Simpson's 1/3rd Rule MATLAB Program Examples

Question


Evaluate the integral x^4 within limits -3 to 3 using Simpson's 1/3 rd rule.

Solution


Let y(x)=x^4

here a=-3 and b=3

therefore (b-a)=6

let ‘n’ be the number of intervals. assume n=6 in this case.

also h=(b-a)/n = 6/6 =1

x: -3  -2  -1  0  1  2  3

y: 81  16  1  0  1  16  81

According to Simpson's 1/3rd  rule:

answer= (h/3)*[(y1+y7)+2*(y3+y5)+4*(y2+y4+y6)]

answer=(1/2)*[(81+81)+2*(1+1)+4*(16+0+16)]

answer=98.

MATLAB Program For Simpson's 1/3rd Rule


%Created by myclassbook.org (Mayuresh)
%Created on 24 May 2013
%Question: Evaluate the integral X^4 within limits 3 to -3

clc;
clear all;
close all;

f=@(x)x^4; %Change here for different function
a=-3;b=3; %Given limits
n=b-a; %Number of intervals
h=(b-a)/n;
p=0;

for i=a:b
p=p+1;
x(p)=i;
y(p)=i^4; %Change here for different function
end

l=length(x);
x
y
answer=(h/3)*((y(1)+y(l))+2*(y(3)+y(5))+4*(y(2)+y(4)+y(6)))

Image Format








[caption id="" align="aligncenter" width="490"]MATLAB code for Simpson's one third rule MATLAB code for Simpson's one-third rule[/caption]





Second Example


Evaluate the integral 1/(1+x) within limits 0 to 6 using Simpson's 1/3 rd rule.

Solution


Let y(x)=1/(1+x)

here a=0 and b=6

therefore (b-a)=6

let ‘n’ be the number of intervals. assume n=6 in this case.

also h=(b-a)/n = 6/6 =1

x: 0                  1                    2              3                  4               5               6

y: 1.0000   0.5000   0.3333   0.2500   0.2000   0.1667   0.1429

According to Simpson's 1/3 rd rule.

answer= (h/3)*[(y1+y7)+2*(y3+y5)+4*(y2+y4+y6)]

answer=1.9587.

MATLAB Code for Simpson's 1/3rd Rule


%Created by myclassbook.org (Mayuresh)
%Created on 24 May 2013
%Question: Evaluate the integral 1/(1+x) within limits 0 to 6

clc;
clear all;
close all;

f=@(x)1/(1+x); %Change here for different function
a=0;b=6; %Given limits
n=b-a; %Number of intervals
h=(b-a)/n;
p=0;

for i=a:b
p=p+1;
x(p)=i;
y(p)=1/(1+i); %Change here for different function
end

l=length(x);
x
y
answer=(h/3)*((y(1)+y(l))+2*(y(3)+y(5))+4*(y(2)+y(4)+y(6)))

Image Format








[caption id="" align="aligncenter" width="490"]MATLAB code for Simpson's one third rule MATLAB code for Simpson's one-third rule[/caption]





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2 Comments

  1. % please I need to help how to integral (sin^3 ).x
    % Generally sin^3 (x) is equal 1/4[3sinx - sin 3x]
    clc;
    clear all;
    close all;

    f=@(x) (sinX)^3 ; %Change here for different function
    a=0;b=pi/2; %Given limits
    n=b-a; %Number of intervals
    N=6;
    h=n/N;
    p=0;

    for i=a:b
    p=p+1;
    x(p)=i;
    y(p)=(sin i)^3 ; %Change here for different function
    end

    l=length(x);
    x
    y
    answer=(h*1/3)*((y(1)+y(l))+2*(y(3)+y(5))+4*(y(2)+y(4)+y(6)))
    plot(x,y);

    ReplyDelete
  2. Please change the first loop I mean
    for i=a:h:b

    this solves your problem and you get a nice plot.

    ReplyDelete