Infinite Loop help Riemann sums

For some reason it's get stuck in an infinite loop. The programs says line 8 is the problem so the "first = first" line so I'm assuming its the bound problem? I don't see the problem any suggestions?

Note: my functions needs to give me an approx that has an difference from the previous estimate less than the tol that's why I'm running a while and for loop.

[Walter Roberson]

I suggest you step through it line by line with the debugger.

[Azzi Abdelmalek]

It's difficult to say if we don not know a,b,tol and your function f.

[Kenny]

the function is myintegral(@(x) (sin(x^2))^2,0,3,.1) but the code needs to be able to approximate any function.

Answer 1

[Youssef Khmou]

hi Kenny, your program works fine for most of cases

Note : no need to declare syms x ,

I tried to to make some change according to your model , like number of samples n, no need to increase through the loop, the theorem states that the Sum converges when N->Inf :

function y = myintegralapprox(f,a,b,tol)

 tol=100*tol;
 n = 1000;
 c = tol + 1;
 first = 0;
 second = 0;
 counter=1;
 while (c > tol)
    for m=a+(((b-a)/n)/2):((b-a)/n):(b-((b-a)/n)/2)
        first = first + ((b-a)/n)*(f(m));
    end

 counter=counter+1;

 if counter>1e+6
    break;
 end
 y = first;    
 c = abs(first - second); 
 second = first;
 end

I made some tests comparing you func with "quad" function :

 >>myintegralapprox(@(x) 1/(1+exp(x)),0,pi,.1)
 >>quad(@(x) 1./(1+exp(x)),0,pi)

 >>myintegralapprox(@(x) cos(x),0,pi,.1)
 >>quad(@(x) cos(x),0,pi)

 >>myintegralapprox(@(x) exp(-x^2),0,pi,.1)
 >>quad(@(x) exp(-x.^2),0,pi,.1)

 >>myintegralapprox(@(x) (x^3)+(x^2)+1,0,pi,1)
 >>quad(@(x) (x.^3)+(x.^2)+1,0,pi)

[Kenny]

Sorry, I dont think this is the right answer. It's not giving me the right numbers according to the example :/. I answer should be 1.17501419