Chapter 13, Problem 48RE

To determine

To find the area of the region that lies under the graph of $f$ over the given interval

Answer to Problem 48RE

The area of the region is $\frac{5}{4}$

Explanation of Solution

Given information:

Consider the given information,

  $f\left(x\right)=x^3-x,1\le x\le 2$

CALCULATION:

Approximate area of the given interval,

  $f\left(x\right)=x^3$

Width $\Delta x=\frac{b-a}{n}=\frac{2-1}{n}=\frac{1}{n}$

Right end point, $x_i=a+k\Delta n=1+k\cdot \frac{1}{n}=1+\frac{k}{n}$

Height, $f\left(x_i\right)=f\left(1+\frac{k}{n}\right)={\left(1+\frac{k}{n}\right)}^3=1+\frac{2k}{n}+\frac{2k^2}{n^2}+\frac{k^3}{n^3}$

  $\text{Area,=}\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(x_i\right)\Delta x}$

  $=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left(1+\frac{2k}{n}+\frac{2k^2}{n^2}+\frac{k^3}{n^3}\right)}\cdot \frac{1}{n}$

  $=\underset{n\to \infty }{\lim }\left(\frac{1}{n}{\displaystyle \sum _{k=1}^n+1\frac{1}{n}{\displaystyle \sum _{k=1}^n\frac{2k}{n}+\frac{1}{n}{\displaystyle \sum _{k=1}^n\frac{2k^2}{n^2}+\frac{1}{n}{\displaystyle \sum _{k=1}^n\frac{k^3}{n^3}}}}}\right)$

  $=\underset{n\to \infty }{\lim }\left(\frac{1}{n}{\displaystyle \sum _{k=1}^{n}1+\frac{2}{n^2}{\displaystyle \sum _{k=1}^{n}k+\frac{2}{n^3}{\displaystyle \sum _{k=1}^n{k}^2+\frac{1}{n^4}{\displaystyle \sum _{k=1}^n{k}^3}}}}\right)$

  $=\underset{n\to \infty }{\lim }\left(\frac{1}{n}.n+\frac{2}{n^2}.\frac{n\left(n+1\right)}{2}+\frac{2}{n^2}.\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{1}{n^2}.\frac{n^2{\left(n+1\right)}^2}{4}\right)$

  $=\underset{n\to \infty }{\lim }\left(1+\frac{2\left(n+1\right)}{n}+\frac{\left(n+1\right)\left(2n+1\right)}{3n^2}+\frac{\left(n+1\right)}{4n^2}\right)$

  $=\underset{n\to \infty }{\lim }\left(1+\left(\frac{2}{n}.\frac{n+1}{n}\right)+\frac{1}{3}\left(\frac{n+1}{n}.\frac{2n+1}{n}\right)+\frac{1}{4}\left(\frac{n+1}{n}.\frac{n+1}{n}\right)\right)$

  $=\underset{n\to \infty }{\lim }\left[1+\frac{2}{n}\left(1+\frac{1}{n}\right)+\frac{1}{3}\left(\left(1+\frac{1}{n}\right).\frac{2}{n}.\left(1+\frac{1}{n}\right)\right)+\frac{1}{4}\left(\left(1+\frac{1}{n}\right).\left(1+\frac{1}{n}\right)\right)\right]$

  $=\left[1+0.1+\frac{1}{3}\mathrm{.1.0.1}+\frac{1}{4}\mathrm{.1.1}\right]$

  $=1+0+\frac{1}{4}=\frac{5}{4}$

Therefore, the area of the shaded region is $\frac{5}{4}$