Chapter 22, Problem 9P

(a)

To determine

To calculate: The given integral ${\displaystyle {\int }_2^{\infty }\frac{1}{x\left(x+2\right)}dx}$ using numerical integration.

Answer to Problem 9P

Solution: The value of ${\displaystyle {\int }_2^{\infty }\frac{1}{x\left(x+2\right)}dx}$ using numerical integration is $0.34633$.

Explanation of Solution

Given Information:

The given integral is,

${\displaystyle {\int }_2^{\infty }\frac{1}{x\left(x+2\right)}dx}$

Formula used:

Simpson’s 1/3 rule.

$I_1=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

Extended Midpoint rule for $h=\frac{1}{8}$.

$I_2=\frac{1}{8}\left(f\left(t_0\right)+f\left(t_1\right)+f\left(t_2\right)+f\left(t_3\right)\right)$

Calculation:

Calculate the analytical value of integral,

$\begin{array}{c}{\displaystyle {\int }_2^{\infty }\frac{1}{x\left(x+2\right)}dx}=\frac{1}{2}{\left[\ln \left(x\right)-\ln \left(x+2\right)\right]}_2^{\infty }\\ =\frac{1}{2}\left(\left[\ln \left(\infty \right)-\ln \left(\infty +2\right)\right]-\left[\ln \left(2\right)-\ln \left(2+2\right)\right]\right)\\ =\frac{1}{2}\left(\left[0\right]-\left[-0.693\right]\right)\\ =0.3465\end{array}$

Rewrite the integral for finite intervals,

$\begin{array}{c}{\displaystyle {\int }_2^{\infty }\frac{1}{x\left(x+2\right)}dx}={\displaystyle {\int }_0^{0.5}\frac{1}{t^2}}\left(t\right)\frac{1}{\left(\frac{1}{t}+2\right)}dt\\ ={\displaystyle {\int }_0^{0.5}\frac{1}{1+2t}dt}\end{array}$

The function values are given in table below,

t	0.0625	0.125	0.1875	0.25	0.3125	0.375	0.4375	0.5
f(t)	0.941176	0.842105	0.761905	0.695652	0.64	0.592593	0.551724	0.516129

Apply numerical integration to simplify,

${\displaystyle {\int }_0^{0.5}\frac{1}{1+2t}}dt$

${\displaystyle {\int }_0^{0.5}\frac{1}{1+2t}dt}=\frac{1}{2\left(8\right)}\left[\begin{array}{l}f\left(0.0625\right)+f\left(0.125\right)+f\left(0.1875\right)+f\left(0.25\right)+f\left(0.3125\right)\\ +f\left(0.375\right)+f\left(0.4375\right)+f\left(0.5\right)\end{array}\right]$

Substitute the values from above table.

$\begin{array}{c}{\displaystyle {\int }_2^{\infty }\frac{1}{x\left(x+2\right)}dx}=\frac{1}{16}\left[\begin{array}{l}0.941176+0.842105+0.761905+\text{0}\text{.695652}+0.64\\ +0.592593+0.551724+0.516129\end{array}\right]\\ =0.34633\end{array}$

Hence, the value of ${\displaystyle {\int }_2^{\infty }\frac{1}{x\left(x+2\right)}dx}$ using numerical integration is $0.34633$.

(b)

To determine

To calculate: The given integral ${\displaystyle {\int }_0^{\infty }{e}^{-y}{\sin }^{2}ydy}$ using numerical integration.

Answer to Problem 9P

Solution: The value of ${\displaystyle {\int }_0^{\infty }{e}^{-y}{\sin }^{2}ydy}$ using numerical integration is $0.393523$.

Explanation of Solution

Given Information:

The given integral is,

${\displaystyle {\int }_0^{\infty }{e}^{-y}{\sin }^{2}ydy}$

Formula used:

Simpson’s 1/3 rule.

$I_1=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

Extended Midpoint rule for $h=\frac{1}{8}$.

$I_2=\frac{1}{8}\left(f\left(t_0\right)+f\left(t_1\right)+f\left(t_2\right)+f\left(t_3\right)\right)$

Calculation:

Calculate the analytical value of integral,

$\begin{array}{c}{\displaystyle {\int }_0^{\infty }{e}^{-y}{\sin }^{2}ydy}={\left[\frac{e^{-y}\left(-2{\sin }^{2}y+\cos \left(2y\right)-5\right)}{10}\right]}_0^{\infty }\\ =\frac{1}{10}{\left[e^{-y}\left(-2{\sin }^{2}y+\cos \left(2y\right)-5\right)\right]}_0^{\infty }\\ =\frac{1}{10}\left(e^{-\infty }\left(-2{\sin }^2\infty +\cos \left(2\cdot \infty \right)-5\right)\right)-\left(e^{-0}\left(-2{\sin }^{2}0+\cos \left(2\cdot 0\right)-5\right)\right)\end{array}$

Since $e^{-\infty }=0$

. So integral is,

$\begin{array}{c}{\displaystyle {\int }_0^{\infty }{e}^{-y}{\sin }^{2}ydy}=\frac{1}{10}\left(\left(0\right)-\left(1\left(0+1-5\right)\right)\right)\\ =\frac{1}{10}\left(\left(0\right)-\left(1\left(-4\right)\right)\right)\\ =\frac{1}{10}\left(4\right)\\ =0.4\end{array}$

The function values are given in table below,

x	0	0.0625	0.125	0.1875	0.25	0.3125	0.375	0.4375	0.5
f(x)	0	0.048	0.139	0.219	0.26	0.258	0.222	0.168	0.112

Apply 4-application Simpson’s 1/3 rule for first part of integral,

$I_1=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

Here, $y=f\left(x\right)$ and $h=\frac{1}{8}$

Substitute the values from above table,

$\begin{array}{c}{I}_1=\left(2-0\right)\frac{0+4\left(0.048+0.219+0.258+0.168\right)+2\left(0.139+0.26+0.222\right)+0.112}{24}\\ =0.344115\end{array}$

Rearrange the integral for calculation of second integral,

$I_2={\displaystyle {\int }_0^{\frac{1}{2}}\frac{1}{t^2}}{e}^{-1/t}{\sin }^2\frac{1}{t}dt$

The function values are changed for rearranged integral which is,

t	0	1	2	3
f(t)	0	0.0908	0.00142	0.303

Apply extended midpoint rule with $h=\frac{1}{8}$.

$I_2=\frac{1}{8}\left(f\left(t_0\right)+f\left(t_1\right)+f\left(t_2\right)+f\left(t_3\right)\right)$

Substitute the values from above table,

$\begin{array}{c}{I}_2=\frac{1}{8}\left(0+0.0908+0.00142+0.303\right)\\ =0.0494\end{array}$

The total integral is,

$I=I_1+I_2$

Substitute the value from above,

$\begin{array}{c}I=0.344115+0.0494\\ =0.3935232\end{array}$

Hence, the value of ${\displaystyle {\int }_0^{\infty }{e}^{-y}{\sin }^{2}ydy}$ using numerical integration is $0.4$.

(c)

To determine

To calculate: The given integral ${\displaystyle {\int }_0^{\infty }\frac{1}{\left(1+y^2\right)\left(1+\frac{y^2}{2}\right)}dy}$ using numerical integration.

Answer to Problem 9P

Solution: The value of ${\displaystyle {\int }_0^{\infty }\frac{1}{\left(1+y^2\right)\left(1+\frac{y^2}{2}\right)}dy}$ using numerical integration is $0.919813$.

Explanation of Solution

Given Information:

The given integral is,

${\displaystyle {\int }_0^{\infty }\frac{1}{\left(1+y^2\right)\left(1+\frac{y^2}{2}\right)}dy}$

Formula used:

Simpson’s 1/3 rule.

$I_1=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

Extended Midpoint rule for $h=\frac{1}{8}$.

$I_2=\frac{1}{8}\left(f\left(t_0\right)+f\left(t_1\right)+f\left(t_2\right)+f\left(t_3\right)\right)$

Calculation:

Calculate the analytical value of the integral,

$\begin{array}{c}{\displaystyle {\int }_0^{\infty }\frac{1}{\left(1+y^2\right)\left(1+\frac{y^2}{2}\right)}dy}={\displaystyle {\int }_0^{\infty }\frac{1}{\left(y^2+1\right)\left(\frac{2+y^2}{2}\right)}dy}\\ ={\displaystyle {\int }_0^{\infty }\frac{2}{\left(y^2+1\right)\left(2+y^2\right)}dy}\\ =2{\displaystyle {\int }_0^{\infty }\frac{1}{\left(y^2+1^2\right)\left(y^2+{\left(\sqrt{2}\right)}^2\right)}dy}\end{array}$

Recall the formula,

$\int \frac{1}{y^2+a^2}dy=\frac{1}{a}{\tan }^{-1}\left(\frac{y}{a}\right)$

Simplify further,

$\begin{array}{c}{\displaystyle {\int }_0^{\infty }\frac{1}{\left(1+y^2\right)\left(1+\frac{y^2}{2}\right)}dy}=2{\left[{\tan }^{-1}\left(y\right)-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{y}{\sqrt{2}}\right)\right]}_0^{\infty }\\ =2\left[\left({\tan }^{-1}\left(\infty \right)-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\infty }{\sqrt{2}}\right)\right)-\left({\tan }^{-1}\left(0\right)-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{0}{\sqrt{2}}\right)\right)\right]\end{array}$

Substitute ${\tan }^{-1}\left(\infty \right)=\frac{\pi }{2}$ and ${\tan }^{-1}\left(0\right)=0$,

$\begin{array}{c}{\displaystyle {\int }_0^{\infty }\frac{1}{\left(1+y^2\right)\left(1+\frac{y^2}{2}\right)}dy}=2\left[\left(\frac{\pi }{2}-\frac{1}{\sqrt{2}}\left(\frac{\pi }{2}\right)\right)-\left(0-\frac{1}{\sqrt{2}}\left(0\right)\right)\right]\\ =2\left(\frac{\pi }{2}-\frac{\pi }{2\sqrt{2}}\right)-\left(0-0\right)\\ =\cancel{2}\times \left(\frac{\pi \left(\sqrt{2}-1\right)}{\cancel{2}\sqrt{2}}\right)\\ =\frac{1.30125}{\sqrt{2}}\end{array}$

Thus, the final value of integral is,

${\displaystyle {\int }_0^{\infty }\frac{1}{\left(1+y^2\right)\left(1+\frac{y^2}{2}\right)}dy}=0.92$

Hence, the value of ${\displaystyle {\int }_0^{\infty }\frac{1}{\left(1+y^2\right)\left(1+\frac{y^2}{2}\right)}dy}$ analytically is $0.92$.

The function values are given in table below,

x	0	0.0625	0.125	0.1875	0.25	0.3125	0.375	0.4375	0.5
f(x)	1	0.9127	0.711	0.4995	0.333	0.2191	0.1448	0.0972	0.0667

Apply 4-application Simpson’s 1/3 rule for first part of integral,

$I_1=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

Here, $y=f\left(x\right)$ and $h=\frac{1}{8}$

Substitute the values from above table,

$\begin{array}{c}{I}_1=\left(2-0\right)\frac{1+4\left(0.9127+0.4995+0.2191+0.0972\right)+2\left(0.711+0.3333+0.1448\right)+0.0667}{24}\\ =0.863262\end{array}$

Rearrange the integral for calculation of second integral,

$I_2={\displaystyle {\int }_0^{\frac{1}{2}}\frac{1}{t^2\left(1+\frac{1}{t^2}\right)\left(1+\frac{1}{2t^2}\right)}}dt$

The function values are changed for rearranged integral which is,

t	0	1	2	3
f(t)	0.007722	0.063462	0.148861	0.232361

Apply extended midpoint rule with $h=\frac{1}{8}$.

$I_2=\frac{1}{8}\left(f\left(t_0\right)+f\left(t_1\right)+f\left(t_2\right)+f\left(t_3\right)\right)$

Substitute the values from above table,

$\begin{array}{c}{I}_2=\frac{1}{8}\left(0.007722+0.063462+0.148861+0.232361\right)\\ =0.056551\end{array}$

The total integral is,

$I=I_1+I_2$

Substitute the value from above,

$\begin{array}{c}I=0.863262+0.056551\\ =0.919813\end{array}$

Hence, the value of ${\displaystyle {\int }_0^{\infty }\frac{1}{\left(1+y^2\right)\left(1+\frac{y^2}{2}\right)}dy}$ using numerical integration is $0.919813$.

(d)

To determine

To calculate: The given integral ${\displaystyle {\int }_{-2}^{\infty }y{e}^{-y}dy}$ using numerical integration.

Answer to Problem 9P

Solution: The value of ${\displaystyle {\int }_{-2}^{\infty }y{e}^{-y}dy}$ using numerical integration is $-7.39695$.

Explanation of Solution

Given Information:

The given integral is,

${\displaystyle {\int }_{-2}^{\infty }y{e}^{-y}dy}$

Formula used:

Simpson’s 1/3 rule.

$I_1=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

Extended Midpoint rule for $h=\frac{1}{8}$.

$I_2=\frac{1}{8}\left(f\left(t_0\right)+f\left(t_1\right)+f\left(t_2\right)+f\left(t_3\right)\right)$

Calculation:

Calculate the analytical value of the integral,

${\displaystyle {\int }_{-2}^{\infty }y{e}^{-y}dy}$

Apply Numerical integration to simplify,

$\begin{array}{c}{\displaystyle {\int }_{-2}^{\infty }y{e}^{-y}dy}={\left[\left(-y-1\right)e^{-y}\right]}_{-2}^{\infty }\\ =\left[\left(-\infty -1\right)e^{-\infty }-\left(-\left(-2\right)-1\right)e^{-\left(-2\right)}\right]\\ =0-\left(2-1\right)e^2\end{array}$

Further simplify,

$\begin{array}{c}0-\left(2-1\right)e^2=-e^2\\ =-7.39\end{array}$

Hence, the value of ${\displaystyle {\int }_{-2}^{\infty }y{e}^{-y}dy}$ analytically is $-7.39$.

The function values are given in table below,

x	0	0.0625	0.125	0.1875	0.25	0.3125	0.375	0.4375	0.5
f(x)	-14.78	-6.72	-2.72	0.824	0	0.303	0.368	0.335	0.2707

Apply 4-application Simpson’s 1/3 rule for first part of integral,

$I_1=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

Here, $y=f\left(x\right)$ and $h=\frac{1}{8}$

Substitute the values from above table,

$\begin{array}{c}{I}_1=\left(2-\left(-2\right)\right)\frac{-14.78+4\left(-6.72-0.824+0.303+0.335\right)+2\left(-2.72+0+0.368\right)+0.2707}{24}\\ =-7.807\end{array}$

Rearrange the integral for calculation of second integral,

$I_2={\displaystyle {\int }_0^{\frac{1}{2}}\frac{1}{t^3}{e}^{\frac{-1}{t}}}dt$

The function values are changed for rearranged integral which is,

t	0	1	2	3
f(t)	0.000461	0.073241	0.1335696	1.214487

Apply extended midpoint rule with $h=\frac{1}{8}$.

$I_2=\frac{1}{8}\left(f\left(t_0\right)+f\left(t_1\right)+f\left(t_2\right)+f\left(t_3\right)\right)$

Substitute the values from above table,

$\begin{array}{c}{I}_2=\frac{1}{8}\left(0.000461+0.0732418+0.1335696+1.214487\right)\\ =0.410383\end{array}$

The total integral is,

$I=I_1+I_2$

Substitute the value from above,

$\begin{array}{c}I=-7.80733+0.410383\\ =-7.39695\end{array}$

Hence, the value of ${\displaystyle {\int }_0^{\infty }\frac{1}{\left(1+y^2\right)\left(1+\frac{y^2}{2}\right)}dy}$ using numerical integration is $-7.39695$.

(e)

To determine

To calculate: The given integral ${\displaystyle {\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx}$ using numerical integration.

Answer to Problem 9P

Solution: The value of ${\displaystyle {\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx}$ using numerical integration is $0.499415$.

Explanation of Solution

Given Information:

The given integral is,

${\displaystyle {\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx}$

Formula used:

Simpson’s 1/3 rule.

$I_1=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

Extended Midpoint rule for $h=\frac{1}{8}$.

$I_2=\frac{1}{8}\left(f\left(t_0\right)+f\left(t_1\right)+f\left(t_2\right)+f\left(t_3\right)\right)$

Calculation:

Calculate the given integral,

${\displaystyle {\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx}$

Rewrite the given integral,

${\displaystyle {\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx}=\frac{1}{\sqrt{2\pi }}{\displaystyle {\int }_0^{\infty }{e}^{\frac{-x^2}{2}}dx}$

Recall the formula,

${\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-a{x}^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}}$

Apply Analytical integration to simplify for $a=\frac{1}{2}$,

$\begin{array}{c}\frac{1}{\sqrt{2\pi }}{\displaystyle {\int }_0^{\infty }{e}^{\frac{-x^2}{2}}dx}=\frac{1}{\sqrt{2\pi }}\left[\frac{1}{2}\sqrt{\frac{\pi }{\frac{1}{2}}}\right]\\ =\frac{1}{\cancel{\sqrt{2\pi }}}\left[\frac{1}{2}\cancel{\sqrt{2\pi }}\right]\\ =\frac{1}{2}\\ =0.5\end{array}$

Hence, the value of ${\displaystyle {\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx}$ analytically is $0.5$.

The function values are given in table below,

x	0	0.0625	0.125	0.1875	0.25	0.3125	0.375	0.4375	0.5
f(x)	0.399	0.387	0.352	0.301	0.242	0.183	0.130	0.086	0.054

Apply 4-application Simpson’s 1/3 rule for first part of integral,

$I_1=\frac{h}{3}\left[y_0+y_n+4\left(y_1+y_3+\mathrm{...}+y_{n-1}\right)+2\left(y_2+y_4+\mathrm{...}+y_{n-2}\right)\right]$

Here, $y=f\left(x\right)$ and $h=\frac{1}{8}$

Substitute the values from above table,

$\begin{array}{c}{I}_1=\left(2-0\right)\frac{0.399+4\left(0.387+0.301+0.183+0.086\right)+2\left(0.352+0.242+0.130\right)+0.054}{24}\\ =0.47725\end{array}$

Rearrange the integral for calculation of second integral,

$I_2={\displaystyle {\int }_0^{\frac{1}{2}}\frac{1}{\sqrt{2\pi }}\frac{1}{t^2}{e}^{\frac{-1}{2t^2}}}dt$

The function values are changed for rearranged integral which is,

t	0	1	2	3
f(t)	0	0	0.024413063	0.152922154

Apply extended midpoint rule with $h=\frac{1}{8}$.

$I_2=\frac{1}{8}\left(f\left(t_0\right)+f\left(t_1\right)+f\left(t_2\right)+f\left(t_3\right)\right)$

Substitute the values from above table,

$\begin{array}{c}{I}_2=\frac{1}{8}\left(0+0+0.024413063+0.152922154\right)\\ =0.02217\end{array}$

The total integral is,

$I=I_1+I_2$

Substitute the value from above,

$\begin{array}{c}I=0.47725+0.02217\\ =0.499415\end{array}$

Hence, the value of ${\displaystyle {\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx}$ using numerical integration is $0.499415$.