To find the displacement of the particle over the time interval $[-3, 6]$, we integrate the velocity function $v(t) = -t^2 + 6t - 8$ with respect to $t$.

Displacement Calculation:

The displacement is given by: $\text{Displacement} = \int_{-3}^{6} v(t) \, dt = \int_{-3}^{6} (-t^2 + 6t - 8) \, dt$

Step-by-step Integration:

1. The integral of $-t^2$ is $-\frac{t^3}{3}$.
2. The integral of $6t$ is $3t^2$.
3. The integral of $-8$ is $-8t$.

So: $\int v(t) \, dt = -\frac{t^3}{3} + 3t^2 - 8t + C$

Apply Limits:

Using the Fundamental Theorem of Calculus: $\int_{-3}^{6} v(t) \, dt = \left[ -\frac{t^3}{3} + 3t^2 - 8t \right]_{-3}^{6}$

1. Evaluate at $t = 6$: $-\frac{(6)^3}{3} + 3(6)^2 - 8(6) = -72 + 108 - 48 = -12$

2. Evaluate at $t = -3$: $-\frac{(-3)^3}{3} + 3(-3)^2 - 8(-3) = -\frac{-27}{3} + 27 + 24 = 9 + 27 + 24 = 60$

3. Subtract: $\text{Displacement} = -12 - 60 = -72$

Thus, the displacement over the interval $[-3, 6]$ is $-72$ units.

Next Steps or Questions?

1. Would you like to calculate the total distance traveled by considering the absolute values of velocity?
2. Do you need clarification on any of the steps?
3. Should I graph the velocity function for visualization?
4. Would you like me to analyze when the velocity changes direction to split the integral?
5. Do you want me to verify the result numerically?

Tip: Always check whether the problem asks for "displacement" (net change) or "total distance" (absolute value integral).