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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?

  1. −3
  2. 23/3
  3. 3
  4. −16/3
Show the answer

A. [t³/3 − 4t]₀³ = 9 − 12 = −3.

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