Chapter 22, Problem 1P

Use order of $h^8$ Romberg integration to evaluate

${\displaystyle {\int }_0^{3}x{e}^{2x}}dx$

Compare ${\epsilon }_a$ and ${\epsilon }_t$.

To determine

To calculate: The value of integral, ${\displaystyle {\int }_{\text{0}}^{3}x{e}^{2x}}dx$ using order of $h^8$ Romberg integration and compare ${\epsilon }_a\text{and}{\epsilon }_t$.

Answer to Problem 1P

Solution:

The value of integral, ${\displaystyle {\int }_{\text{0}}^{3}x{e}^{2x}}dx$ is $504.69318$.

Explanation of Solution

Given:

Integral, ${\displaystyle {\int }_{\text{0}}^{3}x{e}^{2x}}dx$.

Formula used:

(1) Formula for h is given by,

$h=\frac{b-a}{n}$

(2) Formula for true error percent is,

${\epsilon }_t=\left|\frac{\text{analytical value }-I\text{ value}}{\text{analytical value}}\right|\times 100\%$

(3) Formula for relative percent error is,

${\epsilon }_a=\left|\frac{I_{1,k}-I_{2,k-1}}{I_{1,k}}\right|\times 100\%$

Calculation:

Consider the following integral:

$I={\displaystyle {\int }_{\text{0}}^{3}x{e}^{2x}}dx$

This integral is a definite integral.

For analytical solution, simplify it as follows with respect to $x$, $\begin{array}{c}{\displaystyle {\int }_{\text{0}}^{3}x{e}^{2x}}dx={\left[\frac{x{e}^{2x}}{2}-\frac{e^{2x}}{4}\right]}_0^3\\ ={\left[\frac{2x{e}^{2x}-e^{2x}}{4}\right]}_0^3\\ =\left(\frac{2\left(3\right)e^6-e^6}{4}\right)-\left(\frac{0e^0-e^0}{4}\right)\\ =504.2859+0.25\\ =504.5359\end{array}$

Hence, the analytical value of ${\displaystyle {\int }_{\text{0}}^{3}x{e}^{2x}}dx$ is 504.5359.

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

Use the application of the trapezoidal rule:

For $n=1$

, $I=\frac{h}{2}\left[f\left(x_0\right)+f\left(x_1\right)\right]$.

Here, $a=0\text{and}b=3$.

Formula for h is given by,

$h=\frac{b-a}{n}$

Substitute the value of $a=0\text{and}b=3$ and simplify it as follows,

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

Values for $x_0$ and $x_1$ are,

$\begin{array}{c}{x}_0=0\\ x_1=x_0+h\\ =0+3\\ =3\end{array}$

Substitute the values in $I$

$I=\frac{3}{2}\left[f\left(0\right)+f\left(3\right)\right]$.. .. .. (1)

Calculate $f\left(0\right)$ as

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

Calculate $f\left(3\right)$ as

$\begin{array}{c}f\left(3\right)=\left(3\right)e^{2\left(3\right)}\\ =3e^6\\ =1210.286\end{array}$

Substitute the values $f\left(0\right)=0\text{and}f\left(3\right)=1210.286$ in equation (1) and simplify it as follows,

$\begin{array}{c}I=\frac{3}{2}\left[0+1210.286\right]\\ =\frac{3\times 1210.286}{2}\\ =1815.429\end{array}$

For $n=2$

, $I=\frac{h}{2}\left[f\left(x_0\right)+2f\left(x_1\right)+f\left(x_2\right)\right]$.

Formula for h is given by,

$h=\frac{b-a}{n}$

Substitute the value of $a=0,b=3\text{ and }n=2$ and simplify it as follows,

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

Values for $x_0$

, $x_1$ and $x_2$ are,

$\begin{array}{c}{x}_0=0,\\ x_1=0+1.5\\ =1.5\\ x_2=3\end{array}$

Substitute the values in $I$

$I=\frac{1.5}{2}\left[f\left(0\right)+2f\left(1.5\right)+f\left(3\right)\right]$.. .. .. (2)

Calculate $f\left(1.5\right)$ as

$\begin{array}{c}f\left(1.5\right)=\left(1.5\right)e^{2\left(1.5\right)}\\ =1.5e^3\\ =30.1283\end{array}$

Substitute the values $f\left(0\right)=0,f\left(1.5\right)=30.1283\text{and}f\left(3\right)=1210.286$ in equation (2) and simplify it as follows,

$\begin{array}{c}I=\frac{1.5}{2}\left[0+2\left(30.1283\right)+1210.286\right]\\ =\frac{1.5\times 1270.5426}{2}\\ =952.90695\end{array}$

For $n=3$

, $I=\frac{h}{2}\left[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+2f\left(x_3\right)+f\left(x_4\right)\right]$.

Substitute the value of $a=0,b=3\text{ and }n=4$

and simplify it as follows,

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

Values for $x_0$ and $x_1$ are,

$\begin{array}{c}{x}_0=0\\ x_1=0+0.75\\ =0.75\end{array}$

Values for $x_2$ and $x_3$ are,

$\begin{array}{c}{x}_2=0.75+0.75\\ =1.5\\ x_3=1.5+0.75\\ =2.25\end{array}$

And, $x_4=3$

Substitute the values in $I$, $I=\frac{0.75}{2}\left[f\left(0\right)+2f\left(0.75\right)+2f\left(1.5\right)+2f\left(2.25\right)+f\left(3\right)\right]$.. .. .. (3)

Calculate $f\left(0.75\right)$ as

$\begin{array}{c}f\left(0.75\right)=\left(0.75\right)e^{2\left(0.75\right)}\\ =0.75e^{1.5}\\ =3.36127\end{array}$

Calculate $f\left(2.25\right)$ as

$\begin{array}{c}f\left(2.25\right)=\left(2.25\right)e^{2\left(2.25\right)}\\ =2.25e^{4.5}\\ =202.5385\end{array}$

Substitute the values $f\left(0\right)=0,f\left(1.5\right)=30.1283,f\left(0.75\right)=3.36127,f\left(2.25\right)=202.5385\text{and}f\left(3\right)=1210.286$ in equation (3) and simplify it as follows,

$\begin{array}{c}I=\frac{0.75}{2}\left[0+2\left(3.36127\right)+2\left(30.1283\right)+2\left(202.5385\right)+1210.286\right]\\ =\frac{0.75\times 1682.34214}{2}\\ =630.8783\end{array}$

For $n=4$

, $I=\frac{h}{2}\left[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+2f\left(x_3\right)+f\left(x_4\right)\right]$

Formula for h is given by,

$h=\frac{b-a}{n}$

Substitute the value of $a=0,b=3\text{ and }n=4$ and simplify it as follows,

$\begin{array}{c}h=\frac{3-0}{8}\\ =0.375\end{array}$

Values for $x_0$ and $x_1$ are,

$\begin{array}{c}{x}_0=0,\\ x_1=0+0.375\\ =0.375\end{array}$

Values for $x_2$ and $x_3$ are,

$\begin{array}{c}{x}_2=0.375+0.375\\ =0.75\\ x_3=0.75+0.375\\ =1.125\end{array}$

Values for $x_4$ and $x_5$ are,

$\begin{array}{c}{x}_4=1.125+0.375\\ =1.5\\ x_5=1.5+0.375\\ =1.875\end{array}$

Values for $x_6$ and $x_7$ are,

$\begin{array}{c}{x}_6=1.875+0.375\\ =2.25\\ x_7=2.25+0.375\\ =2.625\end{array}$

And,

$\begin{array}{c}{x}_8=2.625+0.375\\ =3\end{array}$

Substitute the values in $I$, $I=\frac{0.75}{2}\left[\begin{array}{l}f\left(0\right)+2f\left(0.375\right)+2f\left(0.75\right)+2f\left(1.125\right)+2f\left(1.5\right)+2f\left(1.875\right)\\ +2f\left(2.25\right)+2f\left(2.625\right)+f\left(3\right)\end{array}\right]$.. .. .. (4)

Calculate $f\left(0.375\right)$ as

$\begin{array}{c}f\left(0.375\right)=\left(0.375\right)e^{2\left(0.375\right)}\\ =0.375e^{0.75}\\ =0.793875\end{array}$

Calculate $f\left(1.125\right)$ as

$\begin{array}{c}f\left(1.125\right)=\left(1.125\right)e^{2\left(1.125\right)}\\ =1.125e^{2.25}\\ =10.6737\end{array}$

Calculate $f\left(1.875\right)$ as

$\begin{array}{c}f\left(1.875\right)=\left(1.875\right)e^{2\left(1.875\right)}\\ =1.875e^{3.75}\\ =79.72702\end{array}$

Calculate $f\left(2.625\right)$ as

$\begin{array}{c}f\left(2.625\right)=\left(2.625\right)e^{2\left(2.625\right)}\\ =2.625e^{5.25}\\ =500.23645\end{array}$

Substitute the values obtained in equation (4) and simplify it as follows,

$\begin{array}{c}I=\frac{0.375}{2}\left[\begin{array}{l}0+2\left(0.793875\right)+2\left(3.36127\right)+2\left(10.6737\right)+\\ 2\left(30.1283\right)+2\left(79.72702\right)+2\left(202.5385\right)+2\left(500.23645\right)+1210.286\end{array}\right]\\ =\frac{0.375\times 2865.20423}{2}\\ =537.22579\end{array}$

Use the Romberg integration method,

For $O\left(h^4\right)$

with $n=1,2$

$I\simeq \frac{4}{3}{I}_m-\frac{1}{3}{I}_l$

The $I_m$

represents more accurate value and $I_l$

represents the less accurate value.

Values for $I_m$ and $I_i$ are,

$\begin{array}{c}{I}_m=952.90695\\ I_l=1815.429\end{array}$

Substitute the values in $I$

and simplify it as follows,

$\begin{array}{c}I\cong \frac{4}{3}\left(952.90695\right)-\frac{1}{3}\left(1815.429\right)\\ \cong 1270.5426-605.143\\ \cong 665.3996\end{array}$

For $O\left(h^4\right)$

with $n=2,4$

Values for $I_m$ and $I_i$ are,

$\begin{array}{c}{I}_m=630.8783\\ I_l=952.90695\end{array}$

Substitute the values in $I$ and simplify it as follows,

$\begin{array}{c}I\cong \frac{4}{3}\left(630.8783\right)-\frac{1}{3}\left(952.90695\right)\\ \cong 841.171067-317.63565\\ \cong 523.535417\end{array}$

For $O\left(h^4\right)$

with $n=4,8$

Values for $I_m$ and $I_i$ are,

$\begin{array}{c}{I}_m=537.22579\\ I_l=630.8783\end{array}$

Substitute the values in $I$

and simplify it as follows,

$\begin{array}{c}I\cong \frac{4}{3}\left(537.22579\right)-\frac{1}{3}\left(630.8783\right)\\ \cong 716.30105-210.29276\\ \cong 506.00829\end{array}$

For $O\left(h^6\right)$

with $n=1,2$

$I\cong \frac{16}{15}{I}_m-\frac{1}{15}{I}_l$

Values for $I_m$ and $I_i$ are,

$\begin{array}{c}{I}_m=523.535417\\ I_l=665.3996\end{array}$

Substitute the values in $I$

and simplify it as follows,

$\begin{array}{c}I\cong \frac{16}{15}\left(523.535417\right)-\frac{1}{15}\left(665.3996\right)\\ \cong 558.437778-44.359973\\ \cong 514.077805\end{array}$

For $O\left(h^6\right)$

with $n=2,4$

Values for $I_m$ and $I_i$ are,

$\begin{array}{c}{I}_m=506.00829\\ I_l=523.535417\end{array}$

Substitute the values in $I$

and simplify it as follows,

$\begin{array}{c}I\simeq \frac{16}{15}\left(506.00829\right)-\frac{1}{15}\left(523.535417\right)\\ \simeq 539.742176-34.902361\\ \simeq 504.839815\end{array}$

For $O\left(h^8\right)$

with $n=4,8$

Values for $I_m$ and $I_i$ are,

$\begin{array}{c}{I}_m=504.839815\\ I_l=514.077805\end{array}$

Substitute the values in $I$

and simplify it as follows,

$\begin{array}{c}I\cong \frac{64}{63}\left(504.839815\right)-\frac{1}{63}\left(514.077805\right)\\ \cong 512.853145-8.159965\\ \cong 504.69318\end{array}$

Formula for true error percent is,

${\epsilon }_t=\left|\frac{\text{analytical value }-I\text{ value}}{\text{analytical value}}\right|\times 100\%$

Consider the following values,

$\begin{array}{c}\text{analytical value}=504.5359\\ I\text{ value}=1815.429\end{array}$

Substitute the values and simplify it as follows,

$\begin{array}{c}{\epsilon }_t=\left|\frac{504.5359-1815.429}{504.5359}\right|\times 100\%\\ =259.821\%\end{array}$

Formula for true error percent for $O\left(h^4\right)$ is,

${\epsilon }_t=\left|\frac{\text{analytical value }-I\text{ value}}{\text{analytical value}}\right|\times 100\%$

Consider the following values,

$\begin{array}{c}\text{analytical value}=504.5359\\ I\text{ value}=665.3996\end{array}$

Substitute the values and simplify it as follows,

$\begin{array}{c}{\epsilon }_t=\left|\frac{504.5359-665.3996}{504.5359}\right|\times 100\%\\ =31.88\%\end{array}$

Formula for true error percent for $O\left(h^6\right)$ is,

${\epsilon }_t=\left|\frac{\text{analytical value }-I\text{ value}}{\text{analytical value}}\right|\times 100\%$

Consider the following values,

$\begin{array}{c}\text{analytical value}=504.5359\\ I\text{ value}=514.077805\end{array}$

Substitute the values and simplify it as follows,

$\begin{array}{c}{\epsilon }_t=\left|\frac{504.5359-514.077805}{504.5359}\right|\times 100\%\\ =1.891\%\end{array}$

Formula for true error percent for $O\left(h^8\right)$ is,

${\epsilon }_t=\left|\frac{\text{analytical value }-I\text{ value}}{\text{analytical value}}\right|\times 100\%$

Consider the following values,

$\begin{array}{c}\text{analytical value}=504.5359\\ I\text{ value}=504.69318\end{array}$

Substitute the values and simplify it as follows,

$\begin{array}{c}{\epsilon }_t=\left|\frac{504.5359-504.69318}{504.5359}\right|\times 100\%\\ =0.031\%\end{array}$

Formula for relative error percent for $O\left(h^4\right)$ is,

${\epsilon }_a=\left|\frac{I_{1,k}-I_{2,k-1}}{I_{1,k}}\right|\times 100\%$

Here, the value of $k=2$ for $O\left(h^4\right)$, Now,

${\epsilon }_a=\left|\frac{I_{1,2}-I_{2,1}}{I_{1,2}}\right|\times 100\%$

Consider the following values,

$\begin{array}{c}{I}_{1,2}=665.3996\\ I_{2,1}=952.90695\end{array}$

Substitute the values and simplify it as follows,

$\begin{array}{c}{\epsilon }_a=\left|\frac{665.3996-952.90695}{665.3996}\right|\times 100\%\\ =43.208\%\end{array}$

Formula for relative error percent for $O\left(h^6\right)$ is,

${\epsilon }_a=\left|\frac{I_{1,k}-I_{2,k-1}}{I_{1,k}}\right|\times 100\%$

Here, the value of $k=3$ for $O\left(h^6\right)$, Now,

${\epsilon }_a=\left|\frac{I_{1,3}-I_{2,2}}{I_{1,3}}\right|\times 100\%$

Consider the following values,

$\begin{array}{c}{I}_{1,3}=514.077805\\ I_{2,2}=523.535417\end{array}$

Substitute the values and simplify it as follows,

$\begin{array}{c}{\epsilon }_a=\left|\frac{514.077805-523.535417}{514.077805}\right|\times 100\%\\ =1.839\%\end{array}$

Formula for relative error percent for $O\left(h^8\right)$ is,

${\epsilon }_a=\left|\frac{I_{1,k}-I_{2,k-1}}{I_{1,k}}\right|\times 100\%$

Here, the value of $k=4$ for $O\left(h^8\right)$, Now,

${\epsilon }_a=\left|\frac{I_{1,4}-I_{2,3}}{I_{1,4}}\right|\times 100\%$

Substitute the value of,

$\begin{array}{c}{I}_{1,4}=504.69318\\ I_{2,3}=504.839815\end{array}$

Substitute the values and simplify it as follows,

$\begin{array}{c}{\epsilon }_a=\left|\frac{504.69318-504.839815}{504.69318}\right|\times 100\%\\ =0.029\%\end{array}$

Consider the following table that shows $O\left(h^4\right),O\left(h^6\right),O\left(h^8\right),{\epsilon }_t,{\epsilon }_a$, $\begin{array}{ccccc}\text{Iteration}& 1& 2& 3& 4\\ {\epsilon }_t& 259.821& 31.88& 1.891& 0.031\\ {\epsilon }_a& & 43.208& 1.839& 0.029\\ 1& 1815.429& 665.3996& 514.077805& 504.69318\\ 2& 952.90695& 523.535417& 504.839815& \\ 4& 630.8783& 506.00829& & \\ 8& 537.22579& & & \end{array}$

Hence, the value of integral, ${\displaystyle {\int }_{\text{0}}^{3}x{e}^{2x}}dx$ is $504.69318$.