Chapter 21, Problem 2P

Evaluate the following integral:

${\displaystyle {\int }_0^3\left(1-e^{-2x}\right)}dx$

(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 ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$ analytically.

Answer to Problem 2P

Solution:

${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.501239$

Explanation of Solution

Given Information:

The integral, ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

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,

${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

The value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}={\left[x+\frac{1}{2}{e}^{-2x}\right]}_0^3\\ =\left[\left(3\right)+\frac{1}{2}{e}^{-2\left(3\right)}-\left(0\right)-\frac{1}{2}{e}^{-2\left(0\right)}\right]\\ =\left(3+0.5e^{-6}-0.5\right)\\ =2.501239\end{array}$

Therefore, the value of the integral is ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.501239$

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$ with the help of single applicationversion of the Trapezoidal rule.

Answer to Problem 2P

Solution:

${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=1.496282$ with percent relative error $40.18\%$

Explanation of Solution

Given Information:

The integral, ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.501239$

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,

$I={\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

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,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.501239=\left(3-0\right)\left[\frac{\left(1-e^{-2\left(3\right)}\right)+\left(1-e^{-2\left(0\right)}\right)}{2}\right]\\ =3\left[\frac{1-e^{-6}}{2}\right]\\ =1.496282\end{array}$

And Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{2.501239-1.496282}{2.501239}\right|\times 100\\ =40.18\%\end{array}$

Therefore, ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=1.496282$ with percent relative error $40.18\%$

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$ with the help of multiple application version of the Trapezoidal rule, with $n=2\text{ and }4$.

Answer to Problem 2P

Solution:

${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.17346\text{ with percent relative error }13.10\%\text{ when }n=2$ and ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.411051\text{ with percent relative error 3}\text{.61}\%\text{ when }n=4$

Explanation of Solution

Given Information:

The integral, $I={\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.501239$

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,

$I={\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

Here, the function is,

$f\left(x\right)=1-e^{-2x}$

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 $n=2$,

$\begin{array}{c}h=\frac{3-0}{2}\\ =1.5\end{array}$

Here, $x_0=0$

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

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

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+1.5\\ =1.5\end{array}$

The value of the function at $x_1=1.5$ is,

$\begin{array}{c}f\left(1.5\right)=1-e^{-2\left(1.5\right)}\\ =1-e^{-3}\\ =0.950213\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =1.5+1.5\\ =3\end{array}$

The value of the function at $x_2=3$,

$\begin{array}{c}f\left(3\right)=1-e^{-2\left(3\right)}\\ =1-e^{-6}\\ =0.997521\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=\frac{h}{2}\left[f\left(x_0\right)+2f\left(x_1\right)+f\left(x_2\right)\right]\\ =\frac{1.5}{2}\left[0+2\left(0.950213\right)+0.997521\right]\\ =2.17346\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{2.501239-2.17346}{2.501239}\right|\times 100\\ =13.10\%\end{array}$

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 $n=4$,

$\begin{array}{c}h=\frac{3-0}{4}\\ =0.75\end{array}$

Here, $x_0=0$

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

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

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+0.75\\ =0.75\end{array}$

The value of the function at $x_1=0.75$ is,

$\begin{array}{c}f\left(0.75\right)=1-e^{-2\left(0.75\right)}\\ =0.77687\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =0.75+0.75\\ =1.5\end{array}$

The value of the function at $x_2=1.5$,

$\begin{array}{c}f\left(1.5\right)=1-e^{-2\left(1.5\right)}\\ =0.950213\end{array}$

The value of $x_3$ is,

$\begin{array}{c}{x}_3=x_2+h\\ =1.5+0.75\\ =2.25\end{array}$

The value of the function at $x_3=2.25$ is,

$\begin{array}{c}f\left(2.25\right)=1-e^{-2\left(2.25\right)}\\ =0.988891\end{array}$

The value of $x_4$ is,

$\begin{array}{c}{x}_4=x_3+h\\ =2.25+0.75\\ =3\end{array}$

The value of the function at $x_4=3$ is,

$\begin{array}{c}f\left(3\right)=1-e^{-2\left(3\right)}\\ =0.997521\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=\frac{h}{2}\left[f\left(x_0\right)+2\left(f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)\right)+f\left(x_4\right)\right]\\ =\frac{0.75}{2}\left[0+2\left(0.77687+0.950213+0.997521\right)+0.997521\right]\\ =2.411051\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{2.501239-2.411051}{2.501239}\right|\times 100\\ =3.61\%\end{array}$

Therefore, the value of the integral when $n=2$ is ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.17346$ with percent relative error $13.10\%$ and the value of the integral when $n=4$ is ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.411051$ with percent relative error is $3.61\%$

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$ with the help of single application version of the Simpson’s ${\frac{1}{3}}^{rd}$ rule.

Answer to Problem 2P

Solution:

${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.399186$ with the percent relative error $4.08\%$

Explanation of Solution

Given Information:

The integral, ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.501239$

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,

$I={\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

Here, the function is,

$f\left(x\right)=1-e^{-2x}$

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, $n=2$,

$\begin{array}{c}h=\frac{3-0}{2}\\ =1.5\end{array}$

Here, $x_0=0$

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

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

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+1.5\\ =1.5\end{array}$

The value of the function at $x_1=1.5$ is,

$\begin{array}{c}f\left(1.5\right)=1-e^{-2\left(1.5\right)}\\ =0.950213\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =1.5+1.5\\ =3\end{array}$

The value of the function at $x_2=3$,

$\begin{array}{c}f\left(3\right)=1-e^{-2\left(3\right)}\\ =0.997521\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-e^{-2x}\right)dx}=\frac{h}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+f\left(x_2\right)\right]\\ =\frac{1.5}{3}\left[0+4\left(0.95213\right)+0.997521\right]\\ =2.399186\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{2.501239-2.399186}{2.501239}\right|\times 100\\ =4.08\%\end{array}$

Therefore, the value of the integral when $n=2$ is ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.399186$ with the percent relative error $4.08\%$

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$ with the help of multiple application version of the Simpson’s ${\frac{1}{3}}^{rd}$ rule.

Answer to Problem 2P

Solution:

${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.490248$ with percent relative error $0.44\%$

Explanation of Solution

Given Information:

The integral, ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.501239$

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,

$I={\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

Here, the function is,

$f\left(x\right)=1-e^{-2x}$

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 $n=4$,

$\begin{array}{c}h=\frac{3-0}{4}\\ =0.75\end{array}$

Here, $x_0=0$

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

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

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+0.75\\ =0.75\end{array}$

The value of the function at $x_1=0.75$ is,

$\begin{array}{c}f\left(0.75\right)=1-e^{-2\left(0.75\right)}\\ =0.77687\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =0.75+0.75\\ =1.5\end{array}$

The value of the function at $x_2=1.5$,

$\begin{array}{c}f\left(1.5\right)=1-e^{-2\left(1.5\right)}\\ =0.950213\end{array}$

The value of $x_3$ is,

$\begin{array}{c}{x}_3=x_2+h\\ =1.50+0.75\\ =2.25\end{array}$

The value of the function at $x_3=2.25$ is,

$\begin{array}{c}f\left(2.25\right)=1-e^{-2\left(2.25\right)}\\ =0.988891\end{array}$

The value of $x_4$ is,

$\begin{array}{c}{x}_4=x_3+h\\ =2.25+0.75\\ =3\end{array}$

The value of the function at $x_4=3$ is,

$\begin{array}{c}f\left(3\right)=1-e^{-2\left(3\right)}\\ =0.997521\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=\frac{h}{3}\left[f\left(x_0\right)+4\left(f\left(x_1\right)+f\left(x_3\right)\right)+2f\left(x_2\right)+f\left(x_4\right)\right]\\ =\frac{0.75}{3}\left[0+4\left(0.77687+0.988891\right)+2\left(0.950213\right)+0.997521\right]\\ =2.490248\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{2.501239-2.490248}{2.501239}\right|\times 100\\ =0.44\%\end{array}$

Therefore, the value of the integral when $n=4$ is ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.490248$ with percent relative error $0.44\%$

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$ with the help of single applicationversion of the Simpson’s ${\frac{3}{8}}^{th}$ rule.

Answer to Problem 2P

Solution:

The value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.451213$ with percent relative error $2.00\%$

Explanation of Solution

Given Information:

The integral, ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.501239$

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,

$I={\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

Here, the function is,

$f\left(x\right)=1-e^{-2x}$

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, $n=3$,

$\begin{array}{c}h=\frac{3-0}{3}\\ =1\end{array}$

Here, $x_0=0$

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

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

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+1\\ =1\end{array}$

The value of the function at $x_1=1$ is,

$\begin{array}{c}f\left(1\right)=1-e^{-2\left(1\right)}\\ =0.864665\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =1+1\\ =2\end{array}$

The value of the function at $x_2=2$,

$\begin{array}{c}f\left(2\right)=1-e^{-2\left(2\right)}\\ =0.981684\end{array}$

The value of $x_3$ is,

$\begin{array}{c}{x}_3=x_2+h\\ =2+1\\ =3\end{array}$

The value of the function at $x_3=3$,

$\begin{array}{c}f\left(3\right)=1-e^{-2\left(3\right)}\\ =0.997521\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=\frac{3h}{8}\left[f\left(x_0\right)+3\left(f\left(x_1\right)+f\left(x_2\right)\right)+f\left(x_3\right)\right]\\ =\frac{3\left(1\right)}{8}\left[0+3\left(0.864665+0.981684\right)+0.997521\right]\\ =2.451213\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{2.501239-2.451213}{2.501239}\right|\times 100\\ =2.00\%\end{array}$

Therefore, the value of the integral when $n=3$ is ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.451213$ with percent relative error $2.00\%$

To determine

To calculate: The value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$ with the help of multiple application version of the Simpson’s ${\frac{3}{8}}^{th}$ rule with $n=5$.

Answer to Problem 2P

Solution:

The value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.495933$ with percent relative error $0.21\%$

Explanation of Solution

Given Information:

The integral, ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

The exact value of the integral ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.501239$

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,

$I={\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}$

Here, the function is,

$f\left(x\right)=1-e^{-2x}$

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, $n=5$,

$\begin{array}{c}h=\frac{3-0}{5}\\ =0.6\end{array}$

Here, $x_0=0$

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

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

The value of $x_1$ is,

$\begin{array}{c}{x}_1=x_0+h\\ =0+0.6\\ =0.6\end{array}$

The value of the function at $x_1=0.6$ is,

$\begin{array}{c}f\left(0.6\right)=1-e^{-2\left(0.6\right)}\\ =0.698806\end{array}$

The value of $x_2$ is,

$\begin{array}{c}{x}_2=x_1+h\\ =0.6+0.6\\ =1.2\end{array}$

The value of the function at $x_2=1.2$,

$\begin{array}{c}f\left(1.2\right)=1-e^{-2\left(1.2\right)}\\ =0.909282\end{array}$

The value of $x_3$ is,

$\begin{array}{c}{x}_3=x_2+h\\ =1.2+0.6\\ =1.8\end{array}$

The value of the function at $x_3=1.8$,

$\begin{array}{c}f\left(1.8\right)=1-e^{-2\left(1.8\right)}\\ =0.972676\end{array}$

The value of $x_4$ is,

$\begin{array}{c}{x}_4=x_3+h\\ =1.8+0.6\\ =2.4\end{array}$

The value of the function at $x_4=2.4$,

$\begin{array}{c}f\left(2.4\right)=1-e^{-2\left(2.4\right)}\\ =0.99177\end{array}$

The value of $x_5$ is,

$\begin{array}{c}{x}_5=x_4+h\\ =2.4+0.6\\ =3\end{array}$

The value of the function at $x_5=3$,

$\begin{array}{c}f\left(3\right)=1-e^{-2\left(3\right)}\\ =0.997521\end{array}$

Thus, the value of the integral is,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=\frac{h}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+f\left(x_2\right)\right]+\\ \frac{3h}{8}\left[f\left(x_2\right)+3\left(f\left(x_3\right)+f\left(x_4\right)\right)+f\left(x_5\right)\right]\\ =\frac{0.6}{3}\left[0+4\left(0.698806\right)+0.909282\right]+\\ \frac{3\left(0.6\right)}{8}\left[0.99282+3\left(0.972676+0.99177\right)+0.997521\right]\\ =0.740901+1.755032\\ =2.495933\end{array}$

Percentage error is,

$\begin{array}{c}\%\text{ error}=\left|\frac{\text{exact value }-\text{ numerical value}}{\text{exact value}}\right|\times 100\\ =\left|\frac{2.501239-2.49533}{2.501239}\right|\times 100\\ =0.21\%\end{array}$

Therefore, the value of the integral when $n=3$ is ${\displaystyle \underset{0}{\overset{3}{\int }}\left(1-e^{-2x}\right)dx}=2.495933$ with percent relative error $0.21\%$