Chapter 5.9, Problem 27E

a.

To determine

To calculate: The area under the graph by using the Trapezoidal Rule .

Answer to Problem 27E

The area under the graph by using the Trapezoidal Rule is $19.9$ .

Explanation of Solution

Given information: Estimate the area under the graph in the figure

  

Formula used: Trapezoidal Rule

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx\approx \frac{\Delta x}{2}}\left[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+\cdot \cdot \cdot \cdot \cdot +2f\left(x_{n-1}\right)+f\left(x_n\right)\right]\\ \text{Where }\Delta x=\frac{b-a}{n}\text{ and }{x}_i=a+i\Delta x.\end{array}$

Calculation: Use a Trapezoidal Rule to find the area under the graph

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ n=6\\ \Delta x=\frac{6-0}{6}\\ \Delta x=1\\ T_6=\frac{\Delta x}{2}\left[f\left(0\right)+2f\left(1\right)+2f\left(2\right)+2f\left(3\right)+2f\left(4\right)+2f\left(5\right)+f\left(6\right)\right]\\ T_6=\frac{1}{2}\left[3+2\left(5\right)+2\left(4\right)+2\left(2\right)+2\left(2.9\right)+2\left(4\right)+1\right]\\ T_6=\frac{1}{2}\left[39.8\right]\\ T_6=19.9\end{array}$

b.

To determine

To calculate: The area under the graph by using theMidpoint Rule .

Answer to Problem 27E

The area under the graph by using theMidpoint Rule is $21.4$ .

Explanation of Solution

Given information: Estimate the area under the graph in the figure

  

Formula used: Midpoint Rule

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx\approx \Delta x}\left[f\left(\overline{x_1}\right)+f\left(\overline{x_2}\right)+\cdot \cdot \cdot \cdot \cdot +f\left(\overline{x_{n-1}}\right)+f\left(\overline{x_n}\right)\right]\\ \text{Where }\Delta x=\frac{b-a}{n}\text{ and }\overline{x_i}=\frac{\left(x_{i-1}+x_i\right)}{2}=\text{ midpoint of }\left[x_{i-1},x_i\right].\end{array}$

Calculation: Use a Midpoint Rule to find the area under the graph

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ n=6\\ \Delta x=\frac{6-0}{6}\\ \Delta x=1\\ M_6=\Delta x\left[f\left(0.5\right)+f\left(1.5\right)+f\left(2.5\right)+f\left(3.5\right)+f\left(4.5\right)+f\left(5.5\right)\right]\\ M_6=\left[4.5+4.8+2.8+2.3+3.7+3.3\right]\\ M_6=21.4\end{array}$

c.

To determine

To calculate: The area under the graph by using theSimpson’s Rule .

Answer to Problem 27E

The area under the graph by using theSimpson’s Rule is $20.39$ .

Explanation of Solution

Given information: Estimate the area under the graph in the figure

  

Formula used: Simpson’s Rule

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx\approx \frac{\Delta x}{3}}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+\cdot \cdot \cdot \cdot \cdot +4f\left(x_{n-1}\right)+f\left(x_n\right)\right]\\ \text{Where }n\text{ is even and }\Delta x=\frac{b-a}{n}.\end{array}$

Calculation: Use a Simpson’s Ruleto find the area under the graph

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ n=6\\ \Delta x=\frac{6-0}{6}\\ \Delta x=1\\ x_0=0\\ x_1=1\\ x_2=2\\ x_3=3\\ x_4=4\\ x_5=5\\ x_6=6\\ f\left(x_0\right)=3\\ f\left(x_1\right)=5\\ f\left(x_2\right)=4\\ f\left(x_3\right)=2\\ f\left(x_4\right)=2.9\\ f\left(x_5\right)=4\\ f\left(x_6\right)=1\\ S_6=\frac{\Delta x}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+4f\left(x_3\right)+2f\left(x_4\right)+4f\left(x_5\right)+f\left(x_6\right)\right]\\ S_6=\frac{1}{3}\left[3+4\left(5\right)+2\left(4\right)+4\left(2\right)+2\left(2.9\right)+4\left(4\right)+\left(1\right)\right]\\ S_6=20.39\end{array}$