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First of all,

If this is a two-terms function I'd be simple. It will produce $$ \mathcal{L}[u(t)] = \frac{1}{s} $$

Except, I'm not sure what to do with $u(4-t)$. If it was $u(t-4)$, it would be simpler.


Since these two terms are not separate. I couldn't find a property to help me deal with it. The closest I found is the 'Time shift' property which is $$ \mathcal{L}[f(t-T)u(t-T)] = e^{-sT}F(s)$$

** So how would you approach**

$$ \mathcal{L}[u(t) u(4-t)] $$

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Dumb question: would $u(t)u(4-t)=u(t)-u(t-4)$? – Mike Jan 10 '13 at 22:59

I think the easiest way is to use the definition of the Laplace transform directly. Note that $$u(t)u(4-t) = \begin{cases} 1, & 0 < t < 4 \\0, & \text{otherwise}, \\ \end{cases}$$ so

$$\mathcal{L}[u(t)u(4-t)](s) = \int_0^4 e^{-st}\,dt,$$

which I am sure you can work out.

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