# Simpson’s 3/8th Rule MATLAB Program Example

## 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:

## 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

## 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:

## 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