To find the particle's displacement over the time interval $[1, 4]$, we compute the definite integral of the velocity function $v(t) = 6t^2 + 2t + 1$ over $[1, 4]$. The displacement is given by:

$\text{Displacement} = \int_{1}^{4} v(t) \, dt$

Step-by-Step Solution:

1. Write the integral expression: $\int_{1}^{4} (6t^2 + 2t + 1) \, dt$

2. Compute the antiderivative:

   • For $6t^2$, the antiderivative is $\frac{6t^3}{3} = 2t^3$.
   • For $2t$, the antiderivative is $\frac{2t^2}{2} = t^2$.
   • For $1$, the antiderivative is $t$.

   Therefore, the antiderivative of $6t^2 + 2t + 1$ is: $2t^3 + t^2 + t$

3. Evaluate the definite integral: Substitute the limits of integration into the antiderivative: $\left[ 2t^3 + t^2 + t \right]_{1}^{4} = \left( 2(4)^3 + (4)^2 + (4) \right) - \left( 2(1)^3 + (1)^2 + (1) \right)$

4. Simplify each term:

   • At $t = 4$: $2(4)^3 + (4)^2 + 4 = 2(64) + 16 + 4 = 128 + 16 + 4 = 148$
   • At $t = 1$: $2(1)^3 + (1)^2 + 1 = 2(1) + 1 + 1 = 2 + 1 + 1 = 4$

5. Subtract the results: $148 - 4 = 144$

Final Answer:

The particle's displacement over the time interval $[1, 4]$ is: $\boxed{144 \, \text{feet}}$

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Would you like further details or have any questions?

Related Questions:

1. How do you interpret the result of a displacement integral in terms of motion?
2. Can velocity be negative, and how would that affect the displacement?
3. What is the difference between displacement and total distance traveled?
4. How would you find the particle's position function given its velocity function?
5. What is the significance of evaluating definite integrals in physics?

Tip:

To find total distance traveled, you need to integrate the absolute value of the velocity function.