Chapter 11, Problem 43CLT

To determine

To calculate:

The area under the following region:

  

Answer to Problem 43CLT

The area under the given region is $2.69$ square unit.

Explanation of Solution

Given information:

  

Formula Used:

The property of the summation is given by the following expression:

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

And

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ {\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{2}\\ {\displaystyle \sum _{i=1}^{n}c}=cn\end{array}$

Calculation:

Assume that:

In the textbook interval $\left(0,2\right)$ the given region is exist:

So, approximate area is the sum of area of $n$

  $A={\displaystyle \sum _{i=1}^{n}f\left(a+\frac{\left(b-a\right)i}{n}\right)}\left(\frac{b-a}{n}\right)$

Now, $a=0,b=2$

  $\begin{array}{l}A={\displaystyle \sum _{i=1}^{n}f\left(a+\frac{\left(b-a\right)i}{n}\right)}\left(\frac{b-a}{n}\right)\\ A={\displaystyle \sum _{i=1}^{n}f\left(0+\frac{\left(2-0\right)i}{n}\right)}\left(\frac{2-0}{n}\right)\\ A=\frac{2}{n}{{\displaystyle \sum _{i=1}^n\frac{1}{4}\left(\left(\frac{2i}{n}\right)+1\right)}}^2\end{array}$

So, number of rectangles: $4$

The sum of areas of four rectangles is given by:

  $\begin{array}{l}A=\frac{2}{n}{{\displaystyle \sum _{i=1}^n\frac{1}{4}\left(\left(\frac{2i}{n}\right)+1\right)}}^2\\ A=\frac{2}{4}{{\displaystyle \sum _{i=1}^4\frac{1}{4}\left(\left(\frac{2i}{4}\right)+1\right)}}^2\\ A=\frac{1}{8}{\displaystyle \sum _{i=1}^4\left(\left(\frac{i^2}{4}\right)+1+i\right)}\\ A=\frac{1}{8}\left[\frac{1}{4}{\displaystyle \sum _{i=1}^4{i}^2+{\displaystyle \sum _{i=1}^{4}1}+{\displaystyle \sum _{i=1}^{4}i}}\right]\\ A=\frac{1}{8}\left[\frac{1}{4}\left(\frac{4\left(4+1\right)\left(8+1\right)}{6}\right)+\left(4\times 1\right)+\frac{4\times \left(4+1\right)}{2}\right]\\ A=\frac{1}{8}\left[\frac{43}{2}\right]\\ A=2.69\end{array}$

Hence, the area under region is calculated as $\frac{2}{4}{{\displaystyle \sum _{i=1}^4\frac{1}{4}\left(\left(\frac{2i}{4}\right)+1\right)}}^2=2.69$