Chapter 12, Problem 100RE

To determine

To find: The area of the region under the given curve on $\left[3,6\right]$ using limit summation formulas.

Answer to Problem 100RE

The required area is $9$ square units.

Explanation of Solution

Given:

The given curve is $f\left(x\right)=2x-6$ .

Concept used:

Suppose $f$ is a continuous function and $f\left(x\right)\ge 0$ through out the interval $\left[a,b\right]$ . Then the area of region under the curve $y=f\left(x\right)$ over the interval is

  $A=\underset{n\to \infty }{\lim }\left[{\displaystyle \sum _{k=1}^{n}f\left(a+\frac{\left(b-a\right)k}{n}\right)\left(\frac{b-a}{n}\right)}\right]$

Formula used:

Summation formula for $k$ is given as follows:

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

Summation formula for $k^3$ is given as follows:

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

Summation formula for constant function is given as follows:

  ${\displaystyle \sum _{k=1}^{n}c}=cn$

Linear rule of summation is as follows:

  ${\displaystyle \sum _{k=1}^n\left(c{a}_k+c{b}_k\right)}=c{\displaystyle \sum _{k=1}^n{a}_k}+c{\displaystyle \sum _{k=1}^n{b}_k}$

Here, $c$ is any constant

Calculation:

Consider the given curve.

  $f\left(x\right)=2x-6$

Here, the interval is $\left[3,6\right]$ .

Therefore, $b=6,a=3$

Therefore, width of each sub-interval will be as shown.

  $\begin{array}{c}\frac{b-a}{n}=\frac{6-3}{n}\\ =\frac{3}{n}\end{array}$

Now, the area of the region under the given curve can be calculated as shown.

  $\begin{array}{c}A=\underset{n\to \infty }{\lim }\left[{\displaystyle \sum _{k=1}^{n}f\left(a+\frac{\left(b-a\right)k}{n}\right)\left(\frac{b-a}{n}\right)}\right]\\ =\underset{n\to \infty }{\lim }\left[{\displaystyle \sum _{k=1}^{n}f\left(3+\frac{3k}{n}\right)\left(\frac{3}{n}\right)}\right]\\ =\underset{n\to \infty }{\lim }\left[{\displaystyle \sum _{k=1}^n\left(6-2\left(3+\frac{3k}{n}\right)\right)\left(\frac{3}{n}\right)}\right]\\ =\underset{n\to \infty }{\lim }\left[{\displaystyle \sum _{k=1}^n\frac{18k}{n^2}}\right]\\ =\underset{n\to \infty }{\lim }\frac{18}{n^2}{\displaystyle \sum _{k=1}^{n}k}\\ =\underset{n\to \infty }{\lim }\frac{18}{n^2}\frac{n\left(n+1\right)}{2}\\ =\underset{n\to \infty }{\lim }9\frac{n\left(n+1\right)}{n^2}\\ =\underset{n\to \infty }{\lim }9\left[\frac{n}{n}\left(\frac{n}{n}+\frac{1}{n}\right)\right]\\ =9\end{array}$

Therefore, the exact area of the region under the given curve is $9$ square units.