laplace \:f\left(t\right)=\left\{t^2:0\le \:t<\:2,\:1-t-3t^2:t\ge \:2\right\}

You are asking to compute the Laplace transform of a piecewise function. Let's break the function \( f(t) \) down and find its Laplace transform step by step.

Given:

\[
f(t) = 
\begin{cases}
t^2 & \text{for } 0 \leq t < 2 \\
1 - t - 3t^2 & \text{for } t \geq 2
\end{cases}
\]

To compute the Laplace transform of this piecewise function, we can use the **second shifting theorem**, which handles functions defined in intervals (like this one). Here's the general strategy:

### 1. Break \( f(t) \) into two separate pieces and find the Laplace transform of each part.

- For \( 0 \leq t < 2 \), the function is \( f_1(t) = t^2 \).
- For \( t \geq 2 \), the function is \( f_2(t) = 1 - t - 3t^2 \).

### 2. Laplace Transform of \( f_1(t) = t^2 \) (for \( 0 \leq t < 2 \)):

The Laplace transform of \( t^n \) is \( \frac{n!}{s^{n+1}} \), so the Laplace transform of \( t^2 \) is:

\[
\mathcal{L}(t^2) = \frac{2!}{s^3} = \frac{2}{s^3}
\]

### 3. Laplace Transform of \( f_2(t) = 1 - t - 3t^2 \) (for \( t \geq 2 \)):

We will apply the **second shifting theorem** to deal with the interval \( t \geq 2 \). The second shifting theorem says that if a function \( g(t) \) is shifted by \( t_0 \), then the Laplace transform is given by:

\[
\mathcal{L}(g(t - t_0)) = e^{-t_0 s} \mathcal{L}(g(t))
\]

So, we must shift \( f_2(t) \) by 2 units. We can write:

\[
f_2(t) = 1 - t - 3t^2 = g(t - 2) \quad \text{where} \quad g(t) = 1 - t - 3t^2
\]

Now, find the Laplace transform of \( g(t) = 1 - t - 3t^2 \):

- The Laplace transform of 1 is \( \frac{1}{s} \)
- The Laplace transform of \( t \) is \( \frac{1}{s^2} \)
- The Laplace transform of \( t^2 \) is \( \frac{2}{s^3} \)

So:

\[
\mathcal{L}(g(t)) = \frac{1}{s} - \frac{1}{s^2} - \frac{6}{s^3}
\]

### 4. Applying the second shifting theorem:

Now we apply the second shifting theorem for \( t_0 = 2 \):

\[
\mathcal{L}(f_2(t)) = e^{-2s} \left( \frac{1}{s} - \frac{1}{s^2} - \frac{6}{s^3} \right)
\]

### 5. Combine the results:

The total Laplace transform of the piecewise function is the sum of the transforms for \( 0 \leq t < 2 \) and \( t \geq 2 \):

\[
\mathcal{L}(f(t)) = \mathcal{L}(f_1(t)) + \mathcal{L}(f_2(t))
\]

\[
\mathcal{L}(f(t)) = \frac{2}{s^3} + e^{-2s} \left( \frac{1}{s} - \frac{1}{s^2} - \frac{6}{s^3} \right)
\]

### Final Result:

\[
\mathcal{L}(f(t)) = \frac{2}{s^3} + e^{-2s} \left( \frac{1}{s} - \frac{1}{s^2} - \frac{6}{s^3} \right)
\]

Would you like further details on any step of the solution? Or have any other questions?

### Related Questions:
1. How do you compute the Laplace transform of a piecewise function with a discontinuity at \( t = 0 \)?
2. What is the second shifting theorem and how is it applied in Laplace transforms?
3. How do you handle a piecewise function with different expressions for \( t \geq 0 \) and \( t < 0 \)?
4. Can the Laplace transform be applied to functions with piecewise continuous but not smooth behavior?
5. How do you handle exponential functions in piecewise Laplace transforms?

### Tip:
When working with piecewise functions, always break down the function into intervals, and apply the second shifting theorem to handle shifts in the function properly.

Learn how to compute the Laplace transform of a piecewise function defined as t^2 for 0 <= t < 2 and 1 - t - 3t^2 for t >= 2. This step-by-step solution applies the second shifting theorem and key Laplace transform formulas. Suitable for advanced high school or college students studying Laplace transforms and piecewise functions.

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