Chapter 5, Problem 49RE

a.

To determine

To calculate: The integral value by using Trapezoidal Rule.

Answer to Problem 49RE

The error value of the integral is $Error=0.250915$ .

The value of n should be n = 5 to get the error of 0.00001 in the above integral.

Explanation of Solution

Given information:

The given expression is ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx$

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

  $\begin{array}{l}△x=\frac{1-0}{10}\\ △x=\frac{1}{10}\\ △x=0.1\end{array}$

$x_0$	0
$x_1$	0.1
$x_2$	0.2
$x_3$	0.3
$x_4$	0.4
$x_5$	0.5
$x_6$	0.6
$x_7$	0.7
$x_9$	0.8
$x_9$	0.9
$x_{10}$	1

The integral can be obtained as,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx=\frac{0.1}{2}\left(\begin{array}{l}f(0)+f\left(0.1\right)+2f\left(0.2\right)+2f\left(0.3\right)+2f\left(0.4\right)+2f\left(0.5\right)\\ +2f\left(0.6\right)+2f\left(0.7\right)+2f\left(0.8\right)+2f\left(0.9\right)+f\left(1\right)\end{array}\right)\\ {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx=0.05\cdot \left(\begin{array}{l}1+1.000045+2.001598+2.008082\\ +2.025438+2.061552+2.25652+2.2271954\\ +2.3745314+2.57379+1.41424\end{array}\right)\\ {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx=0.05\cdot \left(21.9433668\right)\\ {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx=1.04716834\end{array}$

The exact integral can be obtained as,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx\\ ={\left|x\sqrt{1+x^4}\right|}_0^1-{\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{2\sqrt{1+x^4}}4x^3\cdot xdx}\\ =\sqrt{2}-2{\displaystyle \underset{0}{\overset{1}{\int }}\frac{x^4}{\sqrt{1+x^4}}dx}\\ {\text{let 1+x}}^4=z^2\text{ then substituting the value,}\\ \begin{array}{ccc}x& 0& 1\end{array}\\ \begin{array}{ccc}z& 1& \sqrt{2}\end{array}\\ \sqrt{2}-{\displaystyle \underset{1}{\overset{\sqrt{2}}{\int }}\frac{zdz}{\left(z^2-1\right)z}}\\ =\sqrt{2}-{\displaystyle \underset{1}{\overset{\sqrt{2}}{\int }}\frac{dz}{\left(z^2-1\right)}}\\ =\sqrt{2}-{\left|\frac{1}{2\cdot 1}\log \frac{z-1}{z+1}\right|}_1^{\sqrt{2}}\\ =\sqrt{2}-\left(\left|\frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}+1}-\frac{1}{2}\log 0\right|\right)\\ =\sqrt{2}-0.116130\\ =1.298083\end{array}$

The error can be obtained as,

  $\begin{array}{l}\text{Error}=\left|\text{exact integral value}-\text{ integral value}\right|\\ \text{Error}=\left|1.298083-1.047168\right|\\ \text{Error}=0.250915\end{array}$

The value of n should be $n=5$ to get the error of $0.00001$ in the above integral.

b.

To determine

To calculate: The integral value by using Midpoint Rule.

Answer to Problem 49RE

The error value of the integral is $\text{Error}=0.250915$ .

The value of n should be n = 5 to get the error of 0.00001 in the above integral.

Explanation of Solution

Given information:

The given expression is ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx$ .

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

  $\begin{array}{l}△x=\frac{1-0}{10}\\ △x=\frac{1}{10}\\ △x=0.1\end{array}$

$x_0$	0
$x_1$	0.1
$x_2$	0.2
$x_3$	0.3
$x_4$	0.4
$x_5$	0.5
$x_6$	0.6
$x_7$	0.7
$x_9$	0.8
$x_9$	0.9
$x_{10}$	1

The integral value of the given integral can be obtained as,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx=\left[\begin{array}{l}f\left(\frac{0+0.1}{2}\right)\left(0.1\right)+f\left(\frac{0.1+0.2}{2}\right)\left(0.1\right)+f\left(\frac{0.2+0.3}{2}\right)\left(0.1\right)\\ +f\left(\frac{0.3+0.4}{2}\right)\left(0.1\right)+f\left(\frac{0.4+0.5}{2}\right)\left(0.1\right)\\ +f\left(\frac{0.5+0.6}{2}\right)\left(0.1\right)+f\left(\frac{0.6+0.7}{2}\right)\left(0.1\right)\\ +f\left(\frac{0.7+0.8}{2}\right)\left(0.1\right)+f\left(\frac{0.8+0.9}{2}\right)\left(0.1\right)+f\left(\frac{0.9+1}{2}\right)\left(0.1\right)\end{array}\right]\\ {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx=\left[\begin{array}{l}f\left(0.05\right)\left(0.1\right)+f\left(0.15\right)\left(0.1\right)+f\left(0.25\right)\left(0.1\right)\\ +f\left(0.35\right)\left(0.1\right)+f\left(0.45\right)\left(0.1\right)+f\left(0.55\right)\left(0.1\right)\\ +f\left(0.65\right)\left(0.1\right)+f\left(0.75\right)\left(0.1\right)+f\left(0.85\right)\left(0.1\right)+f\left(0.95\right)\left(0.1\right)\end{array}\right]\\ {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx=\left[\begin{array}{l}0.1000005+0.100253+0.10019512+0.1007475+0.102029\\ +0.104475+0.1085590+0.114734+0.123369+0.1347036\end{array}\right]\\ {\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{1+x^4}}dx=1.0890702\end{array}$

The error can be obtained as,

  $\begin{array}{l}\text{Error}=\left|\text{exact integral value}-\text{ integral value}\right|\\ \text{Error}=\left|1.298083-1.047168\right|\\ \text{Error}=0.250915\end{array}$

The value of n should be n = 5 to get the error of 0.00001 in the above integral.