Chapter 21, Problem 1P

Evaluate the following integral:

${\displaystyle {\int }_0^{\pi /2}\left(6+3\cos x\right)}dx$

(a) Analytically;

(b) Single application of the trapezoidal rule;

(c) Multiple-application trapezoidal rule, with $n=2$ and 4;

(d) Single application of Simpson's 1/3 rule;

(e) Multiple-applicationSimpson's 1/3 rule, with $n=4$;

(f) Single application of Simpson's3/8 rule; and

(g) Multiple-application Simpson's rule, with $n=5$.

For each of the numerical estimates (b) through (g), determine the percent relative error based on (a).

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$ analytically.

Answer to Problem 1P

Solution:

The value of integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$ is $12.42478$.

Explanation of Solution

Given:

The integral, ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

Formula used:

$\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}={\left[F\left(x\right)\right]}_a^b\\ =F\left(b\right)-F\left(a\right)\end{array}$

Here, $F\left(x\right)$ is integrand of $f\left(x\right)$.

Calculation:

Consider the integral,

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

The value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}={\left[6x+3\sin x\right]}_0^{\frac{\pi }{2}}\\ =\left[6\left(\frac{\pi }{2}\right)+3\sin \left(\frac{\pi }{2}\right)-6\left(0\right)-3\sin \left(0\right)\right]\\ =\left(3\pi +3\right)\\ =12.42478\end{array}$

Therefore, the value of the integral is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42478$

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$ with the help of single applicationversion of the Trapezoidal rule. Also find percent relative error.

Answer to Problem 1P

Solution:

The value of integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$ is $12.42478$ with percent relative error $5.182\%$

Explanation of Solution

Given:

The integral, ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42478$

Formula used:

Single application version of Trapezoidal rule: If $I={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ any integral, then the value of the integral is,

$I=\left(b-a\right)\left[\frac{f\left(a\right)+f\left(b\right)}{2}\right]$.

Percentage error is,

$\text{\% error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100$

Calculation:

Consider the integral,

$I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

Single application version of Trapezoidal rule is,

$I=\left(b-a\right)\left[\frac{f\left(a\right)+f\left(b\right)}{2}\right]$

And $\text{Percentage error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100$

The value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=\left(\frac{\pi }{2}-0\right)\left[\frac{\left(6+3\cos \left(0\right)\right)+\left(6+3\cos \left(\frac{\pi }{2}\right)\right)}{2}\right]\\ =\frac{\pi }{2}\left[\frac{9+6}{2}\right]\\ =\frac{15\pi }{4}\\ =11.78097\end{array}$

And Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{12.42478-11.78097}{12.42478}\right|\times 100\\ =5.182\%\end{array}$

Therefore, the value of the integral is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=11.78097$ with percent relative error $5.182\%$.

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$ with the help of multiple application version of the Trapezoidal rule, with $n=2\text{ and }4$. Also find percent relative error.

Answer to Problem 1P

Solution:

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.26896\text{ when }n=2$ with percent relative error $1.254\%$ and ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.38613\text{ when }n=4$ with percent relative error $0.311\%$.

Explanation of Solution

Given:

The integral, ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42478$

Formula used:

Multiple application version of Trapezoidal rule: If $I={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ any integral, then the value of the integral is

$I=\frac{h}{2}\left[f\left(x_0\right)+2{\displaystyle \sum _{i=1}^{n-1}f\left(x_i\right)}+f\left(x_n\right)\right]$.

Here, $h=\frac{b-a}{n}$ is the step size and $x_{i+1}=x_i+h\text{ ; }i=0,1,2,\mathrm{...}$

Percentage error is,

$\text{\% error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100$

Calculation:

Consider the integral,

$I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

Here, the function is,

$f\left(x\right)=6+3\cos x$

Multiple application version of Trapezoidal rule is,

$I=\frac{h}{2}\left[f\left(x_0\right)+2{\displaystyle \sum _{i=1}^{n-1}f\left(x_i\right)}+f\left(x_n\right)\right]$

When $n=2$,

$\begin{array}{c}h=\frac{\frac{\pi }{2}-0}{2}\\ =\frac{\pi }{4}\end{array}$

Here, $x_0=0$

The value of the function at $x_0=0$ is,

$\begin{array}{c}f\left(0\right)=6+3\cos \left(0\right)\\ =6+3\\ =9\end{array}$

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+\frac{\pi }{4}\\ =\frac{\pi }{4}\end{array}$

The value of the function at $x_1=\frac{\pi }{4}$ is,

$\begin{array}{c}f\left(\frac{\pi }{4}\right)=6+3\cos \left(\frac{\pi }{4}\right)\\ =6+\frac{3}{\sqrt{2}}\\ =8.12132\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =\frac{\pi }{4}+\frac{\pi }{4}\\ =\frac{\pi }{2}\end{array}$

The value of the function at $x_2=\frac{\pi }{2}$,

$\begin{array}{c}f\left(\frac{\pi }{2}\right)=6+3\cos \left(\frac{\pi }{2}\right)\\ =6\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=\frac{h}{2}\left[f\left(x_0\right)+2f\left(x_1\right)+f\left(x_2\right)\right]\\ =\frac{\pi }{8}\left[9+2\left(8.12132\right)+6\right]\\ =12.26896\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{12.42478-12.26896}{12.42478}\right|\times 100\\ =1.254\%\end{array}$

Multiple application version of Trapezoidal rule is,

$I=\frac{h}{2}\left[f\left(x_0\right)+2{\displaystyle \sum _{i=1}^{n-1}f\left(x_i\right)}+f\left(x_n\right)\right]$

When $n=4$,

$\begin{array}{c}h=\frac{\frac{\pi }{2}-0}{4}\\ =\frac{\pi }{8}\end{array}$

Here, $x_0=0$

The value of the function at $x_0=0$ is,

$\begin{array}{c}f\left(0\right)=6+3\cos \left(0\right)\\ =6+3\\ =9\end{array}$

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+\frac{\pi }{8}\\ =\frac{\pi }{8}\end{array}$

The value of the function at $x_1=\frac{\pi }{8}$ is,

$\begin{array}{c}f\left(\frac{\pi }{8}\right)=6+3\cos \left(\frac{\pi }{8}\right)\\ =8.77164\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =\frac{\pi }{8}+\frac{\pi }{8}\\ =\frac{\pi }{4}\end{array}$

The value of the function at $x_2=\frac{\pi }{4}$,

$\begin{array}{c}f\left(\frac{\pi }{4}\right)=6+3\cos \left(\frac{\pi }{4}\right)\\ =6+\frac{3}{\sqrt{2}}\\ =8.12132\end{array}$

The value of $x_3$ is,

$\begin{array}{c}{x}_3=x_2+h\\ =\frac{\pi }{4}+\frac{\pi }{8}\\ =\frac{3\pi }{8}\end{array}$

The value of the function at $x_3=\frac{3\pi }{8}$ is,

$\begin{array}{c}f\left(\frac{3\pi }{8}\right)=6+3\cos \left(\frac{3\pi }{8}\right)\\ =7.14805\end{array}$

The value of $x_4$ is,

$\begin{array}{c}{x}_4=x_3+h\\ =\frac{3\pi }{8}+\frac{\pi }{8}\\ =\frac{\pi }{2}\end{array}$

The value of the function at $x_4=\frac{\pi }{2}$ is,

$\begin{array}{c}f\left(\frac{\pi }{2}\right)=6+3\cos \left(\frac{\pi }{2}\right)\\ =6\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=\frac{h}{2}\left[f\left(x_0\right)+2\left(f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)\right)+f\left(x_4\right)\right]\\ =\frac{\pi }{16}\left[9+2\left(8.77164+8.12132+7.14805\right)+6\right]\\ =12.38613\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{12.42478-12.38613}{12.42478}\right|\times 100\\ =0.311\%\end{array}$

Therefore, the value of the integral when $n=2$ is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.26896$ with percent relative error $1.254\%$ and the value of the integral when $n=4$ is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.38613$ with percent relative error $0.311\%$.

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$ with the help of single application version of the Simpson’s ${\frac{1}{3}}^{rd}$ rule. Also find percent relative error.

Answer to Problem 1P

Solution:

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.43162$ with percent relative error $0.055\%$.

Explanation of Solution

Given:

The integral, ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42478$

Formula used:

Single application version of Simpson’s ${\frac{1}{3}}^{rd}$ rule: If $I={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ any integral, then the value of the integral is

$I=\frac{h}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+f\left(x_2\right)\right]$.

Here, $h=\frac{b-a}{n}$ is the step size and $x_{i+1}=x_i+h\text{ ; }i=0,1,2,\mathrm{...}$

Percentage error is,

$\text{\% error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100$

Calculation:

Consider the integral,

$I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

Here, the function is,

$f\left(x\right)=6+3\cos x$

Single application version of Simpson’s ${\frac{1}{3}}^{rd}$ rule is,

$I=\frac{h}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+f\left(x_2\right)\right]$.

Here, $n=2$,

$\begin{array}{c}h=\frac{\frac{\pi }{2}-0}{2}\\ =\frac{\pi }{4}\end{array}$

Here, $x_0=0$

The value of the function at $x_0=0$ is,

$\begin{array}{c}f\left(0\right)=6+3\cos \left(0\right)\\ =6+3\\ =9\end{array}$

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+\frac{\pi }{4}\\ =\frac{\pi }{4}\end{array}$

The value of the function at $x_1=\frac{\pi }{4}$ is,

$\begin{array}{c}f\left(\frac{\pi }{4}\right)=6+3\cos \left(\frac{\pi }{4}\right)\\ =6+\frac{3}{\sqrt{2}}\\ =8.12132\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =\frac{\pi }{4}+\frac{\pi }{4}\\ =\frac{\pi }{2}\end{array}$

The value of the function at $x_2=\frac{\pi }{2}$,

$\begin{array}{c}f\left(\frac{\pi }{2}\right)=6+3\cos \left(\frac{\pi }{2}\right)\\ =6\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=\frac{h}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+f\left(x_2\right)\right]\\ =\frac{\pi }{12}\left[9+4\left(8.12132\right)+6\right]\\ =12.43162\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{12.42478-12.43162}{12.42478}\right|\times 100\\ =0.055\%\end{array}$

Therefore, the value of the integral when $n=2$ is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.43162$ with percent relative error $0.055\%$.

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$ with the help of multiple application version of the Simpson’s ${\frac{1}{3}}^{rd}$ rule. Also find percent relative error.

Answer to Problem 1P

Solution:

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42518$ with percent relative error $0.0032\%$.

Explanation of Solution

Given:

The integral, ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42478$

Formula used:

Multiple application version of Simpson’s ${\frac{1}{3}}^{rd}$ rule: If $I={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ any integral, then the value of the integral is

$I=\frac{h}{3}\left[f\left(x_0\right)+4{\displaystyle \sum _{i=1,3,5,\mathrm{...}}^{n-1}f\left(x_i\right)}+2{\displaystyle \sum _{j=2,4,6,\mathrm{...}}^{n-2}f\left(x_j\right)}+f\left(x_n\right)\right]$.

Here, $h=\frac{b-a}{n}$ is the step size and $x_{i+1}=x_i+h\text{ ; }i=0,1,2,\mathrm{...}$

Percentage error is,

$\text{\% error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100$

Calculation:

Consider the integral,

$I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

Here, the function is,

$f\left(x\right)=6+3\cos x$

Multiple application version of Simpson’s ${\frac{1}{3}}^{rd}$ rule is,

$I=\frac{h}{3}\left[f\left(x_0\right)+4{\displaystyle \sum _{i=1,3,5,\mathrm{...}}^{n-1}f\left(x_i\right)}+2{\displaystyle \sum _{j=2,4,6,\mathrm{...}}^{n-2}f\left(x_j\right)}+f\left(x_n\right)\right]$.

When $n=4$,

$\begin{array}{c}h=\frac{\frac{\pi }{2}-0}{4}\\ =\frac{\pi }{8}\end{array}$

Here, $x_0=0$

The value of the function at $x_0=0$ is,

$\begin{array}{c}f\left(0\right)=6+3\cos \left(0\right)\\ =6+3\\ =9\end{array}$

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+\frac{\pi }{8}\\ =\frac{\pi }{8}\end{array}$

The value of the function at $x_1=\frac{\pi }{8}$ is,

$\begin{array}{c}f\left(\frac{\pi }{8}\right)=6+3\cos \left(\frac{\pi }{8}\right)\\ =8.77164\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =\frac{\pi }{8}+\frac{\pi }{8}\\ =\frac{\pi }{4}\end{array}$

The value of the function at $x_2=\frac{\pi }{4}$,

$\begin{array}{c}f\left(\frac{\pi }{4}\right)=6+3\cos \left(\frac{\pi }{4}\right)\\ =6+\frac{3}{\sqrt{2}}\\ =8.12132\end{array}$

The value of $x_3$ is,

$\begin{array}{c}{x}_3=x_2+h\\ =\frac{\pi }{4}+\frac{\pi }{8}\\ =\frac{3\pi }{8}\end{array}$

The value of the function at $x_3=\frac{3\pi }{8}$ is,

$\begin{array}{c}f\left(\frac{3\pi }{8}\right)=6+3\cos \left(\frac{3\pi }{8}\right)\\ =7.14805\end{array}$

The value of $x_4$ is,

$\begin{array}{c}{x}_4=x_3+h\\ =\frac{3\pi }{8}+\frac{\pi }{8}\\ =\frac{\pi }{2}\end{array}$

The value of the function at $x_4=\frac{\pi }{2}$ is,

$\begin{array}{c}f\left(\frac{\pi }{2}\right)=6+3\cos \left(\frac{\pi }{2}\right)\\ =6\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=\frac{h}{3}\left[f\left(x_0\right)+4\left(f\left(x_1\right)+f\left(x_3\right)\right)+2f\left(x_2\right)+f\left(x_4\right)\right]\\ =\frac{\pi }{24}\left[9+4\left(8.77164+7.14805\right)+2\left(8.12132\right)+6\right]\\ =12.42518\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{12.42478-12.42518}{12.42478}\right|\times 100\\ =0.0032\%\end{array}$

Therefore, the value of the integral when $n=4$ is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42518$ with percent relative error $0.0032\%$.

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$ with the help of single applicationversion of the Simpson’s ${\frac{3}{8}}^{th}$ rule. Also find percent relative error.

Answer to Problem 1P

Solution:

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42779$ with percent relative error $0.024\%$.

Explanation of Solution

Given:

The integral, ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42478$

Formula used:

Single application version of Simpson’s ${\frac{3}{8}}^{th}$ rule: If $I={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ any integral, then the value of the integral is

$I=\frac{3h}{8}\left[f\left(x_0\right)+3\left(f\left(x_1\right)+f\left(x_2\right)\right)+f\left(x_3\right)\right]$.

Here, $h=\frac{b-a}{n}$ is the step size and $x_{i+1}=x_i+h\text{ ; }i=0,1,2,\mathrm{...}$

Percentage error is,

$\text{\% error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100$

Calculation:

Consider the integral,

$I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

Here, the function is,

$f\left(x\right)=6+3\cos x$

Single application version of Simpson’s ${\frac{3}{8}}^{th}$ rule is,

$I=\frac{3h}{8}\left[f\left(x_0\right)+3\left(f\left(x_1\right)+f\left(x_2\right)\right)+f\left(x_3\right)\right]$.

Here, $n=3$,

$\begin{array}{c}h=\frac{\frac{\pi }{2}-0}{3}\\ =\frac{\pi }{6}\end{array}$

Here, $x_0=0$

The value of the function at $x_0=0$ is,

$\begin{array}{c}f\left(0\right)=6+3\cos \left(0\right)\\ =6+3\\ =9\end{array}$

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+\frac{\pi }{6}\\ =\frac{\pi }{6}\end{array}$

The value of the function at $x_1=\frac{\pi }{6}$ is,

$\begin{array}{c}f\left(\frac{\pi }{6}\right)=6+3\cos \left(\frac{\pi }{6}\right)\\ =6+\frac{3\sqrt{3}}{2}\\ =8.598076\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =\frac{\pi }{6}+\frac{\pi }{6}\\ =\frac{\pi }{3}\end{array}$

The value of the function at $x_2=\frac{\pi }{3}$,

$\begin{array}{c}f\left(\frac{\pi }{3}\right)=6+3\cos \left(\frac{\pi }{3}\right)\\ =6+\frac{3}{2}\\ =7.5\end{array}$

The value of $x_3$ is,

$\begin{array}{c}{x}_3=x_2+h\\ =\frac{\pi }{3}+\frac{\pi }{6}\\ =\frac{\pi }{2}\end{array}$

The value of the function at $x_3=\frac{\pi }{2}$,

$\begin{array}{c}f\left(\frac{\pi }{2}\right)=6+3\cos \left(\frac{\pi }{2}\right)\\ =6\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=\frac{3h}{8}\left[f\left(x_0\right)+3\left(f\left(x_1\right)+f\left(x_2\right)\right)+f\left(x_3\right)\right]\\ =\frac{\pi }{16}\left[9+3\left(8.598076+7.5\right)+6\right]\\ =12.42779\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{12.42478-12.42779}{12.42478}\right|\times 100\\ =0.024\%\end{array}$

Therefore, the value of the integral when $n=3$ is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42779$ with percent relative error $0.024\%$.

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$ with the help of multiple application version of the Simpson’s ${\frac{3}{8}}^{th}$ rule with $n=5$. Also find percent relative error.

Answer to Problem 1P

Solution:

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42503$ with percent relative error $0.002\%$.

Explanation of Solution

Given:

The integral, ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42478$

Formula used:

Multiple application version of Simpson’s ${\frac{3}{8}}^{th}$ rule for $n=5$: If $I={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ any integral, then the value of the integral is

$I=\text{Simpson's }{\frac{1}{3}}^{rd}\text{ rule for first two segment }+\text{ Simpson's }{\frac{3}{8}}^{th}\text{ rule for last three segment}$.

Single application version of Simpson’s ${\frac{1}{3}}^{rd}$ rule: If $I={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ any integral, then the value of the integral is

$I=\frac{h}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+f\left(x_2\right)\right]$.

Single application of Simpson’s ${\frac{3}{8}}^{th}$ rule: If $I={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ any integral, then the value of the integral is

$I=\frac{3h}{8}\left[f\left(x_0\right)+3\left(f\left(x_1\right)+f\left(x_2\right)\right)+f\left(x_3\right)\right]$.

Here, $h=\frac{b-a}{n}$ is the step size and $x_{i+1}=x_i+h\text{ ; }i=0,1,2,\mathrm{...}$

Percentage error is,

$\text{\% error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100$

Calculation:

Consider the integral,

$I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}$

Here, the function is,

$f\left(x\right)=6+3\cos x$

Multiple application version of Simpson’s ${\frac{3}{8}}^{th}$ rule for $n=5$: If $I={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ any integral, then the value of the integral is

$I=\text{Simpson's }{\frac{1}{3}}^{rd}\text{ rule for first two segment }+\text{ Simpson's }{\frac{3}{8}}^{th}\text{ rule for last three segment}$.

Here, $n=5$,

$\begin{array}{c}h=\frac{\frac{\pi }{2}-0}{5}\\ =\frac{\pi }{10}\end{array}$

Here, $x_0=0$

The value of the function at $x_0=0$ is,

$\begin{array}{c}f\left(0\right)=6+3\cos \left(0\right)\\ =6+3\\ =9\end{array}$

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+\frac{\pi }{10}\\ =\frac{\pi }{10}\end{array}$

The value of the function at $x_1=\frac{\pi }{10}$ is,

$\begin{array}{c}f\left(\frac{\pi }{10}\right)=6+3\cos \left(\frac{\pi }{10}\right)\\ =8.85317\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =\frac{\pi }{10}+\frac{\pi }{10}\\ =\frac{\pi }{5}\end{array}$

The value of the function at $x_2=\frac{\pi }{5}$,

$\begin{array}{c}f\left(\frac{\pi }{5}\right)=6+3\cos \left(\frac{\pi }{5}\right)\\ =8.42705\end{array}$

The value of $x_3$ is,

$\begin{array}{c}{x}_3=x_2+h\\ =\frac{\pi }{5}+\frac{\pi }{10}\\ =\frac{3\pi }{10}\end{array}$

The value of the function at $x_3=\frac{3\pi }{10}$,

$\begin{array}{c}f\left(\frac{3\pi }{10}\right)=6+3\cos \left(\frac{3\pi }{10}\right)\\ =7.763356\end{array}$

The value of $x_4$ is,

$\begin{array}{c}{x}_4=x_3+h\\ =\frac{3\pi }{10}+\frac{\pi }{10}\\ =\frac{2\pi }{5}\end{array}$

The value of the function at $x_4=\frac{2\pi }{5}$,

$\begin{array}{c}f\left(\frac{2\pi }{5}\right)=6+3\cos \left(\frac{2\pi }{5}\right)\\ =6.92705\end{array}$

The value of $x_5$ is,

$\begin{array}{c}{x}_5=x_4+h\\ =\frac{2\pi }{5}+\frac{\pi }{10}\\ =\frac{\pi }{2}\end{array}$

The value of the function at $x_5=\frac{\pi }{2}$,

$\begin{array}{c}f\left(\frac{\pi }{2}\right)=6+3\cos \left(\frac{\pi }{2}\right)\\ =6\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=\frac{h}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+f\left(x_2\right)\right]+\\ \frac{3h}{8}\left[f\left(x_2\right)+3\left(f\left(x_3\right)+f\left(x_4\right)\right)+f\left(x_5\right)\right]\\ =\frac{\pi }{30}\left[9+4\left(8.85317\right)+8.42705\right]+\\ \frac{3\pi }{80}\left[8.42705+3\left(7.763356+6.92705\right)+6\right]\\ =5.533364+6.891665\\ =12.42503\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{12.42478-12.42503}{12.42478}\right|\times 100\\ =0.002\%\end{array}$

Therefore, the value of the integral when $n=3$ is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6+3\cos x\right)dx}=12.42503$ with percent relative error $0.002\%$.