Chapter 5.9, Problem 23E

To determine

To find:The approximations $L_n$ , $R_n$ , $T_n$ , and $M_n$ for $n=5,10,\text{and}20$ , and compute the corresponding errors $E_L$ , $E_R$ , $E_T$ , and $E_M$ .

Answer to Problem 23E

Theapproximations $L_n$ , $R_n$ , $T_n$ , and $M_n$ for $n=5$ are 0.742, 1.287, 0.9903, and 0.993 respectively. The approximations $L_n$ , $R_n$ , $T_n$ , and $M_n$ for $n=10$ are 0.867, 1.139, 0.998, and 0.998 respectively. The approximations $L_n$ , $R_n$ , $T_n$ , and $M_n$ for $n=20$ are 0.932, 1.068, 0.999, and 0.9995 respectively.

Explanation of Solution

Given information:The integral is ${\displaystyle \underset{0}{\overset{1}{\int }}x{e}^{x}dx}$ .

Calculation:

Solve the integral ${\displaystyle \underset{0}{\overset{1}{\int }}x{e}^{x}dx}$ by using integration by parts.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int }}x{e}^{x}dx}={\left[e^x\right]}_0^1-{\displaystyle \underset{0}{\overset{1}{\int }}{e}^{x}dx}\\ =e-e+1\\ =1\end{array}$

To find $M_n$ use the midpoint rule. Use the formula $\Delta x_n=\frac{b-a}{n}$ gives $\Delta x_5=\frac{1}{5}$ , $\Delta x_{10}=\frac{1}{10}$ , and $\Delta x_{20}=\frac{1}{20}$ for $a=0$ , and $b=1$ .

Use the formula $M_n=\Delta x_n\left[{\displaystyle \sum _{k=0}^{n-1}f\left({\overline{x}}_k\right)}\right]$ for $M_n$ .

The value of $M_5$ can be calculated as:

  $\begin{array}{c}{M}_5=\frac{1}{5}\left[\frac{1}{10}\exp \left(\frac{1}{10}\right)+\mathrm{...}+\frac{9}{10}\exp \left(\frac{9}{10}\right)\right]\\ \approx 0.992621\end{array}$

The value of $M_{10}$ can be calculated as:

  $\begin{array}{c}{M}_{10}=\frac{1}{10}\left[\frac{1}{20}\exp \left(\frac{1}{20}\right)+\mathrm{...}+\frac{19}{20}\exp \left(\frac{19}{20}\right)\right]\\ \approx 0.998152\end{array}$

The value of $M_{20}$ can be calculated as:

  $\begin{array}{c}{M}_{20}=\frac{1}{20}\left[\frac{1}{40}\exp \left(\frac{1}{40}\right)+\mathrm{...}+\frac{39}{40}\exp \left(\frac{39}{40}\right)\right]\\ \approx 0.999538\end{array}$

To find the $T_n$ use the trapezoidal rule. Note that the values of $\Delta x$ are the same as above.

Substitute these values into the formula for $M_n$ to obtain the estimates. The point at which evaluate $f\left(x\right)$ are given by $x_i=i\Delta x$ .

Use the formula $T_n=\frac{\Delta x}{2}\left[f\left(x_0\right)+2f\left(x_1\right)+\mathrm{...}+2f\left(x_n-1\right)+f\left(x_n\right)\right]$ .

The value of $T_5$ can be calculated as:

  $\begin{array}{c}{T}_5=\frac{1}{10}\left[\frac{1}{5}\exp \left(\frac{1}{5}\right)+{\displaystyle \sum _{k=2}^4\left(\frac{2k}{5}\right)\exp \left(\frac{k}{5}\right)+\frac{5}{5}\exp \left(\frac{5}{5}\right)}\right]\\ \approx 0.990343\end{array}$

The value of $T_{10}$ can be calculated as:

  $\begin{array}{c}{T}_{10}=\frac{1}{20}\left[\frac{1}{10}\exp \left(\frac{1}{10}\right)+{\displaystyle \sum _{k=2}^9\left(\frac{2k}{10}\right)\exp \left(\frac{k}{10}\right)+\frac{10}{10}\exp \left(\frac{10}{10}\right)}\right]\\ \approx 0.99817\end{array}$

The value of $T_{20}$ can be calculated as:

  $\begin{array}{c}{T}_{20}=\frac{1}{40}\left[\frac{1}{20}\exp \left(\frac{1}{20}\right)+{\displaystyle \sum _{k=2}^{19}\left(\frac{2k}{20}\right)\exp \left(\frac{k}{20}\right)+\frac{20}{20}\exp \left(\frac{20}{20}\right)}\right]\\ \approx 0.99961\end{array}$

Find $L_n$ , and $R_n$ is simpler. To find $L_n$ use the formula on the left and let $x_i=i\Delta x$ , where $i$ ranges from $0$ to $n-1$ . To find $R_n$ let $i$ range from $1$ to $n$ .

The formula to calculate $L_n$ and $R_n$ is given by,

  $\begin{array}{c}{L}_n={\displaystyle \sum _{i=0}^{n-1}f\left(x_i\right)\Delta x}\\ R_n={\displaystyle \sum _{i=1}^{n}f\left(x_i\right)\Delta x}\end{array}$

The value of $L_5$ can be calculated as:

  $\begin{array}{c}{L}_5={\displaystyle \sum _{i=0}^4\left(\frac{i}{5}\right)\left(\exp \left(\frac{i}{5}\right)\right)\left(\frac{1}{5}\right)}\\ \approx 0.742943\end{array}$

The value of $L_{10}$ can be calculated as:

  $\begin{array}{c}{L}_{10}={\displaystyle \sum _{i=0}^9\left(\frac{i}{10}\right)\left(\exp \left(\frac{i}{10}\right)\right)\left(\frac{1}{10}\right)}\\ \approx 0.867782\end{array}$

The value of $L_{20}$ can be calculated as:

  $\begin{array}{c}{L}_{20}={\displaystyle \sum _{i=0}^{19}\left(\frac{i}{20}\right)\left(\exp \left(\frac{i}{20}\right)\right)\left(\frac{1}{20}\right)}\\ \approx 0.932967\end{array}$

The value of $R_5$ can be calculated as:

  $\begin{array}{c}{R}_5={\displaystyle \sum _{i=0}^5\left(\frac{i}{5}\right)\left(\exp \left(\frac{i}{5}\right)\right)\left(\frac{1}{5}\right)}\\ \approx 1.286599\end{array}$

The value of $R_{10}$ can be calculated as:

  $\begin{array}{c}{R}_{10}={\displaystyle \sum _{i=0}^{10}\left(\frac{i}{10}\right)\left(\exp \left(\frac{i}{10}\right)\right)\left(\frac{1}{10}\right)}\\ \approx 1.139610\end{array}$

The value of $R_{20}$ can be calculated as:

  $\begin{array}{c}{R}_{10}={\displaystyle \sum _{i=0}^{20}\left(\frac{i}{20}\right)\left(\exp \left(\frac{i}{20}\right)\right)\left(\frac{1}{20}\right)}\\ \approx 1.068881\end{array}$

Now calculate the errors for each of the estimates. This is done simply by subtracting our estimate from the exact answer.

Errors for $E_M$ are calculated as follows:

  $\begin{array}{c}{E}_{M_5}=0.007379\\ E_{M_{10}}=0.001848\\ E_{M_{20}}=0.000462\end{array}$

Errors for $E_T$ are calculated as follows:

  $\begin{array}{c}{E}_{T_5}=0.009657\\ E_{T_{10}}=0.00183\\ E_{T_{20}}=0.00039\end{array}$

Errors for $E_L$ are calculated as follows:

  $\begin{array}{c}{E}_{L_5}=0.257057\\ E_{L_{10}}=0.132218\\ E_{L_{20}}=0.067033\end{array}$

Errors for $E_R$ are calculated as follows:

  $\begin{array}{c}{E}_{R_5}=-0.286599\\ E_{R_{10}}=-0.139610\\ E_{R_{20}}=-0.068881\end{array}$

Therefore, the approximations $L_n$ , $R_n$ , $T_n$ , and $M_n$ for $n=5$ are 0.742, 1.287, 0.9903, and 0.993respectively. The approximations $L_n$ , $R_n$ , $T_n$ , and $M_n$ for $n=10$ are 0.867, 1.139, 0.998, and 0.998 respectively.The approximations $L_n$ , $R_n$ , $T_n$ , and $M_n$ for $n=20$ are 0.932, 1.068, 0.999, and 0.9995 respectively.