How could I find the integral of a function given this code?

I'm having trouble with a simple Monte Carlo integration code. It is as follows:

%define variables 
Fun = @(x) x^2;
a = 1; % bounds for integration 
b = 2;
c = 0;
d = 10;
n_under = 0; %total number of points that fall below the function
Int = 0; %integral is initially at 0
n_iter = 1000000; %total number of points
for i =1:n_iter
    %random number generator 
    r_x = rand();
    x_t = r_x*(b-a); %for x values 

      r_y = rand();
      y_t = r_y*(d-c); %for y values 

if Fun(x_t) > y_t
    n_under = n_under + 1; %Monte Carlo "walk", if the points fall below the function, then we add, else we
end                        %don't add

end

Int = n_under/n_iter * (b - a) * (d - c) %final function

hold on;

x = 0:.1:10;
y = Fun(x);

plot (a,c,'o',b,d,'o',a,d,'o',b,c,'o')

plot (x,y) %just plotted the function here to make sure the plot is good. it is

I'm having trouble when I change the bounds of integration. Specifically if I use bounds (0,1) for almost any function, it works perfectly (with some error, of course), however if, such as above is use bounds (1,2) or other (anything that is not a = 0) the code outputs the wrong answer. Could anyone help with this issue? Much thanks!

[Birdman]

a and b are the boundaries of x axis, c and d are the boundaries of y axis for corresponding values of a and b from x^2 function , right?

[Gaëtan Poirier]

That is correct

[Birdman]

Then why did you enter (0,10) for y boundaries where it has to be (1,4) for corresponding values of (1,2)?

[Gaëtan Poirier]

It gives roughly the same answer. I've already tried that.

Answer 1

[Roger Stafford]

If a is not zero, then “x_t = r_x*(b-a);” is wrong. It should be:

   x_t = a + r_x*(b-a);

[Gaëtan Poirier]

That's exactly it. Thank you so much!