% truncation_example_3.m
% Truncation error example #3
% A Riemann sum to approximate a limit of Riemann sums

function foo
	xmin = 0;   xmax = 1;
	fprintf('\nNo. of subintervals     Riemann sum\n');
	n=1;
	for k = 1:11
		dx = ( xmax - xmin ) / n; 
	   s = 0;
	   for i = 0:n-1
	      x = xmin + ( i + 0.5 ) * dx;
	      s = s + f(x) * dx;
	   end
	   fprintf('%14i %23.15f\n',n,s);
	   n = n * 2;	% increase the number of subintervals and repeat
	end
	fprintf('\n');
end

function fofx = f(x)
	fofx = x^2;
end