# Scilab/C4/Integration/English

Title of script: Numerical Methods for Integration

Author: Shamika

Keywords: Integration, Numerical Methods, integral

Visual Cue
Narration
Slide 1 Dear Friends,

Welcome to the Spoken Tutorial on “ Composite Numerical Integration

Slide 2,3 -Learning Objective Slide At the end of this tutorial, you will learn how to:
• Develop Scilab code for different Composite Numerical Integration algorithms
• Divide the integral into equal intervals
• Apply the algorithm to each interval
• Calculate the composite value of the integral

Slide 4-System Requirement slide * To record this tutorial, I am using Ubuntu 12.04 as the operating system with Scilab 5.3.3 version

Slide 5- Prerequisites slide * Before practising this tutorial, a learner should have basic knowledge of Scilab and Integration using Numerical Methods
• For Scilab, please refer to the relevant tutorials available on the Spoken Tutorial website.

Slide 6- Numerical Integration Numerical Integration is the:
• Study of how the numerical value of an integral can be found
• It is used when exact mathematical integration is not available
• It approximates a definite integral from values of the
integrand

Slide 7,8- Composite Trapezoidal Rule-I Let us study Composite Trapezoidal Rule. This rule is
• The extension of trapezoidal rule
• We divide the interval a comma b into n equal intervals
• Then,
• h equal to b minus a divided by n is the common length of the intervals
• Then composite trapezoidal rule is given by
• 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

Slide 9- Example

* Let us solve an example using composite trapezoidal rule:
• Assume the number of intervals n is equal to ten.

Switch to Scilab editor

Highlight

function [I1] = Trap_composite(f, a, b, n)

x = linspace(a, b, n+1)

I1 = (h/2)*(2*sum(f(x)) - f(x(1)) - f(x(n+1)))

* Let us look at the code for Composite Trapezoidal Rule on Scilab Editor
• We first define the function with parameters f , a , b , n. f refers to the function we have to solve, a is the lower limit of the integral, b is the upper limit of the integral and n is the number of intervals.
• linspace function is used to create ten equal intervals between zero and one
• We find the value of the integral and store it in I one

Click on Execute on Scilab editor and choose Save and Execute the code * Click on Execute on Scilab editor and choose Save and Execute the code

Switch to Scilab Console

deff ('[y]=f(x)','y=1/(2*x+1)')

Trap_composite(f, 0, 1, 10)

* Define the example function by typing:
• 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
• Press enter
• Type
• Trap underscore composite open paranthesis f comma zero comma one comma ten close paranthesis
• Press enter
• The answer is displayed on the console

Slide 10, 11- Composite Simpson's Rule Next we shall study Composite simpson's rule. In this rule we
• decompose the interval a comma b into n is greater than 1 subintervals of equal length
• Apply Simpson's rule to each interval
• We get the value of the integral to be
• h by three multiplied by the sum of f zero, four into f one , two into f two to f n

Slide 12- Example

* Let us solve an example using Composite Simpson's rule
• We are given a function one by one plus x cube d x in the interval one to two
• Let the number of intervals be twenty

Switch to Scilab Editor and show the code for Simp_composite.sci

Highlight

function I = Simp_composite(f, a, b, n)

for i = 1:(n/2)-1

x1(i) = x(2*i)

end

for j = 2:n/2

x2(i) = x(2*i-1)

end

I = (h/3)*(f(x(1)) + 2*sum(f(x1)) + 4*sum(f(x2)) + f(x(n)))

* Let us look at the code for Composite simpson's rule
• We first define the function with parameters f , a , b , n.
f refers to the function we have to solve, a is the lower limit of the integral, b is the upper limit of the integral and n is the number of intervals.
• We find two sets of points
• We find the value of the function with one set and multiply it with two
• With the other set we find the value and multiply it with four
• We sum these values and multiply it with h by three and store the final value in I
• Let us execute the code

Click on Execute and choose

Save and execute the file

Simp_composite.sci

* Save and execute the file
• Simp underscore composite dot s c i

Switch to Scilab Console

Type

clc

deff ('[y]=f(x)','y=1/(1+x^3)')

Simp_composite( f, 1, 2 20)

* Let me clear the screen first.
• Define the function given in the example by typing
• 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
• Press enter
• Type Simp underscore composite open paranthesis f comma one comma two comma twenty close paranthesis
• Press enter
• The answer is displayed on the console

Slide 13, 14- Composite Midpoint Rule Let us now look at Composite Midpoint Rule. It
• Integrates polynomials of degree one or less
• Divides the interval a comma b into n subintervals of equal width
• Finds the midpoint of each interval indicated by x i
• We find the sum of the values of the integral at each midpoint

Slide 15- Example

Let us solve this problem using Composite Midpoint Rule
• We are given a function one minus x square d x in the interval zero to one point five
• We assume n is equal to twenty

Switch to Scilab Editor

Show the file mid_composite.sci

Highlight

function I = mid_composite(f, a, b, n)

x = linspace(a + h/2, b - h/2, n)

I = h*sum(f(x))

* Let us look at the code for Composite Midpoint rule
• We first define the function with parameters f , a , b , n.
f refers to the function we have to solve, a is the lower limit of the integral, b is the upper limit of the integral and n is the number of intervals.
• We find the midpoint of each interval
• Find the value of integral at each midpoint and then find the sum and store it in I.
• Let us now solve the example

Click on Execute and choose

Save and execute the file mid_composite.sci

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

On the Scilab Console type:

clc

deff ('[y]=f(x)','y=1-x^2')

Type mid_composite(f, 0, 1.5, 20)

* Let me clear the screen
• We define the function given in the example by typing
• 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
• Press enter
• Then type
mid underscore composite open paranthesis f comma zero comma one point five comma twenty close paranthesis
• Press enter
• The answer is displayed on the console

Slide 16- Summary Let us summarize this tutorial. In this tutorial we have learnt to:
• Develop Scilab code for numerical integration
• Find the value of an integral

Show Slide 17

Title: About the Spoken Tutorial Project

• It summarises the Spoken Tutorial project
• If you do not have good bandwidth, you can download and watch it

* Watch the video available at the following link
• It summarises the Spoken Tutorial project
• If you do not have good bandwidth, you can download and watch it

Show Slide 18

Title: Spoken Tutorial Workshops

The Spoken Tutorial Project Team

• Conducts workshops using spoken tutorials
• Gives certificates for those who pass an online test
• For more details, please write to contact@spoken-tutorial.org

The Spoken Tutorial Project Team
• Conducts workshops using spoken tutorials
• Gives certificates for those who pass an online test
• For more details, please write to contact at spoken hyphen tutorial dot org

Show Slide 12

Title: Acknowledgement

• Spoken Tutorial Project is a part of the Talk to a Teacher project
• It is supported by the National Mission on Education through ICT, MHRD, Government of India