Faster method of numerical Integration?

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Howdy,
I'm currently running a for loop containing only the trapezoidal rule and a count variable, i.e. something like:
for n=2:length(a)
computeVel(n-1,1) = trapz(a,b);
n=n+1;
end
I'm running this for 600k iterations, and will be running it often. I need the solution at each successive upper bound in the vector 'b.' I.e., if b=[1,2,3,4] i need the integral on 1 to 2, 1 to 3, and 1 to 4.
Is there a fast(er) way to do this?

Accepted Answer

Andrei Bobrov
Andrei Bobrov on 17 Apr 2012
out = cumtrapz(a,b)

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