Chapter 5.9, Problem 18E

a.

To determine

To estimate ${\displaystyle \underset{1}{\overset{2}{\int }}{e}^{\frac{1}{x}}dx}$ .

Answer to Problem 18E

  $T_8=0.9023\text{and}{M}_4=0.9056$

Explanation of Solution

Given information: ${\displaystyle \underset{1}{\overset{2}{\int }}{e}^{\frac{1}{x}}dx}$

Concept Used:

1. The midpoint rule is defined as:

  $\begin{array}{l}{M}_n=\Delta x[f(\overline{x_1})+f(\overline{x_2})+\mathrm{.....}+f(\overline{x_n})]\\ \text{where,}\Delta x=\frac{b-a}{n}\\ \text{and}\overline{x_i}=\frac{1}{2}\left(x_{i-1}+x_i\right)\end{array}$

2. Trapezoidal rule is given by the formula as:

     $\begin{array}{l}{T}_n=\frac{\Delta x}{2}[(f(x_0)+f(x_n)+2(f(x_1)+\mathrm{........}+f(x_{n-1})]\\ \text{where,}\Delta x=\frac{b-a}{n}\end{array}$

Here, $n=10,a=1,b=2,\Delta x=\frac{1}{10}=0.1,f(x)=e^{\frac{1}{x}}$

1. To estimate the value of the integral by trapezoidal rule:

$x$	1	1.1	1.2	1.3	1.4	1.5	1.6	1.7	1.8	1.9	2
$f(x)$	2.718	2.482	2.300	2.158	2.042	1.947	1.868	1.800	1.742	1.692	1.648

  $\begin{array}{l}{T}_{10}=\frac{0.1}{2}[(2.718+1.648)+2(2.482+2.300+2.158+2.042+1.947+1.868+1.800+1.742+1.692)]\\ =2.0214\end{array}$

2. To estimate the value of the integral by Midpoint rule:

   $x$	1.05	1.15	1.25	1.35	1.45	1.55	1.65	1.75
   $f(x)$	2.591	2.385	2.225	2.097	1.993	1.906	1.833	1.770

  $\begin{array}{l}{M}_{10}=0.1[2.591+2.385+2.225+2.097+1.9993+1.906+1.833+1.770]\\ =1.68063\end{array}$

Hence, $T_{10}=2.0214\text{and}{M}_{10}=1.68063$

b.

To determine

To estimate the error bounds of Trapezoidal and midpoint rule.

Answer to Problem 18E

  $\left|E_T\right|=0.0013\text{and}\left|E_M\right|=0.0007$

Explanation of Solution

Given information: Information in part (a)

Concept Used:

Suppose $\left|f^{″}(x)\right|\le K$ for $a\le x\le b$ . If $E_T$ and $E_M$ be the error in Trapezoidal and Midpoint Rules, then error is defined as:

  $\left|E_T\right|\le \frac{K{(b-a)}^3}{12n^2}\text{and}\left|E_M\right|\le \frac{K{(b-a)}^3}{24n^2}$

Here, $n=10,a=1,b=2,K=1.648$

  $\begin{array}{l}\left|E_T\right|=\frac{1.648}{(12)({10}^2)}=0.0013\\ \left|E_M\right|=\frac{1.648}{(24)({10}^2)}=0.0007\end{array}$

Hence, $\left|E_T\right|=0.0013\text{and}\left|E_M\right|=0.0007$

c.

To determine

To find the value of $n$ for which the accuracy is 0.0001.

Answer to Problem 18E

The value of $n$ Trapezoidal and Midpoint Rules, are greater than 37 and greater than 26.

Explanation of Solution

Given information: $\left|E_T\right|\le 0.0001\text{and}\left|E_M\right|\le 0.0001$

Concept Used:

Suppose $\left|f^{″}(x)\right|\le K$ for $a\le x\le b$ . If $E_T$ and $E_M$ be the error in Trapezoidal and Midpoint Rules, then and point rule is defined as:

  $\left|E_T\right|\le \frac{K{(b-a)}^3}{12n^2}\text{and}\left|E_M\right|\le \frac{K{(b-a)}^3}{24n^2}$

Here, $a=1,b=2,K=1.648$

  $\begin{array}{l}\left|E_T\right|\le \frac{1.648}{12n^2}\le 0.0001\\ n^2\ge 1373.33\end{array}$

By trail and error, $n\ge 37$

  $\begin{array}{l}\left|E_M\right|\le \frac{1.648}{24n^2}\le 0.0001\\ n^2\ge 686.66\end{array}$

By trail and error, $n\ge 26$

Hence, for trapezoidal rule, the value of $n\ge 37$ and for midpoint rule, the value of $n\ge 26$