Simpson's 1/3 Rule Implementation Issue (Code Provided)

I have written the following code to Implement Simpson's 1/3 Rule, but it does not seem to give the correct answer. I am unsure what I have done wrong in my code.

MATLAB CODE:

function Integration(a,b,x,n,true)

a=0; %Lower Integral Bound

b=1; %Upper Integral Bound

x=[a,.25,.5,.75,b];

n=4; %Even Number of Segments

f=zeros(n+1,1);

true=.602298; %True Value of Equation

h=(b-a)/n; %width of each segment

f(1)=(x(1)^0.1)*(1.2-x(1))*(1-exp(20*(x(1)-1))); %Value For Initial Bound

for i=2:n

f(i)=2*(x(i)^0.1)*(1.2-x(i))*(1-exp(20*(x(i)-1))); %Values for In-Between Bounds

end

f(n+1)=(x(n+1)^0.1)*(1.2-x(n+1))*(1-exp(20*(x(n+1)-1))); %Value for Final Bound

q=sum(f,'all');

I=(h/2)*(q); %Width Times Average Height

disp('USING TRAPEZOIDAL')

fprintf('I=')

disp(I)

et=abs((true-I)/true)*100;

fprintf('True Error=')

disp(et)

disp("USING SIMPSON'S 1/3")

f2=zeros(n,1);

f2(1)=(x(1)^0.1)*(1.2-x(1))*(1-exp(20*(x(1)-1)));

for j=2:2:n-2

f2(j)=2*(x(j)^0.1)*(1.2-x(j))*(1-exp(20*(x(j)-1)));

end

for j=3:2:n-1

f2(j)=4*(x(j)^0.1)*(1.2-x(j))*(1-exp(20*(x(j)-1)));

end

f2(n)=(x(n)^0.1)*(1.2-x(n))*(1-exp(20*(x(n)-1)));

v=sum(f2,'all');

I2=(h/3)*v;

fprintf('I=')

disp(I2)

%ea=abs((true-I2)/true)*100;

%printf('True Error=')

%disp(ea)

%Nobody Cares, Work Harder

%Keep Hammering

end

Answer 1

[David Hill]

The following function might help.

function o = simp(a,b,f,n)
x=linspace(b,a,n+1);
o=0;
for k=1:n
    xx=[x(k);(x(k)+x(k+1))/2;x(k+1)];
    y=f(xx);
    c=([xx.^2,xx,ones(3,1)]\y)';
    p=polyint(c);
    o=o+polyval(p,x(k))-polyval(p,x(k+1));
end

Evaluate the function using your equation desiring integration.

f=@(x)x.^.1.*(1.2-x).*(1-exp(20*(x-1)));
output=simp(0,1,f,4);