Taking integral of a continuous pulse??? f: y=1, 0<x<1
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Hello,
I am generating a continuous pulse which is equal to 1 (y=1) while 0<x<1 such that:
t=0:0.001:1;
p1=ones(1,numel(t));
Now, I want to take its square integral (integral(f^2)). How can I do that?
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Walter Roberson
am 17 Mär. 2013
The square of 1 is 1, the square of 0 is 0, so f^2 is the same as f.
The value of the integral is going to depend on the bounds of integration. Over A to B, A <= B, I figure it is
min(B,1) - min(max(A,0),1)
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