Difference between revisions of "Scilab/C4/Integration/English-timed"

From Script | Spoken-Tutorial
Jump to: navigation, search
(Created page with '{| Border=1 || Time || Narration |- | 00.01 |Dear Friends, |- | 00.02. | Welcome to the Spoken Tutorial on '''“ Composite Numerical Integration” ''' |- |00.07 |At the …')
 
Line 1: Line 1:
 
{| Border=1
 
{| Border=1
  
−
|| Time
+
|'''Time'''
−
 
+
|'''Narration'''
−
|| Narration
+
  
 
|-
 
|-
−
| 00.01
+
| 00:01
 
|Dear Friends,  
 
|Dear Friends,  
  
 
|-
 
|-
−
| 00.02.
+
| 00:02.
 
| Welcome to the Spoken Tutorial on '''“ Composite Numerical Integration” '''
 
| Welcome to the Spoken Tutorial on '''“ Composite Numerical Integration” '''
−
 
  
 
|-
 
|-
−
|00.07
+
|00:07
 
|At the end of this tutorial, you will learn how to:  
 
|At the end of this tutorial, you will learn how to:  
  
 
|-
 
|-
−
|00.11
+
|00:11
 
|Develop '''Scilab''' code for different '''Composite Numerical Integration algorithms'''  
 
|Develop '''Scilab''' code for different '''Composite Numerical Integration algorithms'''  
  
 
|-
 
|-
−
| 00.17
+
| 00:17
 
|Divide the '''integral''' into equal '''intervals'''  
 
|Divide the '''integral''' into equal '''intervals'''  
  
 
|-
 
|-
−
|00.21
+
|00:21
 
|Apply the algorithm to each '''interval''' and  
 
|Apply the algorithm to each '''interval''' and  
  
 
|-
 
|-
−
|00.24
+
|00:24
 
|Calculate the '''composite value of the integral'''
 
|Calculate the '''composite value of the integral'''
  
 
|-
 
|-
−
| 00.28
+
| 00:28
 
|To record this tutorial, I am using  
 
|To record this tutorial, I am using  
  
 
|-
 
|-
−
| 00.30
+
| 00:30
 
| '''Ubuntu 12.04''' as the operating system  
 
| '''Ubuntu 12.04''' as the operating system  
  
 
|-
 
|-
−
|00.34
+
|00:34
 
| with '''Scilab 5.3.3''' version  
 
| with '''Scilab 5.3.3''' version  
  
 
|-
 
|-
−
|00.38
+
|00:38
 
||Before practising this tutorial, a learner should have basic knowledge of  
 
||Before practising this tutorial, a learner should have basic knowledge of  
  
 
|-
 
|-
−
| 00.42
+
| 00:42
 
|'''Scilab''' and  
 
|'''Scilab''' and  
−
 
  
 
|-
 
|-
−
|00.44
+
|00:44
 
| '''Integration using Numerical Methods'''
 
| '''Integration using Numerical Methods'''
  
 
|-
 
|-
  
−
| 00.47
+
| 00:47
  
 
| For '''Scilab''', please refer to the relevant tutorials available on the '''Spoken Tutorial''' website.  
 
| For '''Scilab''', please refer to the relevant tutorials available on the '''Spoken Tutorial''' website.  
  
 
|-
 
|-
−
| 00.55
+
| 00:55
 
| '''Numerical Integration''' is the:  
 
| '''Numerical Integration''' is the:  
  
 
|-
 
|-
−
| 00.58
+
| 00:58
 
| Study of how the numerical value of an '''integral''' can be found  
 
| Study of how the numerical value of an '''integral''' can be found  
  
 
|-
 
|-
−
|01.03
+
|01:03
 
| It is used when exact mathematical integration is not available  
 
| It is used when exact mathematical integration is not available  
  
 
|-
 
|-
−
|01.08
+
|01:08
 
|It approximates a definite '''integral''' from values of the '''integrand'''  
 
|It approximates a definite '''integral''' from values of the '''integrand'''  
−
 
  
 
|-
 
|-
  
−
|01.15
+
|01:15
  
 
|Let us study '''Composite Trapezoidal Rule.'''  
 
|Let us study '''Composite Trapezoidal Rule.'''  
Line 90: Line 86:
 
|-
 
|-
  
−
|01.18
+
|01:18
  
 
|This rule is the extension of '''trapezoidal rule'''  
 
|This rule is the extension of '''trapezoidal rule'''  
Line 96: Line 92:
 
|-
 
|-
  
−
| 01.22
+
| 01:22
 
|| We divide the interval '''a comma b '''into '''n''' equal intervals  
 
|| We divide the interval '''a comma b '''into '''n''' equal intervals  
  
 
|-
 
|-
  
−
| 01.29
+
| 01:29
  
 
| Then '''h equal to b minus a divided by n''' is the common length of the intervals  
 
| Then '''h equal to b minus a divided by n''' is the common length of the intervals  
Line 107: Line 103:
 
|-
 
|-
  
−
|01.36
+
|01:36
  
 
| Then '''composite trapezoidal rule''' is given by  
 
| Then '''composite trapezoidal rule''' is given by  
Line 113: Line 109:
 
|-
 
|-
  
−
|01.41
+
|01:41
  
 
|''' 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'''
 
|''' 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'''
Line 119: Line 115:
 
|-
 
|-
  
−
|01.57
+
|01:57
  
 
|| Let us solve an example using '''composite trapezoidal rule.'''  
 
|| Let us solve an example using '''composite trapezoidal rule.'''  
Line 125: Line 121:
 
|-
 
|-
  
−
|02.02
+
|02:02
  
 
| Assume the number of intervals n is equal to ten.  
 
| Assume the number of intervals n is equal to ten.  
−
 
  
 
|-
 
|-
−
|02.09
+
|02:09
 
|Let us look at the code for '''Composite Trapezoidal Rule''' on '''Scilab Editor'''
 
|Let us look at the code for '''Composite Trapezoidal Rule''' on '''Scilab Editor'''
  
 
|-
 
|-
−
| 02.16
+
| 02:16
 
||We first define the function with parameters '''f , a , b , n.'''
 
||We first define the function with parameters '''f , a , b , n.'''
−
 
  
 
|-
 
|-
−
| 02.22
+
| 02:22
 
|'''f '''refers to the function we have to solve,  
 
|'''f '''refers to the function we have to solve,  
 +
 
|-
 
|-
  
−
| 02.25
+
| 02:25
−
|| '''a ''' is the lower limit of the integral,  
+
|| '''a ''' is the lower limit of the integral,
 +
 
|-
 
|-
  
−
|02.28
+
|02:28
  
 
||''' b''' is the upper limit of the integral and  
 
||''' b''' is the upper limit of the integral and  
Line 154: Line 150:
 
|-
 
|-
  
−
|02.31
+
|02:31
  
 
| '''n''' is the number of intervals.  
 
| '''n''' is the number of intervals.  
Line 160: Line 156:
 
|-
 
|-
  
−
|02.34
+
|02:34
  
 
| '''linspace''' function is used to create ten equal intervals between zero and one  
 
| '''linspace''' function is used to create ten equal intervals between zero and one  
Line 166: Line 162:
 
|-
 
|-
  
−
| 02.42
+
| 02:42
  
 
|| We find the value of the integral and store it in ''' I one'''
 
|| We find the value of the integral and store it in ''' I one'''
  
 
|-
 
|-
−
| 02.49
+
| 02:49
 
| Click on '''Execute''' on '''Scilab editor''' and choose '''Save and Execute ''' the code.  
 
| Click on '''Execute''' on '''Scilab editor''' and choose '''Save and Execute ''' the code.  
  
 
|-
 
|-
−
|03.02
+
|03:02
 
|  Define the example function by typing:  
 
|  Define the example function by typing:  
  
 
|-
 
|-
−
| 03.05
+
| 03:05
 
| '''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'''
 
| '''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'''
  
 
|-
 
|-
−
| 03.30
+
| 03:30
 
| Press '''Enter '''
 
| Press '''Enter '''
  
 
|-
 
|-
−
| 03.31
+
| 03:31
 
|  Type '''Trap underscore composite open paranthesis f comma zero comma one comma ten close paranthesis  
 
|  Type '''Trap underscore composite open paranthesis f comma zero comma one comma ten close paranthesis  
 
'''
 
'''
 
|-
 
|-
−
|03.41
+
|03:41
 
| Press '''Enter '''
 
| Press '''Enter '''
−
 
  
 
|-
 
|-
−
|03.43
+
|03:43
 
| The answer is displayed on the '''console '''
 
| The answer is displayed on the '''console '''
  
 
|-
 
|-
−
| 03.47
+
| 03:47
 
| Next we shall study '''Composite simpson's rule.'''
 
| Next we shall study '''Composite simpson's rule.'''
  
 
|-
 
|-
  
−
| 03.51
+
| 03:51
  
 
| In this rule, we decompose the interval ''' a comma b''' into '''n is greater than 1'''  subintervals of equal length  
 
| In this rule, we decompose the interval ''' a comma b''' into '''n is greater than 1'''  subintervals of equal length  
−
 
  
 
|-
 
|-
  
−
| 04.03
+
| 04:03
 
|| Apply '''Simpson's rule''' to each interval  
 
|| Apply '''Simpson's rule''' to each interval  
  
 
|-
 
|-
  
−
| 04.06
+
| 04:06
  
 
| We get the value of the integral to be  
 
| We get the value of the integral to be  
−
 
  
 
|-
 
|-
  
−
|04.10
+
|04:10
  
 
| '''h by three multiplied by the sum of f zero, four into f one , two into f two to f n'''
 
| '''h by three multiplied by the sum of f zero, four into f one , two into f two to f n'''
−
 
  
 
|-
 
|-
  
−
|04.19
+
|04:19
  
 
||Let us solve an example using '''Composite Simpson's rule. '''
 
||Let us solve an example using '''Composite Simpson's rule. '''
Line 237: Line 229:
 
|-
 
|-
  
−
| 04.24
+
| 04:24
  
 
|We are given a '''function one by one plus x cube d x in the interval one to two'''
 
|We are given a '''function one by one plus x cube d x in the interval one to two'''
−
 
  
 
|-
 
|-
  
−
| 04.32
+
| 04:32
  
 
| Let the number of intervals be '''twenty '''
 
| Let the number of intervals be '''twenty '''
−
 
  
 
|-
 
|-
  
−
|04.37
+
|04:37
  
 
| Let us look at the code for  '''Composite simpson's rule'''
 
| Let us look at the code for  '''Composite simpson's rule'''
−
 
−
 
  
 
|-
 
|-
−
|04.42
+
|04:42
 
| We first define the function with parameters '''f , a , b , n. '''
 
| We first define the function with parameters '''f , a , b , n. '''
  
 
|-
 
|-
  
−
| 04.49
+
| 04:49
  
 
| '''f''' refers to the function we have to solve,  
 
| '''f''' refers to the function we have to solve,  
−
 
−
 
  
 
|-
 
|-
  
−
|04.52
+
|04:52
  
 
||'''a'''  is the lower limit of the integral,  
 
||'''a'''  is the lower limit of the integral,  
−
 
  
 
|-
 
|-
  
−
|04.56
+
|04:56
  
 
| '''b''' is the upper limit of the integral and  
 
| '''b''' is the upper limit of the integral and  
−
 
  
 
|-
 
|-
  
−
| 04.58
+
| 04:58
  
 
| '''n''' is the number of intervals.  
 
| '''n''' is the number of intervals.  
−
 
−
 
  
 
|-
 
|-
  
−
| 05.02
+
| 05:02
  
 
|We find two sets of points.  
 
|We find two sets of points.  
−
 
  
 
|-
 
|-
  
−
| 05.04
+
| 05:04
  
 
| We find the value of the function with one set and multiply it with two  
 
| We find the value of the function with one set and multiply it with two  
−
 
−
 
  
 
|-
 
|-
  
−
| 05.10
+
| 05:10
  
 
| With the other set, we find the value and multiply it with four  
 
| With the other set, we find the value and multiply it with four  
−
 
−
 
  
 
|-
 
|-
  
−
| 05.16
+
| 05:16
  
 
||We sum these values and multiply it with '''h by three and store the final value in I '''
 
||We sum these values and multiply it with '''h by three and store the final value in I '''
Line 322: Line 299:
 
|-
 
|-
  
−
| 05.24
+
| 05:24
  
 
||Let us execute the code  
 
||Let us execute the code  
Line 328: Line 305:
 
|-
 
|-
  
−
| 05.28
+
| 05:28
  
 
|| Save and execute the file '''Simp underscore composite dot s c i'''
 
|| Save and execute the file '''Simp underscore composite dot s c i'''
Line 334: Line 311:
 
|-
 
|-
  
−
| 05.39
+
| 05:39
  
 
|Let me clear the screen first.  
 
|Let me clear the screen first.  
Line 340: Line 317:
 
|-
 
|-
  
−
| 05.42
+
| 05:42
  
 
| Define the function given in the example by typing  
 
| Define the function given in the example by typing  
Line 346: Line 323:
 
|-
 
|-
  
−
|05.45
+
|05:45
  
 
|'''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 divided by open paranthesis one plus x cube close paranthesis close quote close paranthesis'''
 
|'''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 divided by open paranthesis one plus x cube close paranthesis close quote close paranthesis'''
−
 
  
 
|-
 
|-
  
−
|06.12
+
|06:12
  
 
| Press '''Enter '''
 
| Press '''Enter '''
−
 
  
 
|-
 
|-
  
−
| 06.14
+
| 06:14
  
 
| Type '''Simp underscore composite open paranthesis f comma one comma two comma twenty close paranthesis'''
 
| Type '''Simp underscore composite open paranthesis f comma one comma two comma twenty close paranthesis'''
−
 
  
 
|-
 
|-
  
−
|06.24
+
|06:24
  
 
||Press '''Enter '''
 
||Press '''Enter '''
−
 
−
 
  
 
|-
 
|-
  
−
| 06.26
+
| 06:26
  
 
| The answer is displayed on the console.  
 
| The answer is displayed on the console.  
−
 
  
 
|-
 
|-
  
−
| 06.31
+
| 06:31
  
 
| Let us now look at '''Composite Midpoint Rule.'''
 
| Let us now look at '''Composite Midpoint Rule.'''
−
 
−
 
  
 
|-
 
|-
  
−
| 06.35
+
| 06:35
  
 
| It integrates polynomials of degree one or less  
 
| It integrates polynomials of degree one or less  
Line 396: Line 365:
 
|-
 
|-
  
−
|06.40
+
|06:40
  
 
| Divides the interval '''a comma b''' into a ''' subintervals'''of equal width  
 
| Divides the interval '''a comma b''' into a ''' subintervals'''of equal width  
Line 402: Line 371:
 
|-
 
|-
  
−
|06.49
+
|06:49
  
 
| Finds the midpoint of each interval indicated by '''x i '''
 
| Finds the midpoint of each interval indicated by '''x i '''
Line 408: Line 377:
 
|-
 
|-
  
−
|06.54
+
|06:54
  
 
| We find the sum of the values of the integral at each midpoint  
 
| We find the sum of the values of the integral at each midpoint  
Line 414: Line 383:
 
|-
 
|-
  
−
|07.00
+
|07:00
  
 
| Let us solve this problem using '''Composite Midpoint Rule'''
 
| Let us solve this problem using '''Composite Midpoint Rule'''
Line 420: Line 389:
 
|-
 
|-
  
−
|07.05
+
|07:05
  
 
| '''We are given a function one minus x square d x in the interval zero to one point five'''
 
| '''We are given a function one minus x square d x in the interval zero to one point five'''
Line 426: Line 395:
 
|-
 
|-
  
−
|07.15
+
|07:15
  
 
| We assume '''n''' is equal to '''twenty '''
 
| We assume '''n''' is equal to '''twenty '''
Line 432: Line 401:
 
|-
 
|-
  
−
|07.18
+
|07:18
  
 
| Let us look at the code for '''Composite Midpoint rule'''
 
| Let us look at the code for '''Composite Midpoint rule'''
Line 438: Line 407:
 
|-
 
|-
  
−
|07.24
+
|07:24
  
 
| We first define the function with parameters '''f , a , b , n. '''
 
| We first define the function with parameters '''f , a , b , n. '''
Line 444: Line 413:
 
|-
 
|-
  
−
|07.30
+
|07:30
  
 
| '''f ''' refers to the function we have to solve,  
 
| '''f ''' refers to the function we have to solve,  
Line 450: Line 419:
 
|-
 
|-
  
−
|07.33
+
|07:33
  
 
| '''a'''  is the lower limit of the integral,  
 
| '''a'''  is the lower limit of the integral,  
Line 456: Line 425:
 
|-
 
|-
  
−
|07.36
+
|07:36
  
 
| '''b ''' is the upper limit of the integral and  
 
| '''b ''' is the upper limit of the integral and  
Line 462: Line 431:
 
|-
 
|-
  
−
|07.39
+
|07:39
  
 
| '''n ''' is the number of intervals.  
 
| '''n ''' is the number of intervals.  
Line 468: Line 437:
 
|-
 
|-
  
−
|07.41
+
|07:41
  
 
| We find the midpoint of each interval  
 
| We find the midpoint of each interval  
Line 475: Line 444:
 
|-
 
|-
  
−
|07.45
+
|07:45
  
 
| Find the value of integral at each midpoint and then find the sum and store it in '''I.'''  
 
| Find the value of integral at each midpoint and then find the sum and store it in '''I.'''  
Line 481: Line 450:
 
|-
 
|-
  
−
|07.53
+
|07:53
  
 
| Let us now solve the example  
 
| Let us now solve the example  
Line 487: Line 456:
 
|-
 
|-
  
−
|07.55
+
|07:55
  
 
| Save and execute the file '''mid underscore composite dot s c i'''  
 
| Save and execute the file '''mid underscore composite dot s c i'''  
Line 493: Line 462:
 
|-
 
|-
  
−
|08.04
+
|08:04
  
 
| Let me clear the screen  
 
| Let me clear the screen  
Line 499: Line 468:
 
|-
 
|-
  
−
|08.08
+
|08:08
  
 
| We define the function given in the example by typing  
 
| We define the function given in the example by typing  
Line 505: Line 474:
 
|-
 
|-
  
−
|08.13
+
|08:13
  
 
| '''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'''
 
| '''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'''
Line 511: Line 480:
 
|-
 
|-
  
−
|08.37
+
|08:37
  
 
| Press '''Enter'''  
 
| Press '''Enter'''  
Line 517: Line 486:
 
|-
 
|-
  
−
|08.39
+
|08:39
  
 
| Then type '''mid underscore composite open paranthesis f comma zero comma one point five comma twenty close paranthesis'''
 
| Then type '''mid underscore composite open paranthesis f comma zero comma one point five comma twenty close paranthesis'''
Line 523: Line 492:
 
|-
 
|-
  
−
|08.53
+
|08:53
  
 
|Press '''Enter '''
 
|Press '''Enter '''
Line 529: Line 498:
 
  |-
 
  |-
  
−
|08.54
+
|08:54
  
 
| The answer is displayed on the '''console'''
 
| The answer is displayed on the '''console'''
Line 535: Line 504:
 
|-
 
|-
  
−
|08.59
+
|08:59
  
 
| Let us summarize this tutorial.  
 
| Let us summarize this tutorial.  
Line 541: Line 510:
 
|-
 
|-
  
−
|09.02
+
|09:02
  
 
| In this tutorial we have learnt to:  
 
| In this tutorial we have learnt to:  
Line 547: Line 516:
 
|-
 
|-
  
−
|09.04
+
|09:04
  
 
| Develop '''Scilab''' code for '''numerical integration'''  
 
| Develop '''Scilab''' code for '''numerical integration'''  
Line 553: Line 522:
 
|-
 
|-
  
−
|09.08
+
|09:08
  
 
| Find the value of an '''integral'''  
 
| Find the value of an '''integral'''  
Line 559: Line 528:
  
 
|-
 
|-
−
|09.11
+
|09:11
  
 
| Watch the video available at the link shown below  
 
| Watch the video available at the link shown below  
Line 565: Line 534:
 
|-
 
|-
  
−
| 09.15
+
| 09:15
  
 
| It summarises the Spoken Tutorial project  
 
| It summarises the Spoken Tutorial project  
−
 
−
 
  
 
|-
 
|-
  
−
|09.18
+
|09:18
  
 
||If you do not have good bandwidth, you can download and watch it  
 
||If you do not have good bandwidth, you can download and watch it  
Line 579: Line 546:
 
|-
 
|-
  
−
|09.23
+
|09:23
  
 
||The spoken tutorial Team
 
||The spoken tutorial Team
Line 585: Line 552:
 
|-
 
|-
  
−
|09.25
+
|09:25
  
 
||Conducts workshops using spoken tutorials  
 
||Conducts workshops using spoken tutorials  
−
 
  
 
|-
 
|-
  
−
|06.29
+
|09:29
  
 
||Gives certificates to those who pass an online test  
 
||Gives certificates to those who pass an online test  
−
 
  
 
|-
 
|-
  
−
|09.32
+
|09:32
  
 
||For more details, please write to contact@spoken-tutorial.org  
 
||For more details, please write to contact@spoken-tutorial.org  
−
 
  
 
|-
 
|-
  
−
|09.40
+
|09:40
  
 
|Spoken Tutorial Project is a part of the Talk to a Teacher project  
 
|Spoken Tutorial Project is a part of the Talk to a Teacher project  
Line 612: Line 576:
 
|-
 
|-
  
−
| 09.45
+
| 09:45
  
 
| It is supported by the National Mission on Eduction through ICT, MHRD, Government of India.  
 
| It is supported by the National Mission on Eduction through ICT, MHRD, Government of India.  
 
|-
 
|-
  
−
| 09.52
+
| 09:52
  
 
|More information on this mission is available at  http://spoken-tutorial.org/NMEICT-Intro
 
|More information on this mission is available at  http://spoken-tutorial.org/NMEICT-Intro
Line 623: Line 587:
 
|-
 
|-
  
−
| 10.03
+
| 10:03
  
 
|This is Ashwini Patil signing off. Thank you for joining.
 
|This is Ashwini Patil signing off. Thank you for joining.

Revision as of 17:46, 10 July 2014

Time Narration
00:01 Dear Friends,
00:02. Welcome to the Spoken Tutorial on “ Composite Numerical Integration”
00:07 At the end of this tutorial, you will learn how to:
00:11 Develop Scilab code for different Composite Numerical Integration algorithms
00:17 Divide the integral into equal intervals
00:21 Apply the algorithm to each interval and
00:24 Calculate the composite value of the integral
00:28 To record this tutorial, I am using
00:30 Ubuntu 12.04 as the operating system
00:34 with Scilab 5.3.3 version
00:38 Before practising this tutorial, a learner should have basic knowledge of
00:42 Scilab and
00:44 Integration using Numerical Methods
00:47 For Scilab, please refer to the relevant tutorials available on the Spoken Tutorial website.
00:55 Numerical Integration is the:
00:58 Study of how the numerical value of an integral can be found
01:03 It is used when exact mathematical integration is not available
01:08 It approximates a definite integral from values of the integrand
01:15 Let us study Composite Trapezoidal Rule.
01:18 This rule is the extension of trapezoidal rule
01:22 We divide the interval a comma b into n equal intervals
01:29 Then h equal to b minus a divided by n is the common length of the intervals
01:36 Then composite trapezoidal rule is given by
01:41 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
01:57 Let us solve an example using composite trapezoidal rule.
02:02 Assume the number of intervals n is equal to ten.
02:09 Let us look at the code for Composite Trapezoidal Rule on Scilab Editor
02:16 We first define the function with parameters f , a , b , n.
02:22 f refers to the function we have to solve,
02:25 a is the lower limit of the integral,
02:28 b is the upper limit of the integral and
02:31 n is the number of intervals.
02:34 linspace function is used to create ten equal intervals between zero and one
02:42 We find the value of the integral and store it in I one
02:49 Click on Execute on Scilab editor and choose Save and Execute the code.
03:02 Define the example function by typing:
03:05 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
03:30 Press Enter
03:31 Type Trap underscore composite open paranthesis f comma zero comma one comma ten close paranthesis

03:41 Press Enter
03:43 The answer is displayed on the console
03:47 Next we shall study Composite simpson's rule.
03:51 In this rule, we decompose the interval a comma b into n is greater than 1 subintervals of equal length
04:03 Apply Simpson's rule to each interval
04:06 We get the value of the integral to be
04:10 h by three multiplied by the sum of f zero, four into f one , two into f two to f n
04:19 Let us solve an example using Composite Simpson's rule.
04:24 We are given a function one by one plus x cube d x in the interval one to two
04:32 Let the number of intervals be twenty
04:37 Let us look at the code for Composite simpson's rule
04:42 We first define the function with parameters f , a , b , n.
04:49 f refers to the function we have to solve,
04:52 a is the lower limit of the integral,
04:56 b is the upper limit of the integral and
04:58 n is the number of intervals.
05:02 We find two sets of points.
05:04 We find the value of the function with one set and multiply it with two
05:10 With the other set, we find the value and multiply it with four
05:16 We sum these values and multiply it with h by three and store the final value in I
05:24 Let us execute the code
05:28 Save and execute the file Simp underscore composite dot s c i
05:39 Let me clear the screen first.
05:42 Define the function given in the example by typing
05:45 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 divided by open paranthesis one plus x cube close paranthesis close quote close paranthesis
06:12 Press Enter
06:14 Type Simp underscore composite open paranthesis f comma one comma two comma twenty close paranthesis
06:24 Press Enter
06:26 The answer is displayed on the console.
06:31 Let us now look at Composite Midpoint Rule.
06:35 It integrates polynomials of degree one or less
06:40 Divides the interval a comma b into a subintervalsof equal width
06:49 Finds the midpoint of each interval indicated by x i
06:54 We find the sum of the values of the integral at each midpoint
07:00 Let us solve this problem using Composite Midpoint Rule
07:05 We are given a function one minus x square d x in the interval zero to one point five
07:15 We assume n is equal to twenty
07:18 Let us look at the code for Composite Midpoint rule
07:24 We first define the function with parameters f , a , b , n.
07:30 f refers to the function we have to solve,
07:33 a is the lower limit of the integral,
07:36 b is the upper limit of the integral and
07:39 n is the number of intervals.
07:41 We find the midpoint of each interval


07:45 Find the value of integral at each midpoint and then find the sum and store it in I.
07:53 Let us now solve the example
07:55 Save and execute the file mid underscore composite dot s c i
08:04 Let me clear the screen
08:08 We define the function given in the example by typing
08:13 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
08:37 Press Enter
08:39 Then type mid underscore composite open paranthesis f comma zero comma one point five comma twenty close paranthesis
08:53 Press Enter
08:54 The answer is displayed on the console
08:59 Let us summarize this tutorial.
09:02 In this tutorial we have learnt to:
09:04 Develop Scilab code for numerical integration
09:08 Find the value of an integral


09:11 Watch the video available at the link shown below
09:15 It summarises the Spoken Tutorial project
09:18 If you do not have good bandwidth, you can download and watch it
09:23 The spoken tutorial Team
09:25 Conducts workshops using spoken tutorials
09:29 Gives certificates to those who pass an online test
09:32 For more details, please write to contact@spoken-tutorial.org
09:40 Spoken Tutorial Project is a part of the Talk to a Teacher project
09:45 It is supported by the National Mission on Eduction through ICT, MHRD, Government of India.
09:52 More information on this mission is available at http://spoken-tutorial.org/NMEICT-Intro
10:03 This is Ashwini Patil signing off. Thank you for joining.

Contributors and Content Editors

Gaurav, PoojaMoolya, Sandhya.np14

Retrieved from "https://script.spoken-tutorial.org/index.php?title=Scilab/C4/Integration/English-timed&oldid=13505"