PoojaMoolya at 06:07, 10 March 2017

2017-03-10T06:07:59Z

Sandhya.np14 at 11:01, 26 February 2015

2015-02-26T11:01:10Z

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Gaurav at 12:16, 10 July 2014

2014-07-10T12:16:50Z

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PoojaMoolya: 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 …'

2014-03-10T09:27:14Z

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 …'

New page

{| Border=1

|| 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:

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|00.11
|Develop '''Scilab''' code for different '''Composite Numerical Integration algorithms'''

|-
| 00.17
|Divide the '''integral''' into equal '''intervals'''

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|00.21
|Apply the algorithm to each '''interval''' and

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|00.24
|Calculate the '''composite value of the integral'''

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| 00.28
|To record this tutorial, I am using

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| 00.30
| '''Ubuntu 12.04''' as the operating system

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|00.34
| with '''Scilab 5.3.3''' version

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|00.38
||Before practising this tutorial, a learner should have basic knowledge of

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| 00.42
|'''Scilab''' and

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|00.44
| '''Integration using Numerical Methods'''

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| 00.47

| For '''Scilab''', please refer to the relevant tutorials available on the '''Spoken Tutorial''' website.

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| 00.55
| '''Numerical Integration''' is the:

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| 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

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|01.08
|It approximates a definite '''integral''' from values of the '''integrand'''

|-

|01.15

|Let us study '''Composite Trapezoidal Rule.'''

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|01.18

|This rule is the extension of '''trapezoidal rule'''

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| 01.22
|| We divide the interval '''a comma b '''into '''n''' equal intervals

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| 01.29

| Then '''h equal to b minus a divided by n''' is the common length of the intervals

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|01.36

| Then '''composite trapezoidal rule''' is given by

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|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'''

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|01.57

|| Let us solve an example using '''composite trapezoidal rule.'''

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|02.02

| Assume the number of intervals n is equal to ten.

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|02.09
|Let us look at the code for '''Composite Trapezoidal Rule''' on '''Scilab Editor'''

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| 02.16
||We first define the function with parameters '''f , a , b , n.'''

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| 02.22
|'''f '''refers to the function we have to solve,
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| 02.25
|| '''a ''' is the lower limit of the integral,
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|02.28

||''' b''' is the upper limit of the integral and

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|02.31

| '''n''' is the number of intervals.

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|02.34

| '''linspace''' function is used to create ten equal intervals between zero and one

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| 02.42

|| We find the value of the integral and store it in ''' I one'''

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| 02.49
| Click on '''Execute''' on '''Scilab editor''' and choose '''Save and Execute ''' the code.

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|03.02
| Define the example function by typing:

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| 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'''

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| 03.30
| Press '''Enter '''

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| 03.31
| Type '''Trap underscore composite open paranthesis f comma zero comma one comma ten close paranthesis
'''
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|03.41
| Press '''Enter '''

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|03.43
| The answer is displayed on the '''console '''

|-
| 03.47
| Next we shall study '''Composite simpson's rule.'''

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| 03.51

| In this rule, we decompose the interval ''' a comma b''' into '''n is greater than 1''' subintervals of equal length

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| 04.03
|| Apply '''Simpson's rule''' to each interval

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| 04.06

| We get the value of the integral to be

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|04.10

| '''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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|04.19

||Let us solve an example using '''Composite Simpson's rule. '''

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| 04.24

|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

| Let the number of intervals be '''twenty '''

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|04.37

| Let us look at the code for '''Composite simpson's rule'''

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|04.42
| We first define the function with parameters '''f , a , b , n. '''

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| 04.49

| '''f''' refers to the function we have to solve,

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|04.52

||'''a''' is the lower limit of the integral,

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|04.56

| '''b''' is the upper limit of the integral and

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| 04.58

| '''n''' is the number of intervals.

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| 05.02

|We find two sets of points.

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| 05.04

| We find the value of the function with one set and multiply it with two

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| 05.10

| With the other set, we find the value and multiply it with four

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| 05.16

||We sum these values and multiply it with '''h by three and store the final value in I '''

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| 05.24

||Let us execute the code

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| 05.28

|| Save and execute the file '''Simp underscore composite dot s c i'''

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| 05.39

|Let me clear the screen first.

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| 05.42

| Define the function given in the example by typing

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|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 '''

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| 06.14

| Type '''Simp underscore composite open paranthesis f comma one comma two comma twenty close paranthesis'''

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|06.24

||Press '''Enter '''

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| 06.26

| The answer is displayed on the console.

|-

| 06.31

| Let us now look at '''Composite Midpoint Rule.'''

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| 06.35

| It integrates polynomials of degree one or less

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|06.40

| Divides the interval '''a comma b''' into a ''' subintervals'''of equal width

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|06.49

| Finds the midpoint of each interval indicated by '''x i '''

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|06.54

| We find the sum of the values of the integral at each midpoint

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|07.00

| Let us solve this problem using '''Composite Midpoint Rule'''

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|07.05

| '''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

| We assume '''n''' is equal to '''twenty '''

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|07.18

| Let us look at the code for '''Composite Midpoint rule'''

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|07.24

| We first define the function with parameters '''f , a , b , n. '''

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|07.30

| '''f ''' refers to the function we have to solve,

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|07.33

| '''a''' is the lower limit of the integral,

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|07.36

| '''b ''' is the upper limit of the integral and

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|07.39

| '''n ''' is the number of intervals.

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|07.41

| We find the midpoint of each interval

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|07.45

| Find the value of integral at each midpoint and then find the sum and store it in '''I.'''

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|07.53

| Let us now solve the example

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|07.55

| Save and execute the file '''mid underscore composite dot s c i'''

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|08.04

| Let me clear the screen

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|08.08

| We define the function given in the example by typing

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|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'''

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|08.37

| Press '''Enter'''

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|08.39

| Then type '''mid underscore composite open paranthesis f comma zero comma one point five comma twenty close paranthesis'''

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|08.53

|Press '''Enter '''

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|08.54

| The answer is displayed on the '''console'''

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|08.59

| Let us summarize this tutorial.

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|09.02

| In this tutorial we have learnt to:

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|09.04

| Develop '''Scilab''' code for '''numerical integration'''

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|09.08

| Find the value of an '''integral'''

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|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

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|09.25

||Conducts workshops using spoken tutorials

|-

|06.29

||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.

@@ Line 6: / Line 6: @@
 |-
 | 00:01
-|Dear Friends,
+|Dear Friends,  Welcome to the Spoken Tutorial on '''Composite Numerical Integration'''.
-−
-|-
-| 00:02.
-| Welcome to the Spoken Tutorial on '''Composite Numerical Integration'''.
 |-
@@ Line 180: / Line 176: @@
 |-
 | 03:30
-| Press '''Enter '''.
+| Press '''Enter '''. Type '''Trap underscore composite open parenthesis f comma zero comma one comma ten close parenthesis'''
-−
-|-
-| 03:31
-|  Type '''Trap underscore composite open parenthesis f comma zero comma one comma ten close parenthesis'''
 |-
@@ Line 440: / Line 432: @@
 | We find the midpoint of each interval.
 |-
@@ Line 494: / Line 485: @@
 |08:53
-|Press '''Enter '''.
+|Press '''Enter '''. The answer is displayed on the '''console'''.
-−
- |-
-−
-|08:54
-−
-| The answer is displayed on the '''console'''.
 |-
@@ Line 525: / Line 510: @@
 | Find the value of an '''integral'''.
 |-

Scilab/C4/Integration/English-timed - Revision history

PoojaMoolya at 06:07, 10 March 2017

Sandhya.np14 at 11:01, 26 February 2015

Gaurav at 12:16, 10 July 2014

PoojaMoolya: 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 …'