Why do I get the wrong value when I increase N?

How can I use the below equation for the function sin(pi*x/10) for epsilon=[ -sqrt(3/5) 0 sqrt(3/5)] and w=[5/9 8/9 5/9]?

This is my code so far. When I increase N to 3 I get the wrong answer.

epsilon = [-sqrt(3/5) 0 sqrt(3/5)]; % absciassas values for 3-point (n=3)
w = [5/9 8/9 5/9];  % weight factors for 3-point (n=3)
N = 2; % Number of segments
a = 0; % Lower Boundary
b = 10; % Upper Boundary
h = (b-a)/2/N; % Width of segment
x = h*epsilon + (b+a)/2/N; % Funtion evaluation point
f = sin(pi*x/10); % Function
Gauss = h*(sum(w.*f)*N) % Answer

Answer 1

[Torsten]

a = 0; % Lower Boundary
b = 10; % Upper Boundary
fun = @(x)sin(pi*x/10);  % Function
zeta = [-sqrt(3/5) 0 sqrt(3/5)]; % absciassas values for 3-point (n=3)
w = [5/9 8/9 5/9];  % weight factors for 3-point (n=3)
x = (a+b)/2 + zeta*(b-a)/2;
f = fun(x);
format long
Gauss = (b-a)/2*w*f.'

Gauss = 
   6.370618772999813

Exact_num = integral(fun,a,b)

Exact_num = 
   6.366197723675814

Exact = 10/pi*(cos(pi*a/10) - cos(pi*b/10))

Exact = 
   6.366197723675814

[Maddy Taylor]

Thank you. How can I apply this when the domain for the integral is diveded into multiple segments. Eg. N=2 (domain divided into 2 segements), For the first segment, the lower and upper limits of the integral are 0 and 5, respectively. For the second segment, the lower and upper limits of the integral are 5 and 10, respectively.

[Torsten]

x = 0:0.1:10;
f = @(x)sin(pi*x/10);  % Function
int_value = 0.0;
for i = 1:numel(x)-1
    a = x(i);
    b = x(i+1);
    int_value = int_value + gauss(a,b,f);
end
format long
int_value

int_value = 
   6.366197723675818

Exact = 10/pi*(cos(pi*x(1)/10) - cos(pi*x(end)/10))

Exact = 
   6.366197723675814

function int_value = gauss(a,b,f)
  zeta = [-sqrt(3/5) 0 sqrt(3/5)]; % absciassas values for 3-point (n=3)
  w = [5/9 8/9 5/9];  % weight factors for 3-point (n=3)
  x = (a+b)/2 + zeta*(b-a)/2;
  int_value = (b-a)/2*w*f(x).';
end