Chapter 12, Problem 95RE

To determine

Calculate the Area of a Region.

Answer to Problem 95RE

  $\boxed{6.75}$ square unit.

Explanation of Solution

Given information:

Calculate the area of the region using the indicated number of rectangles of equal width.

  $f\left(x\right)=4-x$

  

Calculation:

Consider the region.

  $f\left(x\right)=4-x$

  

From the given region the six rectangles are shown with width equal to $3$ units.

So width of one rectangle is $\frac{3}{6}=\frac{1}{2}$ unit.

Height of each rectangle is obtained by evaluating $f$ at the right end point of each interval,

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

The right end point of each interval

  $\begin{array}{l}\frac{1}{2}i\\ i=1,2,3,4,5,6\end{array}$

So that the sum of areas,

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{6}f}\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)\\ ={\displaystyle \sum _{i=1}^6}\left(4-\frac{1}{2}i\right)\left(\frac{1}{2}\right)\\ {\displaystyle \sum _{i=1}^6}\left(2-\frac{1}{4}i\right)\end{array}$

Apply property of submission,

  ${\displaystyle \sum _{i=1}^{n}i=\frac{n\left(n+1\right)\left(2n+1\right)}{6}}and{\displaystyle \sum _{i=1}^{n}c=nc}$

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{6}f\left(\frac{1}{2}i\right)\left(\frac{1}{2}\right)}\\ =2\times 6-\left(\frac{1}{4}\times \frac{6\times 7}{2}\right)\\ =6.75\end{array}$

Hence the area of the region is $\boxed{6.75}$ square unit.