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The unit step function is a fundamental building block for expressing functions that change abruptly at certain points. Often denoted by \( u(t-a) \), this function switches from 0 to 1 at the point \( t = a \). It has the following key properties:

• For \( t < a \), \( u(t-a) = 0 \).
• For \( t \geq a \), \( u(t-a) = 1 \).

This makes the unit step function ideal for representing piecewise-defined functions, enabling us to "turn on" various function components at specific points. As shown in the exercise, using the step function \( u(t-1) \), we can express the given function \( f(t) \) as \( f(t) = t^2 u(t-1) \). This indicates the function remains inactive (or zero) from \( t = 0 \) to \( t = 1 \), and becomes active (or \( t^2 \)) for \( t \geq 1 \). Understanding unit step functions can greatly simplify working with piecewise functions in mathematical analysis.

Piecewise functions allow us to express a function over different intervals with different expressions. They are particularly useful in situations where a function behaves differently across segments of its domain.

Take the given exercise with \( f(t) \). Its piecewise representation is:

• \( 0 \) for the interval \( 0 \leq t < 1 \)
• \( t^2 \) for \( t \geq 1 \)

To work with such functions analytically, converting them using unit step functions simplifies the process. By representing \( f(t) \) as \( t^2 u(t-1) \), we transform the function into a form that can be easily manipulated for operations like the Laplace transform. This involves recognizing the intervals of function change and expressing them in a way that aligns with the mathematics of step functions.

The Laplace transform technique is a powerful tool for solving differential equations and analyzing linear systems. It transforms time-domain functions into a complex frequency domain, making operations like differentiation and convolution much easier.

When using the Laplace transform on functions involving the unit step function, we utilize the formula:
\[ L\{f(t-a)u(t-a)\} = e^{-as}L\{f(t-a)\} \]

This involves shifting the function \( f(t) \) by \( a \) and applying the transform to the shifted function. In our example, the function \( f(t) \) after incorporating the step function is \( t^2 u(t-1) \), leading to the transform focusing on \( (t-1)^2 \). By applying the formula \( L\{(t-a)^n\} = \frac{n!}{s^{n+1}} \), we find:
\[ L\{(t-1)^2 u(t-1) \} = e^{-s} \cdot \frac{2}{s^3} = \frac{2e^{-s}}{s^3} \]

Understanding how to manipulate functions with these transform techniques can greatly enhance one's capability to solve complex problems across various fields.