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To find the Laplace transform of \( f(t) \), we observe that \( f(t) \) is defined as follows:

\[
f(t) = \begin{cases} 
0, & t < 6 \\ 
(t - 6)^5, & t \geq 6 
\end{cases}
\]

This function can be expressed using the unit step function \( u(t - 6) \) as:
\[
f(t) = (t - 6)^5 \cdot u(t - 6)
\]

The Laplace transform of \( f(t) = (t - 6)^5 u(t - 6) \) can be found using the **time-shifting property** of the Laplace transform, which states:

\[
\mathcal{L} \{ f(t - a) u(t - a) \} = e^{-as} \mathcal{L} \{ f(t) \}
\]

In this case, \( f(t - 6) = (t - 6)^5 \) and \( a = 6 \). We first find the Laplace transform of \( t^5 \):

\[
\mathcal{L} \{ t^5 \} = \frac{5!}{s^{6}} = \frac{120}{s^{6}}
\]

Applying the time-shifting property, we get:

\[
\mathcal{L} \{ (t - 6)^5 u(t - 6) \} = e^{-6s} \cdot \frac{120}{s^6}
\]

Thus,

\[
F(s) = \frac{120 e^{-6s}}{s^6}
\]

Would you like a deeper explanation of any step, or have additional questions?

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Here are 5 related questions that expand on this information:

1. How does the time-shifting property work for functions involving the unit step function?
2. What is the significance of the unit step function in Laplace transforms?
3. How would the Laplace transform change if \( f(t) = (t - a)^n u(t - a) \) for any integer \( n \)?
4. Can we use a similar approach if \( f(t) \) were defined piecewise with more intervals?
5. What are other applications of the Laplace transform in solving differential equations?

**Tip:** The time-shifting property is especially useful when dealing with delayed or piecewise-defined functions in Laplace transformations.

Find the Laplace transform of f(t) = {0, t < 6; (t - 6)^5, t >= 6}.

Learn how to find the Laplace transform of a piecewise function f(t) = (t - 6)^5 for t >= 6. This solution explains the use of the time-shifting property and the unit step function, making it ideal for undergraduate students studying Laplace transforms.

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