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Let's go through the problem step by step.

### Part (a)
You need to represent \( f(t) \) using the Heaviside step function \( h(t - c) \).

Given:
\[
f(t) = \begin{cases} 
t^2, & 0 \leq t < 2 \\
7, & t \geq 2 
\end{cases}
\]

We can use the Heaviside function to express \( f(t) \) in a single expression. The Heaviside function \( h(t - 2) \) "activates" at \( t = 2 \), allowing us to switch from \( t^2 \) to 7 at \( t = 2 \).

Thus:
\[
f(t) = t^2 + (7 - t^2) \cdot h(t - 2)
\]

This equation states that for \( t \geq 2 \), the term \( (7 - t^2) \cdot h(t - 2) \) will cancel \( t^2 \) and replace it with 7.

### Part (b)
Now, we need to find the Laplace transform of \( f(t) \), denoted as \( F(s) = \mathcal{L}\{f(t)\} \).

1. **Laplace Transform of \( t^2 \):**
   \[
   \mathcal{L}\{t^2\} = \frac{2}{s^3}
   \]

2. **Laplace Transform of \( (7 - t^2) \cdot h(t - 2) \):**

   Using the **second shifting theorem** for the Laplace transform, which states \( \mathcal{L}\{g(t) h(t - a)\} = e^{-as} \mathcal{L}\{g(t + a)\} \), we need to find the transform of \( g(t) = 7 - t^2 \) when shifted by 2.

   Substitute \( u = t - 2 \), so \( t = u + 2 \):
   \[
   g(t) = 7 - (u + 2)^2 = 7 - (u^2 + 4u + 4) = -u^2 - 4u + 3
   \]

   Thus, we find:
   \[
   \mathcal{L}\{(7 - t^2) \cdot h(t - 2)\} = e^{-2s} \left( \mathcal{L}\{-u^2 - 4u + 3\} \right)
   \]

   Compute each term:
   - \( \mathcal{L}\{u^2\} = \frac{2}{s^3} \)
   - \( \mathcal{L}\{4u\} = \frac{4}{s^2} \)
   - \( \mathcal{L}\{3\} = \frac{3}{s} \)

   So:
   \[
   \mathcal{L}\{(7 - t^2) \cdot h(t - 2)\} = e^{-2s} \left(-\frac{2}{s^3} - \frac{4}{s^2} + \frac{3}{s}\right)
   \]

3. **Combine Results:**

   \[
   F(s) = \frac{2}{s^3} + e^{-2s} \left(-\frac{2}{s^3} - \frac{4}{s^2} + \frac{3}{s}\right)
   \]

This is the Laplace transform of \( f(t) \).

Would you like a deeper explanation or have any questions about this solution?

Here are some related questions to expand on this topic:

1. How does the Heaviside function help in representing piecewise functions?
2. What is the significance of the second shifting theorem in Laplace transforms?
3. How can we interpret the exponential term \( e^{-2s} \) in the Laplace domain?
4. How would the solution change if the function had additional piecewise sections?
5. What are some practical applications of using Laplace transforms for piecewise functions?

**Tip:** When dealing with Heaviside functions, remember that the shifting theorem is a powerful tool for handling shifts in functions within Laplace transforms.

Consider the function f(t) defined as: f(t) = {t^2, 0 ≤ t < 2; 7, t ≥ 2}. Represent f(t) using Heaviside step functions and find the Laplace transform F(s) = L{f(t)}.

Learn how to represent a piecewise function using Heaviside step functions and find its Laplace transform. This problem involves expressing f(t) in terms of Heaviside functions and applying the Second Shifting Theorem. Suitable for college-level students.

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