Taking integral of a continuous pulse??? f: y=1, 0<x<1

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?

Answer 1

[Walter Roberson]

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)