Signals and Systems - Taking Integral of Unit Step Function

I am a taking signals and systems course and my professor posted the solutions to our latest homework and I am trying to understand one of the steps. Below is the solution. Bracketed in red is what I am trying to understand.

Why does u(t) disappear when you move the bounds of the integral to infinity to zero? I am just trying conceptually understand why instead of just accepting it as fact.

• The unit step function changes from 0 to 1 at x=0. The integral of the unit step from -infinity to 0 is 0. Therefore you move the lower limit up to 0 and remove the unit step function. – A.S. Sep 13 '18 at 16:12
• @A.S. simple and efficient...I wish I could upvote your answer wink wink nudge nudge – Simon Marcoux Sep 13 '18 at 16:24

I suspect $x_1(t) = e^{-2t} u(t)$
Since $u(t) = 0$ for $t<0$, the next step simple changes the boundaries of the integration to reflect that the product shown is zero for all $t<0$. Once you do that, you can just drop the $u(t)$, since multiplying a signal by one just returns the original signal.