Answer:
The particle traveled 4 inches over the interval.
Step-by-step explanation:
What distance does the particle travel over the interval 0 ≤ t ≤ 1?
The distance is the integral of the velocity.
So
[tex]d(t) = \int_{0}^{1} v(t) dt[/tex]
[tex]d(t) = \int_{0}^{1} 4t^3 + 3t^2 + 2t + 1 dt[/tex]
We have that:
[tex]\int t^n dn = \frac{t^{n+1}}{n+1}[/tex]
So
[tex]d(t) = t^4 + t^3 + t^2 + t|_{0}^{1}[/tex]
Which is:
[tex]d(1) - d(0) = 1 + 1 + 1 + 1 = 4[/tex]
The particle traveled 4 inches over the interval.
