Difference between revisions of "Scilab/C4/Integration/English-timed"
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| − | | Welcome to the Spoken Tutorial on ''' | + | | Welcome to the Spoken Tutorial on '''Composite Numerical Integration'''. |
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|00:24 | |00:24 | ||
| − | |Calculate the '''composite value of the integral''' | + | |Calculate the '''composite value of the integral'''. |
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|00:34 | |00:34 | ||
| − | | with '''Scilab 5.3.3''' version | + | | with '''Scilab 5.3.3''' version. |
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|00:38 | |00:38 | ||
| − | ||Before | + | ||Before practicing this tutorial, a learner should have basic knowledge of |
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|00:44 | |00:44 | ||
| − | | '''Integration using Numerical Methods''' | + | | '''Integration using Numerical Methods'''. |
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| 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. |
|- | |- | ||
|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 | + | |It approximates a '''definite integral''' from values of the '''integrand'''. |
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|01:18 | |01:18 | ||
| − | |This rule is the extension of '''trapezoidal rule''' | + | |This rule is the extension of '''trapezoidal rule'''. |
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| 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. |
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| 01:29 | | 01:29 | ||
| − | | Then '''h | + | | Then '''h equals to b minus a divided by n''' is the common length of the intervals. |
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|01:36 | |01:36 | ||
| − | | Then '''composite trapezoidal rule''' is given by | + | | Then '''composite trapezoidal rule''' is given by: |
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|02:02 | |02:02 | ||
| − | | Assume the number of intervals n is equal to ten. | + | | Assume the number of intervals n is equal to ten (n=10). |
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|02:09 | |02:09 | ||
| − | |Let us look at the code for '''Composite Trapezoidal Rule''' on '''Scilab | + | |Let us look at the code for '''Composite Trapezoidal Rule''' on '''Scilab editor''' |
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| 02:25 | | 02:25 | ||
| − | || '''a ''' is the lower limit of the integral, | + | || '''a''' is the lower limit of the integral, |
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|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. |
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| 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'''. |
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| 02:49 | | 02:49 | ||
| − | | Click on '''Execute''' on '''Scilab editor''' and choose '''Save and | + | | Click on '''Execute''' on '''Scilab editor''' and choose '''Save and execute ''' the code. |
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| 03:05 | | 03:05 | ||
| − | | '''d e f f open | + | | '''d e f f open parenthesis 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 parenthesis two asterisk x plus one close parenthesis close quote close parenthesis''' |
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| 03:30 | | 03:30 | ||
| − | | Press '''Enter ''' | + | | Press '''Enter '''. |
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| 03:31 | | 03:31 | ||
| − | | Type '''Trap underscore composite open | + | | Type '''Trap underscore composite open parenthesis f comma zero comma one comma ten close parenthesis''' |
| − | ''' | + | |
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|03:41 | |03:41 | ||
| − | | Press '''Enter ''' | + | | Press '''Enter '''. |
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|03:43 | |03:43 | ||
| − | | The answer is displayed on the '''console ''' | + | | The answer is displayed on the '''console '''. |
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| 03:47 | | 03:47 | ||
| − | | Next we shall study '''Composite | + | | Next we shall study '''Composite Simpson's rule.''' |
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| 03:51 | | 03:51 | ||
| − | | In this rule, we decompose the interval ''' a comma b''' into '''n is greater than 1''' | + | | In this rule, we decompose the interval ''' a comma b''' into '''n is greater than 1''' sub-intervals of equal length. |
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| 04:03 | | 04:03 | ||
| − | || Apply '''Simpson's rule''' to each interval | + | || Apply '''Simpson's rule''' to each interval. |
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| 04:06 | | 04:06 | ||
| − | | We get the value of the integral to be | + | | We get the value of the integral to be: |
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|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'''. |
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| 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'''. |
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| 04:32 | | 04:32 | ||
| − | | Let the number of intervals be '''twenty ''' | + | | Let the number of intervals be '''twenty '''. |
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|04:37 | |04:37 | ||
| − | | Let us look at the code for '''Composite | + | | Let us look at the code for '''Composite Simpson's rule'''. |
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| 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. |
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| 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. |
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| 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 '''. |
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| 05:24 | | 05:24 | ||
| − | ||Let us execute the code | + | ||Let us execute the code. |
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| 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'''. |
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| 05:42 | | 05:42 | ||
| − | | Define the function given in the example by typing | + | | Define the function given in the example by typing: |
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|05:45 | |05:45 | ||
| − | |'''d e f f open | + | |'''d e f f open parenthesis 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 parenthesis one plus x cube close parenthesis close quote close parenthesis''' |
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|06:12 | |06:12 | ||
| − | | Press '''Enter ''' | + | | Press '''Enter '''. |
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| 06:14 | | 06:14 | ||
| − | | Type '''Simp underscore composite open | + | | Type '''Simp underscore composite open parenthesis f comma one comma two comma twenty close parenthesis''' |
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|06:24 | |06:24 | ||
| − | ||Press '''Enter ''' | + | ||Press '''Enter '''. |
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| 06:35 | | 06:35 | ||
| − | | It integrates polynomials of degree one or less | + | | It integrates polynomials of degree one or less, |
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|06:40 | |06:40 | ||
| − | | | + | | divides the interval '''a comma b''' into a ''' sub-intervals'''of equal width. |
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|06:49 | |06:49 | ||
| − | | Finds the midpoint of each interval indicated by '''x i ''' | + | | Finds the midpoint of each interval indicated by '''x i '''. |
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|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. |
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|07:00 | |07:00 | ||
| − | | Let us solve this problem using '''Composite Midpoint Rule''' | + | | Let us solve this problem using '''Composite Midpoint Rule'''. |
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|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'''. |
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|07:15 | |07:15 | ||
| − | | We assume '''n''' is equal to '''twenty ''' | + | | We assume '''n''' is equal to '''twenty '''. |
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|07:18 | |07:18 | ||
| − | | Let us look at the code for '''Composite Midpoint rule''' | + | | Let us look at the code for '''Composite Midpoint rule'''. |
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|07:41 | |07:41 | ||
| − | | We find the midpoint of each interval | + | | We find the midpoint of each interval. |
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|07:53 | |07:53 | ||
| − | | Let us now solve the example | + | | Let us now solve the example. |
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|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'''. |
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|08:04 | |08:04 | ||
| − | | Let me clear the screen | + | | Let me clear the screen. |
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|08:08 | |08:08 | ||
| − | | We define the function given in the example by typing | + | | We define the function given in the example by typing: |
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|08:13 | |08:13 | ||
| − | | '''d e f f open | + | | '''d e f f open parenthesis 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 parenthesis''' |
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|08:37 | |08:37 | ||
| − | | Press '''Enter''' | + | | Press '''Enter'''. |
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|08:39 | |08:39 | ||
| − | | Then type '''mid underscore composite open | + | | Then type '''mid underscore composite open parenthesis f comma zero comma one point five comma twenty close parenthesis''' |
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|08:53 | |08:53 | ||
| − | |Press '''Enter ''' | + | |Press '''Enter '''. |
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|08:54 | |08:54 | ||
| − | | The answer is displayed on the '''console''' | + | | The answer is displayed on the '''console'''. |
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|09:08 | |09:08 | ||
| − | | Find the value of an '''integral''' | + | | Find the value of an '''integral'''. |
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|09:11 | |09:11 | ||
| − | | Watch the video available at the link shown below | + | | Watch the video available at the link shown below. |
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| 09:15 | | 09:15 | ||
| − | | It | + | | It summarizes the Spoken Tutorial project. |
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|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. |
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|09:23 | |09:23 | ||
| − | ||The spoken tutorial Team | + | ||The spoken tutorial Team: |
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|09:29 | |09:29 | ||
| − | ||Gives certificates to those who pass an online test | + | ||Gives certificates to those who pass an online test. |
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|09:32 | |09:32 | ||
| − | ||For more details, please write to contact@spoken-tutorial.org | + | ||For more details, please write to contact@spoken-tutorial.org. |
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|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. |
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| 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. |
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| 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 16:31, 26 February 2015
| 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 practicing 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 equals 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 (n=10). |
| 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 parenthesis 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 parenthesis two asterisk x plus one close parenthesis close quote close parenthesis |
| 03:30 | Press Enter . |
| 03:31 | Type Trap underscore composite open parenthesis f comma zero comma one comma ten close parenthesis |
| 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 sub-intervals 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 parenthesis 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 parenthesis one plus x cube close parenthesis close quote close parenthesis |
| 06:12 | Press Enter . |
| 06:14 | Type Simp underscore composite open parenthesis f comma one comma two comma twenty close parenthesis |
| 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 sub-intervalsof 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 parenthesis 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 parenthesis |
| 08:37 | Press Enter. |
| 08:39 | Then type mid underscore composite open parenthesis f comma zero comma one point five comma twenty close parenthesis |
| 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 summarizes 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. |