Scilab/C4/Integration/Khasi

Time	Narration
00:01	Paralok, ngi pdiangsngewbha ia phi sha ka Spoken Tutorial halor ka Composite Numerical Integration.
00:07	Ha kaba kut jong kane ka jinghikai, phin sa nang kumno ban:
00:11	Develop ia ka Scilab code na ka bynta bun jait ki Composite Numerical Integration algorithms
00:17	Divide ia ka integral ha ki equal intervals
00:21	Apply ia ka algorithm sha man kawei ka interval bad
00:24	Khein ia ka composite value of the integral.
00:28	Ban record ia kane ka jinghikai, nga pyndonkam
00:30	Ubuntu 12.04 kum ka operating system
00:34	Ryngkat bad ka Scilab 5.3.3 version.
00:38	Shwa ban pyrshang ia kane ka jinghikai, u ne ka nongpule kidei ban don ia ka jingtip ba donkam jong
00:42	Ka Scilab bad
00:44	Integration using Numerical Methods.
00:47	Na ka bynta ka Scilab, sngewbha peit ia ki jinghikai ba iadei ba don ha ka Spoken Tutorial website.
00:55	Numerical Integration kadei ka
00:58	Jingpule halor kumno ba u numerical value jong ka integral lah ban wad.
01:03	La pyndonkam ia u haba u mathematical integration ba dei thik um don.
01:08	U antad ia u definite integral na ki values jong ka integrand.
01:15	To ngin pule ia ka Composite Trapezoidal Rule.
01:18	Kane ka rule kadei ka tnad baiar jong ka trapezoidal rule.
01:22	Ngi phiah ia ka interval a comma b ha ki n intervals ba iaryngkat.
01:29	Nangta, h equals to b minus a divided by n kadei ka common lengths jong ki intervals.
01:36	Nangta ka composite trapezoidal rule la ai da:
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	To ngin solve ia kawei ka nuksa da kaba pyndonkam ia ka composite trapezoidal rule.
02:02	Shu shim ia u number jong ki intervals n is equal to 10 (n=10)
02:09	To ngin peit ia u code na ka bynta ka Composite Trapezoidal Rule ha ka Scilab editor
02:16	Nyngkong ngi define ia ka function ryngkat bad ki parameters f , a , b , n.
02:22	f u thew ia ka function ba ngi hap ban solve.
02:25	a udei u lower limit jong ka integral,
02:28	b kadei ka upper limit jong ka integral bad
02:31	n udei u number jong ki intervals.
02:34	linspace function la pyndonkam ban create shiphew tylli ki intervals ba iaryngkat hapdeng zero bad one.
02:42	Ngi wad ia u value jong ka integral bad buh ia u ha I one.
02:49	Nion ha Execute ha ka Scilab editor bad jied Save and execute ia u code.
03:02	Define ia ka example function da kaba type:
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	Nion Enter . Type Trap underscore composite open parenthesis f comma zero comma one comma ten close parenthesis
03:41	Nion Enter .
03:43	Ka jubab la pyni ha ka console .
03:47	Nangta ngin sa pule ia ka Composite Simpson's rule.
03:51	Ha kane ka rule, ngi thep ia ka interval a comma b sha ka n is greater than 1 sub-intervals kaba ka jingjrong ka iaryngkat .
04:03	Apply Simpson's rule sha man kawei ka interval.
04:06	Ngi ioh ia u value jong ka integral u ban dei:
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	To ngin solve ia kawei ka nuksa da kaba pyndonkam ia ka Composite Simpson's rule.
04:24	Ngi ai ia ka function one by one plus x cube d x in the interval one to two.
04:32	Ai ba ka number jong ki intervals kan dei twenty .
04:37	To ngin phai sha u code na ka bynta ka Composite Simpson's rule.
04:42	Nyngkong eh ngi define ia ka function ryngkat ki parameters f , a , b , n.
04:49	f ka thew sha ka function ba ngi hap ban solve,
04:52	a udei u lower limit jong ka integral,
04:56	b udei u upper limit jong ka integral bad
04:58	n udei u number jong ki intervals.
05:02	Ngi wad artylli ki sets jong ki points
05:04	Ngi wad ia u value jong ka function ryngkat bad uwei u set bad multiply ia u da two.
05:10	Bad kawei ka set, ngi wad ia u value bad multiply ia u da four.
05:16	Ngi sum ia kine ki values bad muliply ia u da h by three and store the final value in I .
05:24	To ngin execute ia u code.
05:28	Save bad execute ia ka file Simp underscore composite dot s c i.
05:39	Nyngkong to ngan clear ia ka screen.
05:42	Define ia ka function ba la ai ha ka nuksa da kaba type:
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	Nion Enter .
06:14	Type Simp underscore composite open parenthesis f comma one comma two comma twenty close parenthesis
06:24	Nion Enter .
06:26	Ka jubab la pyni ha ka console.
06:31	To mynta ngin peit ia ka Composite Midpoint Rule.
06:35	Ka integrate ia ki polynomials jong ka degree one lane duna ia one.
06:40	Divide ia ka interval a comma b sha ki sub-intervalsba iaryngkat ka width.
06:49	Wad ia u midpoint jong man kawei ka interval ba la kdew da x i .
06:54	Ngi wad ia ka sum jong ki values jong ka integral ha man uwei u midpoint.
07:00	To ngin solve ia kane ka problem da kaba pyndonkam ia ka 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	Ngi shim n is equal to twenty .
07:18	To ngin peit ia u code na ka bynta ka Composite Midpoint rule.
07:24	Nyngkong ngi define ia ka function ryngkat ki parameters f , a , b , n.
07:30	f ka thew sha ka function ba ngi hap ban solve,
07:33	a udei u lower limit jong ka integral,
07:36	b udei u upper limit jong ka integral bad
07:39	n udei u number jong ki intervals.
07:41	Ngi wad ia u midpoint jong man kawei ka interval.
07:45	Wad ia u value jong ka integral ha man uwei u midpoint bad nangta wad ia ka sum bad buh ia ka ha I.
07:53	To mynta ngin ia solve ia ka nuksa.
07:55	Save bad execute ia ka file mid underscore composite dot s c i.
08:04	To ngan clear ia ka screen.
08:08	Ngi define ia ka function ba la ai ha ka nuksa da kaba type:
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	Nion Enter.
08:39	Nangta type mid underscore composite open parenthesis f comma zero comma one point five comma twenty close parenthesis
08:53	Nion Enter. Ka jubab la pyni ha ka console.
08:59	To ngin batai kyllum ia kane ka jinghikai.
09:02	Ha kane ka jinghikai ngi la pule ban:
09:04	Develop ia u Scilab code na ka bynta kanumerical integration
09:08	Wad ia u value jong ka integral.
09:11	Peit ia ka video ba don ha ka link ba la pyni harum.
09:15	Ka kyllum lang ia ka Spoken Tutorial project.
09:18	Lada phim don ia ka bandwidth kaba biang, phi lah ban shu download bad peit ia ka.
09:23	Ka kynhun jong ka Spoken Tutorial
09:25	Ka pynlong ia ki workshops da kaba pyndonkam da ki spoken turorials
09:29	Ka ai certificates sha kito kiba pass ha ka online test.
09:32	Na ka bynta kham bun ki jingtip ba bniah, sngewbha thoh sha ka contact@spoken-tutorial.org.
09:40	Spoken Tutorial Project kadei shibynta jong ka Talk to a Teacher project.
09:45	La kyrshan ia ka da ka National Mission on Eduction lyngba ka ICT, MHRD, Sorkar India
09:52	Kham bun ki jingtip halor kane ka mission kidon ha http://spoken-tutorial.org/NMEICT-Intro.
10:03	Nga i Meboreen na Shillong nga pynkut ia kane. Khublei shibun.

Contributors and Content Editors

Meboreen Mary