Chapter 7.2, Problem 46E

To determine

To find: It needs to be determined the area of the region using a right Reimann sum rule for the table shown in the textbook.

Answer to Problem 46E

The area of the region is 1.745 square meters.

Explanation of Solution

Given information: A table is shown in the textbook where a region is 2 meters tall and the distance from the left boundary to right boundary is taken at various distances from the bottom which are recorded in the table shown in the textbook.

Formula used: The Reimann sum rule is used which states for data set with unequal subintervals one has as $T_n=\frac{1}{2}{\displaystyle \sum _{i=1}^n\left(x_i-x_{i-1}\right)}\left[f\left(x_i\right)+f\left(x_{i-1}\right)\right]$ .

Calculation: The left right distance can be described by the function as $f\left(x\right)$ , so the area of the lot is given by $A={\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)dx}$ .

Now further using trapezoidal sum rule one gets as:

  $\begin{array}{c}A=\frac{1}{2}\left[\begin{array}{l}\left(0.3-0\right)\left(0.6+0.8\right)+\frac{1}{2}\left(0.8-0.3\right)\left(0.7+0.8\right)+\frac{1}{2}\left(1.1-0.8\right)\left(0.7+0.8\right)\\ +\frac{1}{2}\left(1.5-1.1\right)\left(1+0.8\right)+\frac{1}{2}\left(2-1.5\right)\left(1.3+1\right)\end{array}\right]\\ =\frac{1}{2}\left(0.3\right)\left(1.4\right)+\frac{1}{2}\left(0.5\right)\left(1.5\right)+\frac{1}{2}\left(0.4\right)\left(1.8\right)+\frac{1}{2}\left(0.5\right)\left(2.3\right)\\ =1.745\end{array}$

Hence the area of the region is 1.745 square meters.