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To solve this problem, let’s go through each part:

### Part (a): Representing \( f(t) \) with Heaviside Step Functions

The function \( f(t) \) is defined piecewise:
\[
f(t) = \begin{cases} 
t^2, & 0 \leq t < 2 \\
6, & t \geq 2 
\end{cases}
\]

To express this function using Heaviside step functions, \( h(t - c) \), we can rewrite it as follows:

1. For \( t < 2 \), \( f(t) = t^2 \).
2. At \( t = 2 \) and beyond, \( f(t) \) becomes 6.

Using the Heaviside function \( h(t - 2) \), we can rewrite \( f(t) \) as:
\[
f(t) = t^2 + (6 - t^2) h(t - 2)
\]

This form works because, for \( t < 2 \), \( h(t - 2) = 0 \), leaving \( f(t) = t^2 \). For \( t \geq 2 \), \( h(t - 2) = 1 \), so \( f(t) = t^2 + (6 - t^2) = 6 \).

### Part (b): Finding the Laplace Transform \( F(s) = \mathcal{L} \{ f(t) \} \)

Now, let’s find the Laplace transform of \( f(t) \):
\[
f(t) = t^2 + (6 - t^2) h(t - 2)
\]

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

2. **Laplace Transform of \( (6 - t^2) h(t - 2) \)**:
   Using the second shift theorem (or Heaviside shift), we find the transform of \( (6 - t^2) h(t - 2) \):
   \[
   \mathcal{L} \{ (6 - t^2) h(t - 2) \} = e^{-2s} \mathcal{L} \{ 6 - (t + 2)^2 \}
   \]
   
   Expanding \( (t + 2)^2 \) gives:
   \[
   (t + 2)^2 = t^2 + 4t + 4
   \]
   So, \( 6 - (t + 2)^2 = 6 - (t^2 + 4t + 4) = -t^2 - 4t + 2 \).

Now, we need the Laplace transforms of each term:
   \[
   \mathcal{L} \{ 6 \} = \frac{6}{s}, \quad \mathcal{L} \{ -t^2 \} = -\frac{2}{s^3}, \quad \mathcal{L} \{ -4t \} = -\frac{4}{s^2}, \quad \mathcal{L} \{ 2 \} = \frac{2}{s}
   \]

Therefore:
\[
\mathcal{L} \{ (6 - t^2) h(t - 2) \} = e^{-2s} \left( \frac{6}{s} - \frac{2}{s^3} - \frac{4}{s^2} + \frac{2}{s} \right)
\]

3. **Combine Terms**:
   \[
   F(s) = \frac{2}{s^3} + e^{-2s} \left( \frac{6}{s} - \frac{2}{s^3} - \frac{4}{s^2} + \frac{2}{s} \right)
   \]

Would you like further simplification, details on each step, or do you have any specific questions?

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Here are five follow-up questions that could enhance understanding:

1. How does the Heaviside function simplify the representation of piecewise functions in Laplace transforms?
2. Why is the second shifting theorem useful for functions defined in parts?
3. What are the general rules for finding the Laplace transform of polynomial terms like \( t^n \)?
4. How do the terms involving \( e^{-2s} \) affect the Laplace transform in terms of shifting the original function?
5. Can you explain why the Laplace transform helps in solving differential equations involving \( f(t) \)?

**Tip:** Remember, the Laplace transform is a powerful tool to handle differential equations, especially for systems with piecewise functions.

Consider the function:

f(t) = { t^2, 0 ≤ t < 2,
          6,    t ≥ 2 }

a. Represent f(t) using a combination of Heaviside step functions. Use h(t - c) for the Heaviside function hc(t).
b. Find the Laplace transform F(s) = L {f(t)}.

This page provides a solution for finding the Laplace transform of a piecewise function f(t) with different values across intervals. Learn how to represent f(t) using Heaviside step functions and apply the Laplace transform, including the second shifting theorem.

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