# 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.