Question
Evaluate the integral x^4 within limits -3 to 3 using Simpson's 3/8th 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 3/8th Rule:
answer= (3h/8)*[(y1+y7)+3*(y2+y3+y5+y6)+2*(y4)]
answer=(3/8)*[(81+81)+3*(16+1+1+16)+2*(0)]
answer=99.
MATLAB Code for Simpson's 3/8th 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=(3*h/8)*((y(1)+y(l))+3*(y(2)+y(3)+y(5)+y(6))+2*(y(4)))
Image Format
[caption id="" align="aligncenter" width="490"] MATLAB Program for Simpon's Three-Eighth Rule[/caption]
Second Example
Evaluate the integral 1/(1+x) within limits 0 to 6 using Simpson's 3/8th 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 3/8th rule:
answer= (3h/8)*[(y1+y7)+3*(y2+y3+y5+y6)+2*(y4)]
answer=1.9660
MATLAB Code for Simpson's 3/8th 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=(3*h/8)*((y(1)+y(l))+3*(y(2)+y(3)+y(5)+y(6))+2*(y(4)))
Image Format
[caption id="" align="aligncenter" width="490"] MATLAB Program for Simpon's Three-Eighth Rule[/caption]
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