Use the unit step function to express the function and then find the Laplace Transform
Find the Laplace transform of
just use the definition of th laplace transform(or you could memorize a theorem)
for the first one
$\displaystyle \mathcal{L}(f(t))=\int_{0}^{\infty}e^{-st}f(t)dt$
$\displaystyle \int_{0}^{2}e^{-st}\cdot 0 dt +\int_{2}^{\infty}(t+1)e^{-st}dt=-\frac{1}{s}(t+1)e^{-st}+\frac{1}{s^2}e^{-st}|_{2}^{\infty}$
$\displaystyle 0-[-\frac{1}{s}(2+1)e^{-2s}+\frac{1}{s^2}e^{-2s}]=\frac{3}{s}e^{-2s}+\frac{1}{s^2}e^{-2s}$
do the same thing for the 2nd one. Good luck.
First you need to convert the information given in the form of range to heaviside:
f(t) = 0, 0 < t < 2;
→ 0.[H(t-0) - H(t-2)]
= 0
f(t) = (t+1), t > 2;
→ (t+1). H(t+2)
[(t-2)+3].H(t-2)
(t-2)H(t-2) + 3H(t-2)
then you apply laplace transform to each of the terms above:
for the first term you will get: 0
for the second term you will get(using table):
exp(-2s)/(s^2) +3 exp(-2s)/s
for the second part just change (t+1) to 1 and repeat the steps again.
and yeah you can always use the general method posted by TheEmptySet.