AP Calculus AB · Unit 8: Applications of Integration
8.2 Motion with Integrals
Displacement vs. distance
Displacement over [a, b] is ∫[a to b] v(t) dt (net change in position). Total distance is ∫[a to b] |v(t)| dt — the absolute value counts all motion as positive.
Recovering position & velocity
Final position = initial position + ∫ v dt, and integrating acceleration gives the change in velocity.
Key terms
- Displacement
- ∫ v(t) dt (net change in position).
- Total distance
- ∫ |v(t)| dt.
- Final position
- Initial position + ∫ v dt.
- ∫ acceleration
- Gives change in velocity.
- Why |v|
- Counts all motion as positive for distance.
Try an AP-style question
v(t) = t² − 4 on [0, 3]. What is the particle's displacement?
- −3
- 23/3
- 3
- −16/3
Show the answer
A. [t³/3 − 4t]₀³ = 9 − 12 = −3.
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