To determine the correct integral for the displacement of the particle from $t = 0$ to $t = 3$, we need to understand the definition of displacement.

Displacement

The displacement of a particle over a time interval $[a, b]$ is given by the definite integral of the velocity function $v(t)$ over that interval: $\text{Displacement} = \int_a^b v(t) \, dt$

Here, $v(t) = t^2 - t - 2$, so the displacement from $t = 0$ to $t = 3$ is: $\int_0^3 (t^2 - t - 2) \, dt$

Analysis of Options

Looking at the provided options:

• The correct integral for displacement should not include absolute values, as absolute values indicate total distance, not displacement.
• The correct option should match: $\int_0^3 (t^2 - t - 2) \, dt$

Thus, the correct answer is: $\boxed{\int_0^3 (t^2 - t - 2) \, dt}$

Let me know if you would like detailed calculations of the displacement or explanations about the difference between displacement and total distance.

Follow-Up Questions:

1. What is the numerical value of the displacement from $t = 0$ to $t = 3$?
2. How does the total distance differ from displacement for this velocity function?
3. Can you explain the role of the definite integral in computing physical quantities?
4. What happens if the velocity function changes sign during the interval?
5. How do absolute values impact the computation of total distance?

Tip: Always double-check whether the problem asks for displacement or total distance, as they often require different interpretations of the integral.