Scilab/C4/Integration/English-timed
From Script | Spoken-Tutorial
Time | Narration |
00:01 | Dear Friends, 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 . 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 . 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. |