x-x^3$

  $\begin{array}{c}f\left(x\right)=4\left(\frac{2i}{n}\right)-{\left(\frac{2i}{n}\right)}^3\\ =\frac{8i}{n}-\frac{8i^3}{n^3}\end{array}$

Approximate the areas as the sum of the areas of $n$ rectangles.

  $\begin{array}{c}\text{A=}\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{\text{i=1}}^{\text{n}}\text{f}\left(\text{a+}\frac{\left(\text{b-a}\right)\text{i}}{\text{n}}\right)\left(\frac{\text{b-a}}{\text{n}}\right)}\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\left(\frac{8i}{n}-\frac{8i^3}{n^3}\right)\left(\frac{2}{n}\right)}\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\frac{16i}{n^2}-\frac{16i^3}{n^4}}\right)\end{array}$

Apply Summation formula and properties.

  ${\displaystyle \sum _{i=1}^{n}i=\frac{n\left(n+1\right)}{2}}$

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

Now,

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left(\frac{16}{n^2}{\displaystyle \sum _{i=1}^{n}i}-\frac{16}{n^4}{\displaystyle \sum _{i=1}^n{i}^3}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{16}{n^2}\times \frac{n\left(n+1\right)}{2}-\frac{16}{n^4}\left(\frac{n^2{\left(n+1\right)}^2}{4}\right)\right)\\ =\underset{x\to \infty }{\lim }\left(\left(\frac{8\left(n+1\right)}{n}\right)-\left(4\times \frac{{\left(n+1\right)}^2}{n^2}\right)\right)\end{array}$

Finally the exact area by taking the limit as $n$ approaches $\infty$

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left[\left(8\times \frac{\left(n+1\right)}{n}\right)-\left(4\times \frac{{\left(n+1\right)}^2}{n^2}\right)\right]\\ =\left[8-4\right]\\ =4\end{array}$

Therefore, the exact area is $4\text{square}\text{units}$