Chapter 12, Problem 96RE

To determine

Calculate the Area of a Region.

Answer to Problem 96RE

  $\boxed{3.53}$ square unit.

Explanation of Solution

Given information:

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

  $f\left(x\right)=4-x^2$

  

Calculation:

Consider the region.

  $f\left(x\right)=4-x^2$

  

From the given region the $4$ rectangles are shown with width equal to $1$ units.

So width of one rectangle is $\frac{1}{4}$ unit.

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

  $\left(0,\frac{1}{4}\right),\left(\frac{1}{4},\frac{1}{2}\right),\left(\frac{1}{2},\frac{3}{4}\right)and\left(\frac{3}{4},1\right)$

The right end point of each interval

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

So that the sum of areas,

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{4}f}\left(\frac{1}{4}i\right)\left(\frac{1}{4}\right)\\ ={\displaystyle \sum _{i=1}^4}\left(1-\frac{1}{64}{i}^2\right)\\ {\displaystyle \sum _{i=1}^4}\left(1-\frac{1}{64}{\displaystyle \sum _{i=1}^4{i}^2}\right)\end{array}$

Apply property of submission,

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

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{4}f\left(\frac{1}{4}i\right)\left(\frac{1}{4}\right)}\\ =4-\frac{1}{64}\left(\frac{4\times 5\times 9}{6}\right)\\ =4-\frac{180}{384}\\ =\boxed{3.53}\end{array}$

Hence the area of the region is $\boxed{3.53}$ square unit.