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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.

enter image description here

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.

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  • \$\begingroup\$ 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. \$\endgroup\$ – A.S. Sep 13 '18 at 16:12
  • \$\begingroup\$ @A.S. simple and efficient...I wish I could upvote your answer wink wink nudge nudge \$\endgroup\$ – Simon Marcoux Sep 13 '18 at 16:24
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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.

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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.

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