Chapter 12.5, Problem 43E

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 43E

The area of region is $4.25\text{square}\text{units}$ s

Explanation of Solution

Given information:

The given function is,

  $g\left(x\right)=8-x^3$

Given intervals $\left[1,2\right]$

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,

  $g\left(x\right)=8-x^3$

The area of the region bounded by the graph of $g\left(x\right)=8-x^3$ and the x-axis between $x=1$ and $x=2$

The dimensions of the rectangles;

$\begin{array}{c}width;\frac{b-a}{n}=\frac{2-1}{n}\\ =\frac{1}{n}\end{array}$

And,

  $\begin{array}{c}\text{Height}\text{;}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(1+\frac{\left(2-1\right)i}{n}\right)\\ =f\left(1+\frac{i}{n}\right)\end{array}$

Given function; $g\left(x\right)=8-x^3$

  $\begin{array}{c}f\left(x\right)=8-{\left(1+\frac{i}{n}\right)}^3\\ =8-\left(\frac{n^3+i^3+3ni\left(n+i\right)}{n^3}\right)\\ =\frac{8n^3-n^3-i^3-3n^{2}i-3n{i}^2}{n^3}\\ =\frac{7n^3-i^3-3n^{2}i-3n{i}^2}{n^3}\\ \end{array}$

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

  $\begin{array}{c}\text{A=}\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{\text{i=1}}^{\text{n}}\text{f}\left(\text{a+}\frac{\left(\text{b-a}\right)\text{i}}{\text{n}}\right)\left(\frac{\text{b-a}}{\text{n}}\right)}\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\left(\frac{7n^3-i^3-3n^{2}i-3n{i}^2}{n^3}\right)\left(\frac{1}{n}\right)}\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\frac{7}{n}-\frac{i^3}{n^4}-\frac{3i}{n^2}-\frac{3i^2}{n^3}}\right)\end{array}$

Apply Summation formula and properties.

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

  ${\displaystyle \sum _{i=1}^n{i}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}$

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

Now,

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left(\frac{7}{n}{\displaystyle \sum _{i=1}^{n}1}-\frac{1}{n^4}{\displaystyle \sum _{i=1}^n{i}^3}-\frac{3}{n^2}{\displaystyle \sum _{i=1}^{n}i}-\frac{3}{n^3}{\displaystyle \sum _{i=1}^n{i}^2}\right)\\ =\underset{x\to \infty }{\lim }\left(7-\frac{1}{n^4}\times \frac{n^2{\left(n+1\right)}^2}{4}-\frac{3}{n^2}\left(\frac{n\left(n+1\right)}{2}\right)-\frac{3}{n^3}\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)\right)\\ =\underset{x\to \infty }{\lim }\left(7-\left(\frac{1}{4}\times \frac{n^2+2n+1}{n^2}\right)-\left(\frac{3}{2}\times \frac{n^2+n}{n^2}\right)-\left(\frac{1}{2}\times \frac{2n^2+3n+1}{n^3}\right)\right)\end{array}$

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

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left[7-\left(\frac{1}{4}\times \frac{n^2+2n+1}{n^2}\right)-\left(\frac{3}{2}\times \frac{n^2+n}{n^2}\right)-\left(\frac{1}{2}\times \frac{2n^2+3n+1}{n^4}\right)\right]\\ =\left[7-\frac{1}{4}-\frac{3}{2}-1\right]\\ =\frac{17}{4}\\ =4.25\end{array}$

Therefore, the exact area is $4.25\text{square}\text{units}$