Chapter 12, Problem 49CLT

To determine

Find the area of the region

Answer to Problem 49CLT

The area of region is $\boxed{1.56}$ square units.

Explanation of Solution

Given information:

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

  

Calculation:

Consider the given region,

  

According to the given figure there are eight rectangles in interval of two units.

So, width of each rectangle is,

  $width=\frac{2}{8}=\frac{1}{4}$

Height of rectangles is obtained by evaluating $f$ at the right end point of each interval.

The six intervals are,

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

The right end point of each interval is,

  $\begin{array}{l}\frac{1}{4}i\\ i=-2,-1,0,1,2,3,4\end{array}$

Now sum of area of six rectangles are,

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{4}f}\left(\frac{1}{4}i\right)\left(\frac{1}{4}\right)={\displaystyle \sum _{i=1}^{0}f\left(\frac{1}{4}i\right)}\left(\frac{1}{4}\right)+{\displaystyle \sum _{i=1}^{4}f\left(\frac{1}{4}i\right)}\left(\frac{1}{4}\right)\\ ={\displaystyle \sum _{i=-2}^0\frac{1}{{\left(\frac{1}{4}i\right)}^2+1}}\left(\frac{1}{4}\right)+{\displaystyle \sum _{i=1}^4\frac{1}{{\left(\frac{1}{4}i\right)}^2+1}}\left(\frac{1}{4}\right)\\ =8{\displaystyle \sum _{i=1}^4\frac{1}{{\left(i\right)}^2+16}}\end{array}$

Apply property of summation,

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

  $\begin{array}{l}{\displaystyle \sum _{i=1}^8\left(\frac{1}{i^2+16}\right)}=1.56\\ =1.56\end{array}$

Hence the area of region is $\boxed{1.56}$ square units.