Chapter 15.3, Problem 18E

To determine

Find the area using the limit definition and sigma notation.

Answer to Problem 18E

The area of the region is $\frac{15}{2}$ square units.

Explanation of Solution

Given:

The given expressions are

  $y=x^2-x+1$ from $x=0$ to $x=3$ .

Calculation:

The exact area of the region under the graph of the function is

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)\Delta x}$ , where

  $\Delta x=\frac{b-a}{n}$

  $x_i=a+i\Delta x$

Graph of the given region is given below.

  

For the given region

  $\Delta x=\frac{3-0}{n}=\frac{3}{n}$

  $x_i=0+i\cdot \frac{3}{n}=\frac{3i}{n}$

  $\begin{array}{l}{\displaystyle {\int }_0^3\left(x^2-x+1\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left[{\left(\frac{3i}{n}\right)}^2-\left(\frac{3i}{n}\right)+1\right]\frac{3}{n}}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{27i^2}{n^3}-\frac{9i}{n^2}+\frac{3}{n}\right)}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{27i^2}{n^3}\right)}-\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{9i}{n^2}\right)}+\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{3}{n}\right)}\end{array}$

  $\begin{array}{l}=27\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i^2}{n^3}\right)}-9\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i}{n^2}\right)}+3\\ =27\underset{n\to \infty }{\lim }\left(\frac{1^2+2^2+3^2+---+n^2}{n^3}\right)-9\underset{n\to \infty }{\lim }\left(\frac{1+2+3+---+n}{n^2}\right)+3\\ =27\underset{n\to \infty }{\lim }\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6n^3}\right)-9\underset{n\to \infty }{\lim }\left(\frac{n\left(n+1\right)}{2n^2}\right)+3\end{array}$

  $\begin{array}{l}=\frac{27}{6}\underset{n\to \infty }{\lim }\left(\frac{2n^2+3n+1}{n^2}\right)-\frac{9}{2}\underset{n\to \infty }{\lim }\left(\frac{n+1}{n}\right)+3\\ =\frac{9}{2}\underset{n\to \infty }{\lim }\left(\frac{2n^2}{n^2}+\frac{3n}{n^2}+\frac{1}{n^2}\right)-\frac{9}{2}\underset{n\to \infty }{\lim }\left(\frac{n}{n}+\frac{1}{n}\right)+3\\ =\frac{9}{2}\underset{n\to \infty }{\lim }\left(2+\frac{3}{n}+\frac{1}{n^2}\right)-\frac{9}{2}\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)+3\end{array}$

  $\begin{array}{l}=\frac{9}{2}\left(\underset{n\to \infty }{\lim }2+\underset{n\to \infty }{\lim }\frac{3}{n}+\underset{n\to \infty }{\lim }\frac{1}{n^2}\right)-\frac{9}{2}\left(\underset{n\to \infty }{\lim }1+\underset{n\to \infty }{\lim }\frac{1}{n}\right)+3\\ =\frac{9}{2}\left(2+0+0\right)-\frac{9}{2}\left(1+0\right)+3\end{array}$

  $\begin{array}{l}=\frac{18}{2}-\frac{9}{2}+3\\ =\frac{18-9+6}{2}\\ =\frac{15}{2}\end{array}$

Hence the area of the region is $\frac{15}{2}$ square units.