Chapter 5, Problem 48RE

a.

To determine

To calculate: The integral value by using Trapezoidal Rule.

Answer to Problem 48RE

The integral value of the integral is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=0.3560773$ .

Explanation of Solution

Given information:

The given expression is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx$

Formula used:

Assume that f(x) is continuous on [a, b].

Let n be a positive Integer and $△x=\frac{b-a}{n}.$

If [a, b] is divided into n sub intervals, each of length $△x$ with endpoint at $P=\left\{x_0,x_1,x_2,\mathrm{...},x_n\right\}.$

  $\begin{array}{l}{T}_n=\frac{△x}{2}\left(f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+2f\left(x_{n-1}\right)+f\left(x_n\right)\right).\\ then\underset{n\to \infty }{\lim }{\text{ T}}_n={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx\end{array}$

Calculation:

Each subinterval has length

  $\begin{array}{l}△x=\frac{\frac{\pi }{2}-0}{10}\\ △x=\frac{\pi }{20}\end{array}$

$x_0$	$0$
$x_1$	$\frac{\pi }{20}$
$x_2$	$\frac{\pi }{10}$
$x_3$	$\frac{3\pi }{20}$
$x_4$	$\frac{\pi }{5}$
$x_5$	$\frac{\pi }{4}$
$x_6$	$\frac{2\pi }{10}$
$x_7$	$\frac{7\pi }{20}$
$x_9$	$\frac{2\pi }{5}$
$x_9$	$\frac{9\pi }{20}$
$x_{10}$	$\frac{\pi }{2}$

The integral can be obtained as,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=\frac{\frac{\pi }{20}}{2}\left(\begin{array}{l}f\left(0\right)+2f\left(\frac{\pi }{20}\right)+2f\left(\frac{\pi }{10}\right)+2f\left(\frac{3\pi }{20}\right)+2f\left(\frac{\pi }{5}\right)\\ +2f\left(\frac{\pi }{4}\right)+2f\left(\frac{2\pi }{10}\right)+2f\left(\frac{7\pi }{20}\right)+2f\left(\frac{2\pi }{5}\right)+f\left(\frac{9\pi }{20}\right)+f\left(\frac{\pi }{2}\right)\end{array}\right)\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=\frac{\pi }{40}\cdot \left(\begin{array}{l}0+0.0740270+0.1047187+0.798448+0.1480574\\ +1.189207+0.148094+0.1935096+0.661641+0.2176450+1\end{array}\right)\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=\frac{\pi }{40}\cdot 4.5337174\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=0.3560773\end{array}$

b.

To determine

To calculate: The integral value by using Midpoint Rule.

Answer to Problem 48RE

The integral value of the integral is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=0.3560773$ .

Explanation of Solution

Given information:

The given expression is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx$ .

Formula used:

Assume that f(x) is continuous on [a, b].

Let n be a positive Integer and $△x=\frac{b-a}{n}.$

If [a, b] is divided into n sub intervals, each of length $△x$ and $m_i$ is the midpoint of the $i^{th}$ sub interval, set

  $\begin{array}{l}{M}_n={\displaystyle \sum _{i=1}^{n}f\left(m_i\right)}△x.\\ then\underset{n\to \infty }{\lim }{M}_n={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx\end{array}$

Calculation:

Each subinterval has length

  $\begin{array}{l}△x=\frac{\frac{\pi }{2}-0}{10}\\ △x=\frac{\pi }{20}\end{array}$

$x_0$	$0$
$x_1$	$\frac{\pi }{20}$
$x_2$	$\frac{\pi }{10}$
$x_3$	$\frac{3\pi }{20}$
$x_4$	$\frac{\pi }{5}$
$x_5$	$\frac{\pi }{4}$
$x_6$	$\frac{2\pi }{10}$
$x_7$	$\frac{7\pi }{20}$
$x_9$	$\frac{2\pi }{5}$
$x_9$	$\frac{9\pi }{20}$
$x_{10}$	$\frac{\pi }{2}$

The integral can be obtained as,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=\frac{\frac{\pi }{20}}{2}\left(\begin{array}{l}f\left(0\right)+f\left(\frac{\pi }{20}\right)+f\left(\frac{\pi }{10}\right)+2f\left(\frac{3\pi }{20}\right)+2f\left(\frac{\pi }{5}\right)\\ +f\left(\frac{\pi }{4}\right)+f\left(\frac{2\pi }{10}\right)+f\left(\frac{7\pi }{20}\right)+f\left(\frac{2\pi }{5}\right)+f\left(\frac{9\pi }{20}\right)+f\left(\frac{\pi }{2}\right)\end{array}\right)\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=\left(\begin{array}{l}0+0.0740270+0.1047187+0.798448+0.1480574\\ +1.189207+0.148094+0.1935096+0.661641+0.2176450+1\end{array}\right)\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=0.3560773\end{array}$

c.

To determine

To calculate: The integral value by using Simpson’s Rule.

Answer to Problem 48RE

The integral value of the integral is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=0.183137$ .

Explanation of Solution

Given information:

The given expression is ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx$ .

Formula used:

Assume that f(x) is continuous on [a, b].

Let n be a positive Integer and $△x=\frac{b-a}{n}.$

If [a, b] is divided into n sub intervals, each of length $△x$ with endpoint at $P=\left\{x_0,x_1,x_2,\mathrm{...},x_n\right\}.$

  $\begin{array}{l}{S}_n=\frac{△x}{3}\left(f\left(x_0\right)+4\{f\left(x_1\right)+f\left(x_3\right)+f\left(x_5\right)+\mathrm{...}+f\left(x_{n-1}\right)\}+2\{f\left(x_2\right)+f\left(x_4\right)+\mathrm{...}+f\left(x_{n-2}\right)\}+f\left(x_n\right)\right).\\ then\underset{n\to \infty }{\lim }{\text{ S}}_n={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx.\end{array}$

Calculation:

The integral can be obtained as,

Each subinterval has length

  $\begin{array}{l}△x=\frac{\frac{\pi }{2}-0}{10}\\ △x=\frac{\pi }{20}\end{array}$

$x_0$	$0$
$x_1$	$\frac{\pi }{20}$
$x_2$	$\frac{\pi }{10}$
$x_3$	$\frac{3\pi }{20}$
$x_4$	$\frac{\pi }{5}$
$x_5$	$\frac{\pi }{4}$
$x_6$	$\frac{3\pi }{10}$
$x_7$	$\frac{7\pi }{20}$
$x_8$	$\frac{2\pi }{5}$
$x_9$	$\frac{9\pi }{20}$
$x_{10}$	$\frac{\pi }{2}$

The integral can be obtained as,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=\frac{\frac{\pi }{20}}{3}\left[\begin{array}{l}f\left(0\right)+4\left\{f\left(\frac{\pi }{20}\right)+f\left(\frac{3\pi }{20}\right)+f\left(\frac{\pi }{4}\right)+f\left(\frac{7\pi }{20}\right)+f\left(\frac{9\pi }{20}\right)\right\}\\ +2\left\{f\left(\frac{\pi }{10}\right)+f\left(\frac{\pi }{5}\right)+f\left(\frac{3\pi }{10}\right)+f\left(\frac{2\pi }{5}\right)\right\}+f\left(\frac{\pi }{2}\right)\end{array}\right]\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=\frac{\pi }{60}\cdot \left[\begin{array}{l}0+4\left(0.0027401+0.0081876+0.117050+0.136832+0.153898\right)\\ +2\left(0.0740472+0.104693+0.127965+0.104718\right)+1\end{array}\right]\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=\frac{\pi }{60}\cdot \left(1.674831+0.822884+1\right)\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=\frac{\pi }{60}\cdot \left(3.497677\right)\\ {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\sin x}}dx=0.183137\end{array}$