Difference between revisions of "Scilab/C4/Integration/Gujarati"

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(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 th...")
 
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|Dear Friends,  
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| નમસ્તે મિત્રો,
  
 
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| 00:02.
 
| 00:02.
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| Welcome to the Spoken Tutorial on '''Composite Numerical Integration'''.
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| '''Composite Numerical Integration''' પરના આ સ્પોકન ટ્યુટોરિયલમાં તમારું સ્વાગત છે.
  
 
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|At the end of this tutorial, you will learn how to:  
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| આ ટ્યુટોરીયલના અંતમાં આપણે શીખીશું કેવી રીતે:  
  
 
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|Develop '''Scilab''' code for different '''Composite Numerical Integration algorithms'''  
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| વિવિધ  '''Composite Numerical Integration algorithms''' ના લીધે સાઈલેબ કોડ કેવી રીતે બનાવે છે
  
 
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|Divide the '''integral''' into equal '''intervals'''
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| '''integral''' (ઈંટીગ્રલ) ને સમાન અંતરાલે કેવી રીતે વિભાજીત થાય છે. 
  
 
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|Apply the algorithm to each '''interval''' and
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| પ્રત્યેક અંતરાલે અલગોરિધમ ને કેવી રીતે લાગુ કરાય છે.અને
  
 
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|Calculate the '''composite value of the integral'''.
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| ઈંટીગ્રલ ની કોમ્પોઝિટ વેલ્યુ ની ગણના કેવી રીતે કરાય છે.
  
 
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|To record this tutorial, I am using
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| આ ટ્યુટોરિયલ રિકોર્ડ કરવા માટે હું ઉપયોગ કરી રહી છું,
  
 
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| '''Ubuntu 12.04''' as the operating system
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|   '''Scilab 5.3.3''' વર્જનના સાથે.
  
 
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| with '''Scilab 5.3.3''' version.
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| '''Ubuntu 12.04''' ઓપરેટીંગ સીસ્ટમ
  
 
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||Before practicing this tutorial, a learner should have basic knowledge of
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|| આ ટ્યુટોરિયલ ના અભ્યાસ પહેલા શીખનારને
  
 
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|'''Scilab''' and
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|'''Scilab''' સાઈલેબ અને
  
 
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| '''Integration using Numerical Methods'''.
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| '''Integration using Numerical Methods''' નું સામન્ય જ્ઞાન હોવું જોઈએ.  
  
 
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| For '''Scilab''', please refer to the relevant tutorials available on the '''Spoken Tutorial''' website.  
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| સાઈલેબ માટે સ્પોકન ટ્યુટોરિયલ વેબ સાઈટ પર ઉપલબ્ધ સંબંધિત ટ્યુટોરિયલ જુઓ.
  
 
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| '''Numerical Integration''' is the
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| '''Numerical Integration''' છે. કે
  
 
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| study of how the numerical value of an '''integral''' can be found.  
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| '''integral''' (ઈંટીગ્રલ) ની  ન્યુમેરીક્લ વેલ્યુ ને કેવી રીતે મેળવી શકાય છે.  
  
 
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| It is used when exact mathematical integration is not available.  
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| આ ઉપયોગ થાય છે જયારે ચોક્કસ મેથેમેટિકલ ઇન્ટીગ્રેશન ઉપલબ્ધ નથી થતું.
  
 
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|It approximates a '''definite integral''' from values of the '''integrand'''.  
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| આ  '''integrand''' (ઈંટીગ્રેંડ) ની વેલ્યુ થી  '''definite integral''' નો અંદાજ લગાડે છે.
  
 
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Revision as of 11:30, 10 December 2015

Time Narration
00:01 નમસ્તે મિત્રો,
00:02. Composite Numerical Integration પરના આ સ્પોકન ટ્યુટોરિયલમાં તમારું સ્વાગત છે.
00:07 આ ટ્યુટોરીયલના અંતમાં આપણે શીખીશું કેવી રીતે:
00:11 વિવિધ Composite Numerical Integration algorithms ના લીધે સાઈલેબ કોડ કેવી રીતે બનાવે છે
00:17 integral (ઈંટીગ્રલ) ને સમાન અંતરાલે કેવી રીતે વિભાજીત થાય છે.
00:21 પ્રત્યેક અંતરાલે અલગોરિધમ ને કેવી રીતે લાગુ કરાય છે.અને
00:24 ઈંટીગ્રલ ની કોમ્પોઝિટ વેલ્યુ ની ગણના કેવી રીતે કરાય છે.
00:28 આ ટ્યુટોરિયલ રિકોર્ડ કરવા માટે હું ઉપયોગ કરી રહી છું,
00:30 Scilab 5.3.3 વર્જનના સાથે.
00:34 Ubuntu 12.04 ઓપરેટીંગ સીસ્ટમ
00:38 આ ટ્યુટોરિયલ ના અભ્યાસ પહેલા શીખનારને
00:42 Scilab સાઈલેબ અને
00:44 Integration using Numerical Methods નું સામન્ય જ્ઞાન હોવું જોઈએ.
00:47 સાઈલેબ માટે સ્પોકન ટ્યુટોરિયલ વેબ સાઈટ પર ઉપલબ્ધ સંબંધિત ટ્યુટોરિયલ જુઓ.
00:55 Numerical Integration છે. કે
00:58 integral (ઈંટીગ્રલ) ની ન્યુમેરીક્લ વેલ્યુ ને કેવી રીતે મેળવી શકાય છે.
01:03 આ ઉપયોગ થાય છે જયારે ચોક્કસ મેથેમેટિકલ ઇન્ટીગ્રેશન ઉપલબ્ધ નથી થતું.
01:08 આ integrand (ઈંટીગ્રેંડ) ની વેલ્યુ થી definite integral નો અંદાજ લગાડે છે.
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.

Contributors and Content Editors

Jyotisolanki, PoojaMoolya

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