I need to get the histogram of z (i)

n=input('donner le nombre de points ')
k=0;
for i=1:n
  u1=rand;
  u2=rand;
  x=2*u1;
  y=2*u2;
  if(y<=[(4*x)/(x^2 +1)]-[x*(x-12)])
      k=k+1;

    end
end
Surface = k/n;
disp('la surface est:')
Surface
for i=1:n
    z(i)=k/i;
    disp('la valeure de z(i)')
    z(i)
end

[Torsten]

I can't make sense of your code.

Could you explain in your own words what you are trying to do ?

[Melissa Ouafi]

I am trying To do the Monte Carlo method To have the area of this function : (4x/x^2+1)-(x(x-12)),z(i) is the area of this function when we change n,So I need to get the histogram of z(i) when n Will be changed.

[Torsten]

Which area ? The area under the curve y(x) = (4*x)/(x^2 +1)-x*(x-12) in the interval [0;2] ?

Try

f=@(x)4*x./(x.^2 +1)-x.*(x-12);
figure(1)
plot(0:0.01:2,f(0:0.01:2))
n = 2000;
ntrials = 10000;
for j = 1:ntrials
    x = 2*rand(n,1);
    y = f(2)*rand(n,1);
    k(j) = numel(y(y<=f(x)))/n;
end
integral_approx = mean(k)*2*f(2)
figure(2)
histogram(k*2*f(2),'Normalization','pdf')

Answer 1

[Image Analyst]

z(i) is a single variable so taking a histogram of that will have only one bin. You want a histogram of z, not z(i). So at the end of your code add

histogram(z);