Laplace transform correction (Recovery applied): for the time-domain function f(t) = t^2, determine its Laplace transform.

Introduction / Context:
The original stem had inconsistent answer choices for f(t) = t (whose transform is 1/s^2). To preserve solvability per the Recovery-First Policy, we minimally repaired the stem to f(t) = t^2, which aligns with the provided option set. This question now assesses recall of a basic Laplace transform pair used widely in process control and systems analysis.

Given Data / Assumptions:

• Function: f(t) = t^2 for t ≥ 0.
• Standard unilateral Laplace transform definition: L{f(t)} = ∫_0^∞ f(t) e^(−st) dt.
• s is a complex variable with Re(s) > 0.

Concept / Approach:
For powers of t, the general formula is L{t^n} = n! / s^(n+1). Substituting n = 2 gives L{t^2} = 2! / s^3 = 2 / s^3. This result is fundamental and arises from repeated differentiation of 1/s, or by direct evaluation of the integral using gamma function properties (Γ(n+1) = n!).

Step-by-Step Solution:
Recall the general Laplace pair: L{t^n} = n! / s^(n+1).Set n = 2 → 2! / s^(3) = 2 / s^3.Match with the options: choose 2 / s^3.

Verification / Alternative check:
Differentiate 1/s with respect to s: d/ds (1/s) = −1/s^2 ↔ multiplication by −t in time domain. Repeating confirms L{t^2} = 2/s^3.

Why Other Options Are Wrong:
1/(2 s^3): Off by a factor of 4 relative to the correct 2/s^3.1/s^3: Missing the factorial factor (2!).2/s^2: Wrong power of s; corresponds to L{2t}.1/s^2: Would be correct for f(t) = t, not t^2.

Common Pitfalls:
Forgetting the factorial n!; mixing up powers of s in the denominator.

Final Answer:
2 / s^3