Unit Step Function Question: What is $\mathcal{L}\{u(t-1)u(t-2)\}$? 
My calculations
$e^{-2}s \mathcal{L}\{u(t+2)-1\}$
$e^{-2}s \mathcal{L}\{t+1\}$
$e^{-2}s (s^{-2}+\frac{1}{2})$
I'm confused, I gotten the wrong answer.. Please point out my mistake
 A: First let us recall that, $$\mathcal{L} \{ f(t) u(t - a) \} = e^{-as}\mathcal{L} \{ f(t+a)\}.$$
Let's do this step by step.
Note that for any $f$ defined on $t \geq 0$, the above identity gives us
$$\mathcal{L} \{ f(t) u(t - 2) \} = e^{-2s}\mathcal{L} \{ f(t+2)\}.$$
Therefore if we let $f(t) = u(t -1)$ now, then we get
\begin{align}\mathcal{L} \{ u(t-1) u(t - 2) \} &= e^{-2s}\mathcal{L} \{ u(t-1 + 2)\} \\
&= e^{-2s}\mathcal{L} \{ u(t+1)\}\\
&=e^{-2s} \frac{1}{s} \\
&=  \frac{e^{-2s}}{s}. \end{align}
Alternatively, you may use the fact that $\mathcal{L} \{ u(t-1) u(t - 2) \}= \mathcal{L} \{  u(t - 2) \}$ and apply the identity when $f(t) = 1.$
A: The Time Displacement Theorem (reference link) says:

If $F(s)=\mathcal{L}\{f(t)\}$, then $\mathcal{L}\{u(t−a)⋅g(t−a)\}=e^{−as}G(s)$ for some real number $a$.

Applying this theorem, 
\begin{align} \mathcal{L}\{\underbrace{u(t-1)}_{u(t-a)}\underbrace{u(t-2)}_{g(t-a)}\}&=e^{-s}\mathcal{L}\{\underbrace{u(t-2)}_{g(t-a)}\} \\
&= e^{-s} \underbrace{\frac{e^{-2s}}{s}}_{G(s)} \\
&=\frac{e^{-3s}}{s}
\end{align}
A: Resorting to the definition, in my eyes, is a lot easier. The definition of the unit step is 
$$
\mathcal{U}(t-a)=\begin{cases}
0,& t<a\\
1,& t>a
\end{cases}
$$
and the definition of Laplace transform is
$$
\mathcal{L}\{f(t)\}(s) = \int_0^{\infty}f(t)e^{-st}dt
$$
Now, let's consider your $f(t)$.
\begin{align}
\mathcal{L}\{f(t)\}(s)&=\int_0^{\infty}\mathcal{U}(t-1)\mathcal{U}(t-2)e^{-st}dt\\
&=\int_1^{\infty}\mathcal{U}(t-2)e^{-st}dt\\
&=\int_2^{\infty}e^{-st}dt\\
&=\frac{e^{-2s}}{s}
\end{align}
From the first unit step, we had that argument of the integral was zero up until $t=1$ and then it was $1\cdot\mathcal{U}(t-2)e^{-2s}$, but from the second unit step, we get the argument is zero up until $t=2$ which lead to evaluating 
$$
\int_2^{\infty}e^{-st}dt
$$
