Difference between revisions of "Scilab/C4/Integration/English"

Welcome to the Spoken Tutorial on “ Composite Numerical Integration”

• 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

• Ubuntu 12.04 as the operating system
• and Scilab 5.3.3 version

• Scilab and
• Integration using Numerical Methods

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

• 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

• 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

Assume the number of intervals n is equal to ten.

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)))

• 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

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

Trap_composite(f, 0, 1, 10)

• 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

• 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

• 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

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)))

• 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

Type clc

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

Simp_composite( f, 1, 2 20)

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.

• 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

• 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

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))

• 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

Save and execute the file mid_composite.sci

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

Type mid_composite(f, 0, 1.5, 20)

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

• Develop Scilab code for numerical integration
• Find the value of an integral

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