Chapter 5.9, Problem 9E

a.

To determine

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

Answer to Problem 9E

approximate the given integral ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{\ln x}{1+x}dx\approx 0.146879}$ .

Explanation of Solution

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

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

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 approximate the given integral with the specified value of n.

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

  $\begin{array}{l}\Delta x=\frac{1}{10}\text{, }{x}_i=1+i\frac{1}{10}\\ {\displaystyle \underset{1}{\overset{2}{\int }}\frac{\ln x}{1+x}dx\approx \frac{\Delta x}{2}}\left[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+\cdot \cdot \cdot \cdot \cdot +2f\left(x_{n-1}\right)+f\left(x_n\right)\right]\\ {\displaystyle \underset{1}{\overset{2}{\int }}\frac{\ln x}{1+x}dx\approx \frac{1}{20}}\left[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+\cdot \cdot \cdot \cdot \cdot +2f\left(x_{n-1}\right)+f\left(x_n\right)\right]\\ {\displaystyle \underset{1}{\overset{2}{\int }}\frac{\ln x}{1+x}dx\approx 0.146879}\end{array}$

b.

To determine

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

Answer to Problem 9E

approximate the given integral ${\displaystyle \underset{0}{\overset{1}{\int }}\frac{\ln x}{1+x}dx}\approx 0.1473916$ .

Explanation of Solution

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

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

Formula used: MidpointRule

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

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

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

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{\ln x}{1+x}dx\approx \frac{1}{10}\left(.0238+0.065006+0.099175+0.127704+0.151659+0.171865+0.188972+0.203497+0.215855+0.226383\right)}\\ {\displaystyle \underset{0}{\overset{1}{\int }}\frac{\ln x}{1+x}dx}\approx \frac{1.473916}{10}\\ {\displaystyle \underset{0}{\overset{1}{\int }}\frac{\ln x}{1+x}dx}\approx 0.1473916\end{array}$

c.

To determine

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

Answer to Problem 9E

approximate the given integral ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{\ln x}{1+x}dx}\approx 0.147219$ .

Explanation of Solution

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

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

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’sRule to approximate the given integral with the specified value of n.

0.114071

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{\ln x}{1+x}dx}\approx \frac{\Delta x}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+\cdot \cdot \cdot \cdot \cdot +4f\left(x_{n-1}\right)+f\left(x_n\right)\right]\\ \Delta x=\frac{1}{10}\\ {\displaystyle \underset{1}{\overset{2}{\int }}\frac{\ln x}{1+x}dx}\approx \frac{1}{30}\left[4(0.045386)+2(0.082873)+4(0.114071)+2(0.14019)+4(0.162186)+2(.180771)+4(0.196529)+2(0.209924)+4(0.221329)+0.232049\right]\\ {\displaystyle \underset{1}{\overset{2}{\int }}\frac{\ln x}{1+x}dx}\approx \frac{4.416583}{30}\\ {\displaystyle \underset{1}{\overset{2}{\int }}\frac{\ln x}{1+x}dx}\approx 0.147219\end{array}$