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
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
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