Chapter 13.5, Problem 20E

To determine

To find : The area of the region under the graph of f over the given interval.

Answer to Problem 20E

The area of the region is $\frac{22}{3}$ square units.

Explanation of Solution

Given information:

The given function is $f\left(x\right)=20-2x^2$ , $2\le x\le 3$.

Calculation :

The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles:

  $\begin{array}{l}A=\underset{n\to \infty }{\lim }{R}_n=\underset{n\to \infty }{\lim }\left[f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+\mathrm{......}+f\left(x_n\right)\Delta x\right]\text{ or }\\ A=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k\right)\Delta x\end{array}$

The function is $f\left(x\right)=20-2x^2$ , $2\le x\le 3$.

Here,

Width:

  $\Delta x=\frac{b-a}{n}=\frac{3-2}{n}=\frac{1}{n}$

Right end point:

  $\begin{array}{l}{x}_k=a+k\Delta x\\ x_k=2+k\cdot \frac{1}{n}\\ x_k=2+\frac{k}{n}\end{array}$

Height:

  $\begin{array}{l}f\left(x_k\right)=20-2x_k^2\\ f\left(x_k\right)=20-2{\left(2+\frac{k}{n}\right)}^2\\ f\left(x_k\right)=20-2\left(4+\frac{4k}{n}+\frac{k^2}{n^2}\right)\\ f\left(x_k\right)=12-\frac{8k}{n}-\frac{2k^2}{n^2}\end{array}$

Now, according to definition area is

  $A=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k\right)\Delta x$

  $A=\underset{n\to \infty }{\lim }\sum _{k=1}^n\left[\left(12-\frac{8k}{n}-\frac{2k^2}{n^2}\right)\cdot \frac{1}{n}\right]$

  $A=\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(\frac{12}{n}-\frac{8k}{n^2}-\frac{2k^2}{n^3}\right)$

  $A=\underset{n\to \infty }{\lim }\left(\frac{1}{n}\sum _{k=1}^{n}12-\frac{8}{n^2}\sum _{k=1}^{n}k-\frac{2}{n^3}\sum _{k=1}^n{k}^2\right)$

  $A=\underset{n\to \infty }{\lim }\left[\frac{12n}{n}-\frac{8}{n^2}\cdot \frac{n\left(n+1\right)}{2}-\frac{2}{n^3}\cdot \frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]$

  $A=\underset{n\to \infty }{\lim }\left[12-4\left(\frac{n+1}{n}\right)-\frac{1}{3}\left(\frac{2n^2+3n+1}{n^2}\right)\right]$

  $A=12-4\underset{n\to \infty }{\lim }\left(\frac{1+\frac{1}{n}}{1}\right)-\frac{1}{3}\underset{n\to \infty }{\lim }\left(2+\frac{3}{n}+\frac{1}{n^2}\right)$

  $A=12-4\left(\frac{1+\frac{1}{\infty }}{1}\right)-\frac{1}{3}\left(2+\frac{3}{\infty }+\frac{1}{{\infty }^2}\right)$

  $A=12-4\left(1\right)-\frac{1}{3}\left(2\right)\text{ (Since }\frac{a}{\infty }=0\text{)}$

  $A=\frac{22}{3}$

Hence,

The area of the region is $\frac{22}{3}$ square units.