To find the Laplace transform of the piecewise function $f(t)$, we need to handle it in segments:

1. Function Definition: \begin{cases} e^{-t}, & 0 \leq t < 1, \\ e^{-2t}, & t \geq 1. \end{cases}$$
2. Laplace Transform Formula: The Laplace transform of a function $f(t)$ is defined as: $\mathcal{L}\{f(t)\} = \int_0^{\infty} f(t) e^{-st} \, dt.$

Splitting the Integral:

We split the Laplace transform into two integrals based on the piecewise definition of $f(t)$: $\mathcal{L}\{f(t)\} = \int_0^1 e^{-t} e^{-st} \, dt + \int_1^\infty e^{-2t} e^{-st} \, dt.$

Solve Each Integral:

1. For $0 \leq t < 1$:

$\int_0^1 e^{-t} e^{-st} \, dt = \int_0^1 e^{-(s+1)t} \, dt.$ Using the integral formula $\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C$: $\int_0^1 e^{-(s+1)t} \, dt = \left[ \frac{e^{-(s+1)t}}{-(s+1)} \right]_0^1 = \frac{1}{s+1} (1 - e^{-(s+1)}).$

2. For $t \geq 1$:

$\int_1^\infty e^{-2t} e^{-st} \, dt = \int_1^\infty e^{-(s+2)t} \, dt.$ Using the same formula: $\int_1^\infty e^{-(s+2)t} \, dt = \left[ \frac{e^{-(s+2)t}}{-(s+2)} \right]_1^\infty.$ At $t = \infty$, $e^{-(s+2)t} \to 0$. At $t = 1$: $\int_1^\infty e^{-(s+2)t} \, dt = \frac{e^{-(s+2)}}{s+2}.$

Combine Results:

Adding the results of the two integrals: $\mathcal{L}\{f(t)\} = \frac{1}{s+1} (1 - e^{-(s+1)}) + \frac{e^{-(s+2)}}{s+2}.$

This is the Laplace transform of the given piecewise function.

Do you want a step-by-step derivation for any specific part? Let me know!

Related Questions:

1. How does the Laplace transform handle piecewise functions in general?
2. Can we use the Heaviside function to simplify the piecewise Laplace transform process?
3. What are the implications of changing the boundaries in the Laplace transform?
4. How does the Laplace transform help in solving differential equations involving such functions?
5. What happens to the Laplace transform when $s \to \infty$?

Tip: Always verify continuity and smoothness at the transition points of piecewise functions to avoid miscalculations in transforms.