Chapter 5.9, Problem 17E

a.

To determine

To estimate ${\displaystyle \underset{0}{\overset{1}{\int }}\cos (x^2)dx}$ .

Answer to Problem 17E

  $T_8=0.9023\text{and}{M}_8=0.9056$

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{1}{\int }}\cos (x^2)dx}$

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, $n=8,a=0,b=1,\Delta x=\frac{1}{8}=0.125,f(x)=\cos (x^2)$

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

$x$	0	0.125	0.25	0.375	0.5	0.625	0.75	0.875	1
$f(x)$	1	0.9998	0.9980	0.9901	0.9689	0.9246	0.8459	0.7209	0.5403

  $\begin{array}{l}{T}_8=\frac{0.125}{2}[(f(0)+f(1))+2(f(0.25)+f(0.375)+f(0.5)+f(0.625)+f(0.75)+f(0.875))]\\ =0.0625[(1+0.5403)+2(0.9980+0.9901+0.9689+0.9246+0.8459+0.7209)]=0.9023\end{array}$

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

   $x$	0.0625	0.1875	0.3125	0.4375	0.5625	0.6875	0.8125	0.9375
   $f(x)$	0.9999	0.9993	0.9952	0.9817	0.9503	0.8903	0.7898	0.6379

  $\begin{array}{l}{M}_8=0.125[f(0.625)+f(0.1875)+f(0.3125)+f(0.4375)+f(0.5625)+f(0.6875)+f(0.8125)+f(0.9375)]\\ =0.125[0.9999+0.9993+0.9952+0.9817+0.9503+0.8903+0.7898+0.6379]=0.9056\end{array}$

Hence, $T_8=0.9023\text{and}{M}_8=0.9056$

b.

To determine

To estimate the error bounds of Trapezoidal and midpoint rule.

Answer to Problem 17E

  $\left|E_T\right|=0.0078\text{and}\left|E_M\right|=0.0039$

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, $n=8,a=0,b=1,K=6$

  $\begin{array}{l}\left|E_T\right|=\frac{6}{(12)(8^2)}=0.0078\\ \left|E_M\right|=\frac{6}{(24)(8^2)}=0.0039\end{array}$

Hence, $\left|E_T\right|=0.0078\text{and}\left|E_M\right|=0.0039$

c.

To determine

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

Answer to Problem 17E

The value of $n$ Trapezoidal and Midpoint Rules, are greater than 71 and greater than 50.

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, $a=0,b=1,K=6$

  $\begin{array}{l}\left|E_T\right|\le \frac{6}{12n^2}\le 0.0001\\ n^2\ge 5000\end{array}$

By trail and error, $n\ge 71$

  $\begin{array}{l}\left|E_M\right|\le \frac{6}{24n^2}\le 0.0001\\ n^2\ge 2500\end{array}$

By trail and error, $n\ge 50$

Hence, for trapezoidal rule, the value of $n\ge 71$ and for midpoint rule, the value of $n\ge 50$