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find laplace transfrom of it and what is the period

The function \( f(t) \) given in the image is a piecewise function defined as follows:

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

### 1. **Finding the Laplace Transform of \( f(t) \):**

The Laplace transform \( \mathcal{L}\{f(t)\} \) is defined as:

\[
\mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t) \, dt
\]

Since \( f(t) \) is piecewise, we compute the Laplace transform over the different intervals.

#### 1.1. Interval \( 0 \leq t < 1 \):

For this interval, \( f(t) = 0 \), so the contribution to the Laplace transform is:

\[
\mathcal{L}_1 = \int_0^1 e^{-st} \cdot 0 \, dt = 0
\]

#### 1.2. Interval \( 1 \leq t < 2 \):

For this interval, \( f(t) = t - 1 \), so:

\[
\mathcal{L}_2 = \int_1^2 e^{-st} (t - 1) \, dt
\]

This can be solved by integration by parts. Let \( u = t - 1 \) and \( dv = e^{-st} dt \). We find:

\[
\mathcal{L}_2 = \left[ \frac{-(t - 1) e^{-st}}{s} \right]_1^2 + \frac{1}{s} \int_1^2 e^{-st} dt
\]

The remaining integral \( \int_1^2 e^{-st} dt \) is straightforward:

\[
\mathcal{L}_2 = \frac{e^{-2s}}{s} - \frac{1}{s} + \frac{e^{-2s} - e^{-s}}{s^2}
\]

#### 1.3. Interval \( t \geq 2 \):

For this interval, \( f(t) = t + 1 \):

\[
\mathcal{L}_3 = \int_2^\infty e^{-st} (t + 1) \, dt
\]

Again, apply integration by parts:

\[
\mathcal{L}_3 = \left[ \frac{-(t + 1)e^{-st}}{s} \right]_2^\infty + \frac{1}{s} \int_2^\infty e^{-st} dt
\]

The integral can be simplified as:

\[
\mathcal{L}_3 = \frac{(2 + 1)e^{-2s}}{s} + \frac{e^{-2s}}{s^2}
\]

### 2. **Period of \( f(t) \):**

The function \( f(t) \) as defined does not repeat itself over regular intervals. Therefore, \( f(t) \) does not have a period. A periodic function repeats its values at regular intervals \( T \), and in this case, there is no such interval for which \( f(t + T) = f(t) \) for all \( t \).

### Summary:
- The Laplace transform of \( f(t) \) is found by integrating piecewise over the intervals defined by \( f(t) \).
- The function \( f(t) \) is not periodic.

Would you like me to calculate the exact expression for the Laplace transform of \( f(t) \), or do you have any other questions?

### Questions for Further Exploration:
1. What is the general method to find the Laplace transform of a piecewise function?
2. How can you determine if a function is periodic?
3. Can the Laplace transform be applied to discontinuous functions?
4. What is the significance of the Laplace transform in solving differential equations?
5. How does the Laplace transform handle initial conditions in differential equations?
6. What are some practical applications of the Laplace transform in engineering?
7. How does the shifting property of the Laplace transform apply to this problem?
8. How can inverse Laplace transforms be used to solve time-domain problems?

### Tip:
When dealing with piecewise functions, it's crucial to carefully consider each interval separately when applying operations like integration or transformation.

Understand the Laplace transform of a piecewise function and explore its periodicity. Learn how to compute the Laplace transform over different intervals and why the function is non-periodic. Suitable for advanced undergraduate students and beyond.

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