x = 0:0.5:4
f = [3 1 2 0 5 -2 3 1 -1]
% plot(x,f,'*')
function T = trapez(x,f)
  h = x(2)-x(1)
  m = length(x)
  % intervallumok szama:
  n = m-1
  %% itt nem baj, ha paratlan sok az intervallum
  so = 2*sum(f) - f(1) - f(end)
  T = so*h/2
endfunction
function T = trapez_hibaval(x,f)
  h = x(2)-x(1)
  m = length(x)
  % intervallumok szama:
  n = m-1
  so = 2*sum(f) - f(1) - f(end)
  T = so*h/2
  if mod(n,2) == 1
    %% ha n paratlan, ott nem tudunk T_(n/2)-t szamolni!
    hiba = Inf
  else
    f_uj = f(1:2:end)
    so_uj = 2*sum(f_uj) - f_uj(1) - f_uj(end)
    T_uj = so_uj*h %% so_uj*h*2/2
    hiba = abs(T-T_uj)
  endif
  hiba
endfunction
function S = Simpson(x,f)
  h = x(2)-x(1)
  n = length(x) - 1
  if mod(n,2) == 1
    %% ha n paratlan, Simpson formula nem ertelmezheto
    S = Inf
  else
    %% ha n paros, szamolunk
    so = f(1) + f(end)
    so += 4* sum( f(2:2:(end-1)) )
    so += 2* sum( f(3:2:(end-2)) )
    S = so*h/3
  endif
endfunction
function S = Simpson_hibaval(x,f)
  h = x(2)-x(1)
  n = length(x) - 1
  if mod(n,2) == 1
    %% ha n paratlan, Simpson formula nem ertelmezheto
    S = Inf
  else
    %% ha n paros, szamolunk
    so = f(1) + f(end)
    so += 4* sum( f(2:2:(end-1)) )
    so += 2* sum( f(3:2:(end-2)) )
    S = so*h/3
  endif
  if mod(n,4) == 0
    %% ha n oszthato 4-gyel, akkor meg n-2 is paros,
    % arra is lehet Simpson, es van utolagos hibabecsles
    %% szamoljuk ki S_(n/2)-t:
    f_uj = f(1:2:end)
    so_uj = f_uj(1) + f_uj(end)
    so_uj += 4* sum( f_uj(2:2:(end-1)) )
    so_uj += 2* sum( f_uj(3:2:(end-2)) )
    S_uj = so_uj*h*2/3 % vagy: h_uj = 2*h -val
    hiba = abs(S - S_uj)
  else
    hiba = Inf
  endif
  hiba
endfunction