Scilab/C4/Integration/Gujarati

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