Answer
$$\dfrac{1-2 e^{-2s}}{s} $$
Work Step by Step
The Laplace Transform can be written as:
$L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt $
Now, $L[F(t)]=\int_{0}^{2} e^{-st} f(t) dt+\int_{2}^{\infty} e^{-st} f(t) dt\\=\int_{0}^{2} e^{-st} dt+\int_{2}^{\infty} e^{-st} dt\\=\dfrac{e^{-2s}}{-s}+\dfrac{1}{s}+ \dfrac{1}{s} (e^{-\infty}-e^{-2s})\\=\dfrac{1-2 e^{-2s}}{s} $
