Chapter 5.9, Problem 16E

a.

To determine

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

Answer to Problem 16E

approximate the given integral ${\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx\approx }9.649753$ .

Explanation of Solution

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

  ${\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx,n=10}$

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.

  $\begin{array}{l}{\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx,n=10}\\ {\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx\approx }\frac{\Delta x}{2}\left[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+\cdot \cdot \cdot \cdot \cdot +2f\left(x_{n-1}\right)+f\left(x_n\right)\right]\\ \text{Where }\Delta x=\frac{b-a}{n}\text{ and }{x}_i=a+i\Delta x.\\ \Delta x=\frac{6-4}{10}\\ \Delta x=\frac{1}{5}\\ {\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx\approx }\frac{1}{10}\left[f\left(4\right)+2f\left(4.2\right)+2f\left(4.4\right)+2f\left(4.6\right)+2f\left(4.8\right)+2f\left(5\right)+2f\left(5.2\right)+2f\left(5.4\right)+2f\left(5.6\right)+2f\left(5.8\right)+f\left(6\right)\right]\\ {\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx\approx }9.649753\end{array}$

b.

To determine

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

Answer to Problem 16E

approximate the given integral ${\displaystyle \underset{1}{\overset{5}{\int }}\frac{\cos x}{x}dx}\approx -0.543321$ .

Explanation of Solution

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

  ${\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx,n=10}$

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.

  $\begin{array}{l}{\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx,n=10}\\ {\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx}\approx \Delta x\left[f\left(\overline{x_1}\right)+f\left(\overline{x_2}\right)+\cdot \cdot \cdot \cdot \cdot +f\left(\overline{x_{n-1}}\right)+f\left(\overline{x_n}\right)\right]\\ \text{Where }\Delta x=\frac{b-a}{n}\text{ and }\overline{x_i}=\frac{\left(x_{i-1}+x_i\right)}{2}=\text{ midpoint of }\left[x_{i-1},x_i\right]\\ \Delta x=\frac{2}{10}\\ \Delta x=\frac{1}{5}\\ {\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx}\approx \frac{1}{5}\left[f\left(4.1\right)+f\left(4.3\right)+f\left(4.5\right)+f\left(4.7\right)+f\left(4.9\right)+f\left(5.1\right)+f\left(5.3\right)+f\left(5.5\right)+f\left(5.7\right)+f\left(5.9\right)\right]\\ {\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx}\approx \frac{1}{5}\left[4.261566+4.400688+4.533942+4.661767+4.784562+4.902686+5.016464+5.126193+5.232140+5.334548\right]\\ {\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx}\approx 9.650911\end{array}$

c.

To determine

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

Answer to Problem 16E

approximate the given integral ${\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx}\approx 9.650526$ .

Explanation of Solution

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

  ${\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx,n=10}$

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.

  $\begin{array}{l}{\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx,n=10}\\ {\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx}\approx \frac{\Delta x}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+4f\left(x_3\right)+2f\left(x_4\right)+4f\left(x_5\right)+2f\left(x_6\right)+4f\left(x_7\right)+2f\left(x_8\right)+4f\left(x_9\right)+f\left(x_{10}\right)\right]\\ \Delta x=\frac{1}{5}\\ {\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx}\approx \frac{1}{15}\left[f\left(1\right)+4f\left(1.5\right)+2f\left(2\right)+4f\left(2.5\right)+2f\left(3\right)+4f\left(2\right)+2f\left(3.5\right)+4f\left(4\right)+2f\left(4.5\right)+f\left(5\right)\right]\\ {\displaystyle \underset{4}{\overset{6}{\int }}\ln \left(x^3+2\right)dx}\approx 9.650526\end{array}$