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The image contains a mathematical problem that involves finding the Laplace transform of a piecewise-defined function \( f(t) \). Here's the transcription and detailed steps to solve it.

### Function Definition
The function \( f(t) \) is defined as:

\[
f(t) =
\begin{cases}
1, & 0 < t \leq 1, \\
t, & 1 < t \leq 2, \\
0, & t > 2.
\end{cases}
\]

The goal is to compute the Laplace transform of \( f(t) \).

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### Laplace Transform Formula
The Laplace transform of a function \( f(t) \) is given by:
\[
F(s) = \int_0^\infty f(t) e^{-st} \, dt.
\]

For this piecewise function, we compute the integral over each interval separately and sum the results.

---

### Step-by-Step Solution

#### Interval 1: \( 0 < t \leq 1 \)
Here, \( f(t) = 1 \). The Laplace transform contribution is:
\[
F_1(s) = \int_0^1 1 \cdot e^{-st} \, dt = \int_0^1 e^{-st} \, dt.
\]
The integral of \( e^{-st} \) is:
\[
\int e^{-st} \, dt = \frac{-e^{-st}}{s}.
\]
Evaluate from 0 to 1:
\[
F_1(s) = \left[ \frac{-e^{-st}}{s} \right]_0^1 = \frac{-e^{-s}}{s} - \frac{-1}{s} = \frac{1 - e^{-s}}{s}.
\]

---

#### Interval 2: \( 1 < t \leq 2 \)
Here, \( f(t) = t \). The Laplace transform contribution is:
\[
F_2(s) = \int_1^2 t \cdot e^{-st} \, dt.
\]
Using integration by parts (\( u = t, dv = e^{-st} dt \)):
\[
u = t, \quad du = dt, \quad v = \frac{-e^{-st}}{s}.
\]
The integral becomes:
\[
F_2(s) = \left[ \frac{-t e^{-st}}{s} \right]_1^2 + \int_1^2 \frac{e^{-st}}{s} \, dt.
\]
The first term evaluates to:
\[
\left[ \frac{-t e^{-st}}{s} \right]_1^2 = \frac{-2 e^{-2s}}{s} - \frac{-1 e^{-s}}{s} = \frac{-2 e^{-2s} + e^{-s}}{s}.
\]
For the second term:
\[
\int_1^2 \frac{e^{-st}}{s} \, dt = \frac{1}{s} \int_1^2 e^{-st} \, dt = \frac{1}{s} \left[ \frac{-e^{-st}}{s} \right]_1^2 = \frac{1}{s} \left( \frac{-e^{-2s}}{s} - \frac{-e^{-s}}{s} \right).
\]
Simplify:
\[
\int_1^2 \frac{e^{-st}}{s} \, dt = \frac{1}{s^2} \left( e^{-s} - e^{-2s} \right).
\]
Combine the results for \( F_2(s) \):
\[
F_2(s) = \frac{-2 e^{-2s} + e^{-s}}{s} + \frac{1}{s^2} \left( e^{-s} - e^{-2s} \right).
\]

---

#### Interval 3: \( t > 2 \)
Here, \( f(t) = 0 \). The Laplace transform contribution is:
\[
F_3(s) = \int_2^\infty 0 \cdot e^{-st} \, dt = 0.
\]

---

### Combine Results
The total Laplace transform is:
\[
F(s) = F_1(s) + F_2(s) + F_3(s).
\]
Substitute:
\[
F(s) = \frac{1 - e^{-s}}{s} + \frac{-2 e^{-2s} + e^{-s}}{s} + \frac{1}{s^2} \left( e^{-s} - e^{-2s} \right).
\]

---

### Simplified Expression
Combine terms and simplify:
\[
F(s) = \frac{1}{s} - \frac{e^{-s}}{s} - \frac{2 e^{-2s}}{s} + \frac{e^{-s}}{s} + \frac{e^{-s}}{s^2} - \frac{e^{-2s}}{s^2}.
\]

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Would you like me to further simplify or explain any part of the process? Let me know!

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### Related Questions
1. How is the Laplace transform useful for piecewise functions in engineering applications?
2. What is the role of the Heaviside step function in simplifying Laplace transform computations?
3. How do you handle Laplace transforms of more complex piecewise functions?
4. Can this method be extended to compute Laplace transforms of periodic functions?
5. What are the common challenges in solving Laplace transforms with integration by parts?

---

### Tip
To simplify piecewise Laplace transforms, always consider expressing the function using step functions for easier integration!

Find the Laplace transform of the given piecewise function f(t).

Explore the detailed solution to finding the Laplace transform of a piecewise-defined function. This problem involves integration by parts and properties of Laplace transforms. Ideal for undergraduate students studying differential equations or engineering mathematics.

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