Chapter 13, Problem 47RE

To determine

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

Answer to Problem 47RE

The area of the region is $\frac{1}{2}$

Explanation of Solution

Given information:

Consider the given information,

  $f\left(x\right)=x^2-x,1\le x\le 2$

CALCULATION:

Approximate area of the given interval,

  $f\left(x\right)=2x+3$

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, $\begin{array}{l}f\left(x_i\right)=f\left(1+\frac{k}{n}\right)={\left(1+\frac{k}{n}\right)}^2-\left(1+\frac{k}{n}\right)\\ =1+\frac{2k}{n}+\frac{k^2}{n^2}-1-\frac{k}{n}\\ =\frac{k^2}{n^2}-\frac{k}{n}\end{array}$

  $\begin{array}{l}\text{Area }\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(x_1\right)}\Delta x\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left(\frac{k^2}{n^2}-\frac{1}{n}\right)}\left(\frac{1}{n}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{1}{n}{\displaystyle \sum _{k=1}^n\frac{k^2}{n^2}-\frac{1}{n}{\displaystyle \sum _{k=1}^n\frac{k}{n}}}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{1}{n^3}{\displaystyle \sum _{k=1}^n{k}^2-\frac{1}{n^2}{\displaystyle \sum _{k=1}^{n}k}}\right)\\ =\underset{n\to \infty }{\lim }\left\{\frac{1}{n^3}\cdot \frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{1}{n^2}\cdot \frac{n\left(n+1\right)}{2}\right\}\\ =\underset{n\to \infty }{\lim }\left\{\frac{1}{n^2}\cdot \frac{\left(n+1\right)\left(2n+1\right)}{6}-\frac{\left(n+1\right)}{2n}\right\}\\ =\underset{n\to \infty }{\lim }\left\{\frac{1}{6}\left(\frac{1}{n}+\frac{1}{n^2}\right)\left(\frac{2}{n}+\frac{1}{n^2}\right)-\frac{1}{2}\left(1+\frac{1}{n}\right)\right\}\\ =\frac{1}{6}\left(0\right)\times 0-\frac{1}{2}\times 1=-\frac{1}{2}=\frac{1}{2}\end{array}$

As area cannot be negative, therefore the area of the region is $\frac{1}{2}$