Scilab/C4/Integration/English - Revision history

Nancyvarkey at 02:37, 22 December 2013

2013-12-22T02:37:40Z

Nancyvarkey at 02:34, 22 December 2013

2013-12-22T02:34:51Z

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Lavitha Pereira at 04:46, 18 December 2013

2013-12-18T04:46:50Z

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Lavitha Pereira: Created page with ''''Title of script''': Numerical Methods for Integration '''Author: Shamika''' '''Keywords: Integration, Numerical Methods, integral''' {| style="border-spacing:0;" !
2013-06-13T05:22:30Z

Created page with ''''Title of script''': Numerical Methods for Integration '''Author: Shamika''' '''Keywords: Integration, Numerical Methods, integral''' {| style="border-spacing:0;" ! <center…'

New page
'''Title of script''': Numerical Methods for Integration

'''Author: Shamika'''

'''Keywords: Integration, Numerical Methods, integral'''

{| style="border-spacing:0;"
! <center>Visual Cue</center>
! <center>Narration</center>

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 1
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Dear Friends,

Welcome to the Spoken Tutorial on “''' Composite Numerical Integration'''”''' '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 2,3 -Learning Objective Slide
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| At the end of this tutorial, you will learn how to:

* Develop '''Scilab''' code for different '''Composite Numerical Integration algorithms'''
* Divide the''' integral''' into equal '''intervals'''
* Apply the algorithm to each '''interval'''
* Calculate the '''composite value of the integral'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 4-System Requirement slide
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * To record this tutorial, I am using '''Ubuntu 12.04''' as the operating system with '''Scilab 5.3.3''' version

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 5- Prerequisites slide
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Before practising this tutorial, a learner should have basic knowledge of '''Scilab and Integration using Numerical Methods'''

* For Scilab, please refer to the relevant tutorials available on the '''Spoken Tutorial '''website.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 6- Numerical Integration
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| '''Numerical Integration''' is the:
* Study of how the numerical value of an '''integral '''can be found
* It is used when exact mathematical integration is not available
* It approximates a definite '''integral '''from values of the <br/> '''integrand '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 7,8- Composite Trapezoidal Rule-I
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| '''Composite Trapezoidal Rule''' is

* The extension of '''trapezoidal rule'''
* We divide the interval '''a comma b '''into n equal intervals
* Then,
* '''h equal to b minus a divided by n''' is the common length of the intervals
* Then '''composite trapezoidal rule '''is given by <br/> <nowiki>[</nowiki>'''The integral of the function F of x in the interval a to b is approximately equal to h multiplied by the sum of the values of the function at x zero to x n'''<br/>

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 9- Example

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Let us solve an example using '''composite trapezoidal rule''':
* Assume the number of intervals n is equal to 10.

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Show the code for Trap_composite.sci on Scilab Editor
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Let us look at the code for C'''omposite Trapezoidal Rule '''on''' Scilab Editor'''
* We first define the function with parameters''' f , a , b , n. f refers to the function we have to solve, a is the lower limit of the integral, b is the upper limit of the integral and n is the number of intervals. '''
* '''linspace''' function is used to create 10 equal intervals between 0 and 1
* '''We find the value of the integral and store it in I1'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Click on Execute on Scilab editor and choose Save and Execute the code
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Click on '''Execute''' on '''Scilab editor''' and choose '''Save and Execute''' the code

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Switch to Scilab Console

'''<nowiki>deff ('[y]=f(x)','y=1/(2x+1)')</nowiki>'''

'''Trap_composite(f, 0, 1, 10)'''
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Define the example function by typing:
* '''d e f f open paranthesis open single quote open square bracket y close square bracket is equal to f of x close quote comma open quote y is equal to one by open paranthesis two asterisk x plus one close paranthesis close quote close paranthesis'''
* Press enter
* Type
* '''Trap underscore composite open paranthesis f comma zero comma one comma ten close paranthesis'''
* Press enter
* The answer is displayed on the console

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 10, 11- Composite Simpson's Rule
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| In '''Composite simpson's rule''', we

* decompose the interval''' '''<nowiki>[a comma b]</nowiki> into '''''n is greater than 1 '''''subintervals of equal length
* Apply '''Simpson's rule''' to each interval
* We get the value of the integral to be<br/> <nowiki>[</nowiki>'''h by 3 multiplied by the sum of f zero, 4 into f one , 2 into f two to f n''']

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 12- Example

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Let us solve an example using '''Composite Simpson's rule'''
* We are given a '''function one by one plus x cube d x in the interval one to two'''
* Let the number of intervals be 20

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Switch to Scilab Editor and show the code for Simp_composite.sci
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Let us look at the code for '''Composite simpson's rule'''
* '''We first define the function with parameters f , a , b , n. '''<br/> '''f refers to the function we have to solve, a is the lower limit of the integral, b is the upper limit of the integral and n is the number of intervals.'''
* We find two sets of points
* We find the value of the function with one set and multiply it with 2
* With the other set we find the value and multiply it with 4
* We sum these values and multiply it with h by 3 and store the final value in I
* Let us execute the code

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Click on Execute and choose

Save and execute the file

Simp_composite.sci
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Save and execute the file
* '''Simp underscore composite dot s c i'''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Switch to Scilab Console

'''Type '''

'''clc'''

'''<nowiki>deff ('[y]=f(x)','y=sinx+sin(2x)')</nowiki>'''

'''Simp_composite( f, 0, %pi, 20)'''
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let me clear the screen first.
* Define the function given in the example by typing
* '''<nowiki>[d e f f open paranthesis open single quote open square bracket y close square bracket is equal to f of x close quote comma open quote y is equal to sine asterisk x plus sine asterisk open paranthesis two asterisk x close paranthesis close quote close paranthesis]</nowiki>'''
* Press enter
* '''Type Simp underscore composite open paranthesis f comma one comma two comma twenty close paranthesis'''
* Press enter
* The answer is displayed on the console

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 13, 14- Composite Midpoint Rule
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us now look at '''Composite Midpoint Rule. It'''
* Integrates polynomials of degree one or less
* Divides the interval <nowiki>[ a comma b ]into n subintervals of equal width</nowiki>
* Finds the '''midpoint '''of each interval indicated by x i
* We find the sum of the values of the integral at each midpoint <br/>

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 15- Example

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us solve this problem using '''Composite Midpoint Rule'''
* '''We are given a function one minus x square d x in the interval zero to one point five'''
* We assume n is equal to 20

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Switch to Scilab Editor

Show the file mid_composite.sci

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Let us look at the code for '''Composite Midpoint rule'''
* '''We first define the function with parameters f , a , b , n. '''<br/> '''f refers to the function we have to solve, a is the lower limit of the integral, b is the upper limit of the integral and n is the number of intervals.'''
* We find the '''midpoint '''of each interval
* Find the value of '''integral''' at each '''midpoint''' and then find the sum and store it in I.
* Let us now solve the example

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Click on Execute and choose

Save and execute the file mid_composite.sci
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Save and execute the file '''mid underscore composite dot s c i '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| On the Scilab Console type:

'''clc'''

'''<nowiki>deff ('[y]=f(x)','y=1-x^2')</nowiki>'''

Type '''mid_composite(f, 0, 1.5, 20)'''
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Let me clear the screen
* We define the function given in the example by typing
* '''<nowiki>[d e f f open paranthesis open single quote open square bracket y close square bracket is equal to f of x close quote comma open quote y is equal to one minus x square close quote close paranthesis]</nowiki>'''
* Press enter
* Then type <br/> '''<nowiki>[mid underscore composite open paranthesis f comma zero comma one point five comma twenty close paranthesis]</nowiki>'''
* Press enter
* The answer is displayed on the console

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| Slide 16- Summary
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| Let us summarize this tutorial. In this tutorial we have learnt to:

* Develop '''Scilab''' code for '''numerical integration'''
* Find the value of an '''integral '''

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| '''Show Slide 17'''

'''Title: About the Spoken Tutorial Project'''

* Watch the video available at [http://spoken-tutorial.org/What_is_a_Spoken_Tutorial http://spoken-tutorial.org/What_is_a_Spoken_Tutorial]

* It summarises the Spoken Tutorial project

* If you do not have good bandwidth, you can download and watch it

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Watch the video available at the following link

* It summarises the Spoken Tutorial project

* If you do not have good bandwidth, you can download and watch it

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| '''Show Slide 18'''

'''Title: Spoken Tutorial Workshops'''

The Spoken Tutorial Project Team

* Conducts workshops using spoken tutorials

* Gives certificates for those who pass an online test

* For more details, please write to contact@spoken-tutorial.org

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| The Spoken Tutorial Project Team

* Conducts workshops using spoken tutorials

* Gives certificates for those who pass an online test

* For more details, please write to contact at spoken hyphen tutorial dot org

|-
| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:none;padding:0.097cm;"| '''Show Slide 19'''

'''Title: Acknowledgement'''

* Spoken Tutorial Project is a part of the Talk to a Teacher project

* It is supported by the National Mission on Education through ICT, MHRD, Government of India

* More information on this Mission is available at

* [http://spoken-tutorial.org/NMEICT-Intro http://spoken-][http://spoken-tutorial.org/NMEICT-Intro tutorial.org/NMEICT-][http://spoken-tutorial.org/NMEICT-Intro Intro]

| style="border-top:none;border-bottom:1pt solid #000000;border-left:1pt solid #000000;border-right:1pt solid #000000;padding:0.097cm;"| * Spoken Tutorial Project is a part of the Talk to a Teacher project

* It is supported by the National Mission on Education through ICT, MHRD, Government of India

* More information on this Mission is available at

* spoken hyphen tutorial dot org slash NMEICT hyphen Intro

|}