Chapter 11, Problem 41CLT

To determine

To calculate:

The area under the following region:

  

Answer to Problem 41CLT

The area under the given region is $10.5$ square unit.

Explanation of Solution

Given information:

  

Formula Used:

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

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

Calculation:

From the given figure, understand that:

Along the $x$ axis, the length of the interval is $3units$

Numbers of rectangles: $6$

So, the width of the rectangles is given by:

  $\frac{3}{6}=\frac{1}{2}$

By evaluating the function $f$ at right end of each interval, the height of each rectangle is estimated.

So, the six intervals are given as:

  $\left(0,\frac{1}{2}\right),\left(\frac{1}{2},1\right),\left(1,\frac{3}{2}\right),\left(\frac{3}{2},2\right),\left(2,\frac{5}{2}\right),\left(\frac{5}{2},3\right)$

That means:

The right end of each interval is $\frac{1}{2}i,i=0,1,2,3,4,5,6$

The sum of areas of six rectangles is given by:

  $\begin{array}{l}{\displaystyle \sum _{k=1}^{6}f\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)}={\displaystyle \sum _{k=1}^6\left(2\times \frac{1}{2}i\right)\left(\frac{1}{2}\right)}\\ {\displaystyle \sum _{k=1}^{6}f\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)}={\displaystyle \sum _{k=1}^6\left(\frac{i}{2}\right)}\end{array}$

Use the property of the summation:

  $\begin{array}{l}{\displaystyle \sum _{k=1}^{6}f\left(\frac{1}{2}i\right)}={\displaystyle \sum _{k=1}^6\left(\frac{i}{2}\right)}\\ {\displaystyle \sum _{k=1}^{6}f\left(\frac{1}{2}i\right)}=\left(\frac{1}{2}\right)\frac{n\left(n+1\right)}{1}\\ {\displaystyle \sum _{k=1}^{6}f\left(\frac{1}{2}i\right)}=\left(\frac{1}{2}\right)\times \frac{6\left(6+1\right)}{1}\\ {\displaystyle \sum _{k=1}^{6}f\left(\frac{1}{2}i\right)}=10.5\end{array}$

Hence, the area under region is calculated as ${\displaystyle \sum _{k=1}^{6}f\left(\frac{1}{2}i\right)}=10.5$