Chapter 12, Problem 96RE

To determine

Approximate the area of the region using the rectangles of equal width shown.

Answer to Problem 96RE

The area under the region is 3.53 square units.

Explanation of Solution

Given:

The given function:

  $f(x)=4-x^2$

  

Here length of the interval along x -axis is 3 units and there are six rectangles, the width of the rectangle is $\frac{1}{4}$

Height of each rectangle can be obtained by evaluating $f$ at the right end point of each interval

The six intervals are as follows:

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

The right end point of each interval is $\frac{1}{4}i$ for $i=1,2,3,4.$

Therefore the sum of areas of four rectangles is

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

Use properties of summation ${\displaystyle \sum }_{i=1}^{\infty }i=\frac{n(n+1)(2n+1)}{6}$ and ${\displaystyle \sum }_{i=1}^{n}c=nc$

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

Therefore, the area under the region is 3.53 square units.