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Integral of heaviside, why multiply the result with a heavside function again?

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data


    ∫(H(t-2)t)dt from t=0 to t.

    2. Relevant equations



    3. The attempt at a solution

    It's same as ∫t=2 t (t-2)dt, which is
    t^2/2-2t+2.

    But do we have to multiply the answer with H(t-2)?
     
  2. jcsd
  3. Nov 8, 2012 #2
    It's probably more for practice than anything else. You are correct that when you have a simple case like this, it's easy to look at it as a piecewise function and break the integral into the relevant pieces. That said, if you have a more complicated situation, it will be much easier to evaluate it as (I assume your textbook/teacher) says, without eliminating the unit step functions.
     
  4. Nov 8, 2012 #3

    LCKurtz

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    Gold Member

    No, it isn't the same as that.

    First of all, in order to not confuse the dummy variable inside the integral with the t in the limits you should write the problem like this:$$
    \int_0^t sH(s-2)\, ds$$Then think about where ##H(s-2) = 1## or ##H(s-2)=0## when working the integral.
     
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