Let's solve the Laplace transform step by step: **Given:** \[ f(t) = \begin{cases} 0, & t < 2 \\ (t-2)^3, & t \geq 2 \end{cases} \] --- ### **Step 1: Express \( f(t) \) using the unit step function** The unit step (Heaviside) function is \( u_a(t) \), which is 0 for \( t < a \) and 1 for \( t \geq a \). So, rewrite \( f(t) \) as: \[ f(t) = (t-2)^3 \cdot u_2(t) \] **Explanation:** The unit step function \( u_2(t) \) turns on the function \( (t-2)^3 \) at \( t = 2 \), matching the original piecewise definition. --- ### **Step 2: Apply the Laplace transform using the shifting theorem** Recall the Laplace transform property: \[ \mathcal{L}\{f(t-a)u_a(t)\} = e^{-as} F(s) \] where \( F(s) \) is the Laplace transform of \( f(t) \). Here, \( f(t-2) = (t-2)^3 \) and \( a = 2 \). The Laplace transform of \( t^3 \) is \( \frac{6}{s^4} \). So, \[ \mathcal{L}\{(t-2)^3 u_2(t)\} = e^{-2s} \cdot \frac{6}{s^4} \] **Explanation:** The shifting theorem lets us handle the step function. We take the Laplace of \( t^3 \) and multiply it by \( e^{-2s} \) because of the shift. --- ## **Final Answer:** \[ \boxed{ \mathcal{L}\{f(t)\} = \frac{6 e^{-2s}}{s^4} } \]