Chapter 5.9, Problem 19E

a.

To determine

To estimate ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$ .

Answer to Problem 19E

  $T_{10}=1.983,M_{10}=2.0089\text{and}{\text{S}}_{10}=1.9998$

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$

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:

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

3. Simpson rule is given by the formula as:

     $\begin{array}{l}{S}_n=\frac{\Delta x}{3}[(f(x_0)+f(x_n)+4(f(x_1)+\mathrm{........}+f(x_{n-1})+2((f(x_2)+\mathrm{........}+f(x_{n-2}))]\\ \text{where,}\Delta x=\frac{b-a}{n}\end{array}$

Here, $n=10,a=0,b=\pi ,\Delta x=\frac{\pi }{10}=0.3141,f(x)=\sin x$

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

$x$	0	0.314	0.628	0.942	1.256	1.57	1.884	2.198	2.512	2.826	3.14
$f(x)$	0	0.3088	0.5875	0.8087	0.9508	1	0.9513	0.8096	0.5888	0.3103	0

  $\begin{array}{l}{T}_{10}=\frac{0.314}{2}[(0+0)+2(0.3088+0.5875+0.8087+0.9508+1+0.9513+0.8096+0.5888+0.3103)]\\ =1.983\end{array}$

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

   $x$	0.157	0.471	0.785	1.099	1.413	1.727	2.041	2.355	2.669	2.983	0.157
   $f(x)$	0.1563	0.4537	0.7068	0.8907	0.9875	0.9878	0.8914	0.7079	0.4551	0.1579	0.1563

  $\begin{array}{l}{M}_{10}=0.314[0.1563+0.4537+0.7068+0.8907+0.9875+0.9878+0.8914+0.7079+\\ 0.4551+0.1579+0.1563)]\\ =2.0089\end{array}$

3.  To estimate the value of the integral by simpson rule:

   $x$	0	0.314	0.628	0.942	1.256	1.57	1.884	2.198	2.512	2.826	3.14
   $f(x)$	0	0.3088	0.5875	0.8087	0.9508	1	0.9513	0.8096	0.5888	0.3103	0

  $\begin{array}{l}{S}_{10}=\frac{0.314}{3}[(0+0)+4(0.3088+0.8087+1+0.8096+0.3103)+\\ 2(0.5875++0.9508+0.9513+0.5888)]\\ =1.9998\end{array}$

Hence, $T_{10}=1.983,M_{10}=2.0089\text{and}{\text{S}}_{10}=1.9998$

b.

To determine

To estimate the error bounds of Trapezoidal, Midpoint and Simpson rule.

Answer to Problem 19E

  $\left|E_T\right|=0.02583;\left|E_M\right|=0.01291\text{and}\left|E_S\right|=0.00017$

Explanation of Solution

Given information: Information in part (a)

Concept Used:

Suppose $\left|f^{″}(x)\right|\le K$ for trapezoidal and midpoint rule and $\left|f^{(4)}(x)\right|\le K$ for simpson rule, for $a\le x\le b$ . If $E_T$ , $E_M$ and $E_S$ be the error in Trapezoidal, Midpoint and Simpson Rules, then error is defined as:

  $\left|E_T\right|\le \frac{K{(b-a)}^3}{12n^2};\left|E_M\right|\le \frac{K{(b-a)}^3}{24n^2}\text{and}\left|E_S\right|\le \frac{K{(b-a)}^5}{18n^4}$

Here, $n=10,a=0,b=\pi ,K=1$

  $\begin{array}{l}\left|E_T\right|=\frac{{\pi }^3}{(12)({10}^2)}=0.02583\\ \left|E_M\right|=\frac{{\pi }^3}{(24)({10}^2)}=0.01291\\ \left|E_S\right|=\frac{{\pi }^5}{(180)({10}^4)}=0.00017\end{array}$

Hence, $\left|E_T\right|=0.02583;\left|E_M\right|=0.01291\text{and}\left|E_S\right|=0.00017$

c.

To determine

To find the value of $n$ for which the accuracy is 0.00001.

Answer to Problem 19E

The value of $n$ Trapezoidal, Midpoint and simpson Rules, are greater than 509 , greater than 359 and greater than 21.

Explanation of Solution

Given information: $\left|E_T\right|=0.02583;\left|E_M\right|=0.01291\text{and}\left|E_S\right|=0.00017$

Concept Used:

Suppose $\left|f^{″}(x)\right|\le K$ for trapezoidal and midpoint rule and $\left|f^{(4)}(x)\right|\le K$ for simpson rule, for $a\le x\le b$ . If $E_T$ , $E_M$ and $E_S$ be the error in Trapezoidal, Midpoint and Simpson Rules, then error is defined as:

  $\left|E_T\right|\le \frac{K{(b-a)}^3}{12n^2};\left|E_M\right|\le \frac{K{(b-a)}^3}{24n^2}\text{and}\left|E_S\right|\le \frac{K{(b-a)}^5}{18n^4}$

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

  $\begin{array}{l}\left|E_T\right|\le \frac{{\pi }^3}{12n^2}\le 0.00001\\ n^2\ge 258385.63\end{array}$

By trail and error, $n\ge 509$

  $\begin{array}{l}\left|E_M\right|\le \frac{{\pi }^3}{24n^2}\le 0.0001\\ n^2\ge 129192.8\end{array}$

By trail and error, $n\ge 359$

  $\begin{array}{l}\left|E_S\right|\le \frac{{\pi }^5}{180n^4}\le 0.0001\\ n^4\ge 170010.09\end{array}$

By trail and error, $n\ge 21$

Hence, for trapezoidal rule, the value of $n\ge 509$ , for midpoint rule, the value of $n\ge 359$ and for simpson rule, the value of $n\ge 21$