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

1. Nov 8, 2012

kougou

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. Nov 8, 2012

bossman27

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.

3. Nov 8, 2012

LCKurtz

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.