Chapter 12, Problem 50CLT

To determine

Find the area of the region.

Answer to Problem 50CLT

The area of region is $\boxed{\frac{3}{4}}$ square units.

Explanation of Solution

Given information:

Use the limit process to find the area of the region bounded by the graph of the function and the $X$ -axis over the specified interval.

  $\begin{array}{l}f\left(x\right)=1-x^3\\ Interval\\ \left[0,1\right]\end{array}$

Calculation:

Consider the given function and interval,

  $\begin{array}{l}f\left(x\right)=1-x^3\\ Interval\\ \left[0,1\right]\end{array}$

The graph of the function $f\left(x\right)=1-x^3$ is as shown below,

  

Now estimate the dimensions of the rectangle,

From

  $\begin{array}{l}Interval\\ \left[0,1\right]\\ a=0\\ b=1\end{array}$

  $\begin{array}{l}width=\frac{b-a}{n}\\ =\frac{1-0}{n}=\frac{1}{n}\end{array}$

  $\begin{array}{l}height=f\left(a+\frac{\left(b-a\right)i}{n}\right)\\ =f\left(0+\frac{\left(1-0\right)i}{n}\right)\\ =f\frac{i}{n}\\ =1-{\left(\frac{i}{n}\right)}^3\end{array}$

Now calculate area of $n$ rectangles

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

Apply property of summation,

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

Now the area of region.

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

Hence the area of region is $\boxed{\frac{3}{4}}$ square units.