Chapter 5.9, Problem 5E

a.

To determine

To calculate: The given integral with the specified value of n.

Answer to Problem 5E

approximate the given integral ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx}\approx 0.806596$ .

Explanation of Solution

Given information: Use a Midpoint Rule to approximate the given integral with the specified value of n.

  ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx,n=10}$

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 approximate the given integral with the specified value of n.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx,n=10}\\ {\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx}\approx \Delta x\left[f\left(\overline{x_1}\right)+f\left(\overline{x_2}\right)+\cdot \cdot \cdot \cdot \cdot +f\left(\overline{x_{n-1}}\right)+f\left(\overline{x_n}\right)\right]\\ \Delta x=\frac{b-a}{n}\\ \Delta x=\frac{2-0}{10}\\ \Delta x=0.2\\ {\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx}\approx 0.2\left[f\left(0.10\right)+f\left(0.30\right)+f\left(0.50\right)+f\left(0.70\right)+f\left(0.90\right)+f\left(1.10\right)+f\left(1.30\right)+f\left(1.50\right)+f\left(1.70\right)+f\left(1.90\right)\right]\\ {\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx}\approx 0.2\left[0.099009+0.275229+0.4000000+0.469798+0.497237+0.497737+0.483271+0.461538+0.437018+0.412147\right]\\ {\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx}\approx 0.806596\end{array}$

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx=\frac{1}{2}}{\left[\ln \left(1+x^2\right)\right]}_0^2\\ {\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx=\frac{1}{2}}\left[\ln \left(1+4\right)-\ln 1\right]\\ {\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx\approx 0.804719}\end{array}$

The error of the approximation is about

  $0.804719-0.806596=-0.001877$

b.

To determine

To calculate: The given integral with the specified value of n.

Answer to Problem 5E

approximate the given integral ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx\approx }0.804778$ .

Explanation of Solution

Given information: Use a Simpson’s Rule to approximate the given integral with the specified value of n.

  ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx,n=10}$

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 Rule to approximate the given integral with the specified value of n.

  $\begin{array}{l}\Delta x=\frac{2-0}{10}\\ \Delta x=\frac{1}{5}\\ x_0=0\\ x_1=0.20\end{array}$

  $\begin{array}{l}{x}_2=0.40\\ x_3=0.60\\ x_4=0.80\\ x_5=1.00\end{array}$

  $\begin{array}{l}{x}_6=1.20\\ x_7=1.40\\ x_8=1.60\\ x_9=1.80\\ x_{10}=2.00\end{array}$

  $\begin{array}{l}f\left(x_0\right)=0.0\\ f\left(x_1\right)=0.192307\\ f\left(x_2\right)=0.344827\\ f\left(x_3\right)=0.441176\end{array}$

  $\begin{array}{l}f\left(x_4\right)=0.487804\\ f\left(x_5\right)=0.500000\\ f\left(x_6\right)=0.491803\\ f\left(x_7\right)=0.472973\\ f\left(x_8\right)=0.449438\end{array}$

  $\begin{array}{l}f\left(x_9\right)=0.424528\\ f\left(x_{10}\right)=0.400000\end{array}$

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx\approx }\frac{\Delta x}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+4f\left(x_3\right)+2f\left(x_4\right)+4f\left(x_5\right)+2f\left(x_6\right)+4f\left(x_7\right)+2f\left(x_8\right)+4f\left(x_9\right)+f\left(x_{10}\right)\right]\\ {\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx\approx }0.804778\end{array}$

  $\begin{array}{l}\text{And}\\ {\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx=\frac{1}{2}}{\left[\ln \left(1+x^2\right)\right]}_0^2\\ {\displaystyle \underset{0}{\overset{2}{\int }}\frac{x}{1+x^2}dx\approx 0.804719}\end{array}$

The error of the approximation is about

  $0.804719-0.804778=-0.000059$