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To solve part (v), we need to evaluate the Laplace Transform of \( F(t) \), which is defined as a piecewise function:

\[
F(t) = \begin{cases} 
0, & 0 < t < 2 \\
4, & t \geq 2 
\end{cases}
\]

### Step-by-Step Solution

1. **Define the Laplace Transform for a Piecewise Function:**

   The Laplace Transform of a function \( F(t) \) is defined by
   \[
   \mathcal{L} \{ F(t) \} = \int_{0}^{\infty} F(t) e^{-st} \, dt.
   \]
   For piecewise functions, we split the integral at the point where the function changes (in this case, \( t = 2 \)).

   \[
   \mathcal{L} \{ F(t) \} = \int_{0}^{2} 0 \cdot e^{-st} \, dt + \int_{2}^{\infty} 4 \cdot e^{-st} \, dt.
   \]

2. **Evaluate the First Integral:**

   Since \( F(t) = 0 \) for \( 0 < t < 2 \), the first integral is:
   \[
   \int_{0}^{2} 0 \cdot e^{-st} \, dt = 0.
   \]

3. **Evaluate the Second Integral:**

   For \( t \geq 2 \), \( F(t) = 4 \), so we compute:
   \[
   \int_{2}^{\infty} 4 \cdot e^{-st} \, dt = 4 \int_{2}^{\infty} e^{-st} \, dt.
   \]

4. **Integrate \( e^{-st} \):**

   The integral of \( e^{-st} \) is:
   \[
   \int e^{-st} \, dt = \frac{-e^{-st}}{s}.
   \]
   Applying this from \( t = 2 \) to \( t = \infty \):
   \[
   4 \cdot \left[ \frac{-e^{-st}}{s} \right]_{2}^{\infty} = 4 \cdot \left( 0 - \frac{-e^{-2s}}{s} \right) = \frac{4e^{-2s}}{s}.
   \]

### Final Answer

Thus,
\[
\mathcal{L} \{ F(t) \} = \frac{4e^{-2s}}{s}.
\]

Would you like a detailed explanation of any step, or do you have further questions?

Here are some related questions for deeper understanding:

1. How does the Laplace Transform handle other types of piecewise functions?
2. What is the role of the Heaviside function in representing piecewise functions for Laplace Transforms?
3. How would the solution change if \( F(t) \) had more intervals or different values?
4. Why does the integral from \( t = 0 \) to \( t = 2 \) contribute zero to the Laplace Transform?
5. Can we use a Laplace Transform table to simplify evaluations of such piecewise-defined functions?

**Tip:** Piecewise functions are often handled in Laplace Transforms by using the Heaviside step function, which simplifies the transform of functions defined on different intervals.

Please solve \( \mathcal{L} \{ F(t) \} \), where \( F(t) \) is defined as a piecewise function:

\[ F(t) = \begin{cases} 0, & 0 < t < 2 \\ 4, & t \geq 2 \end{cases} \]


Explore the solution for the Laplace Transform of a piecewise function, \( F(t) \), where \( F(t) = 0 \) for \( 0 < t < 2 \) and \( F(t) = 4 \) for \( t \geq 2 \). This problem covers advanced concepts in Laplace Transforms, integral calculus, and piecewise functions. Ideal for university-level students.

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