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AP® Calculus AB · Unit 8 · Topic 8.2

Connecting Position, Velocity, and Acceleration Using Integrals: Revision Notes (Calculus AB 8.2)

One-page recap of motion with integrals: position from velocity, velocity from acceleration, displacement versus total distance, and average velocity versus average speed.

Short on time? This page is the recap. For explanations, figures and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.

Notation: ∫ (a to b) v(t) dt means the definite integral of v(t) from t = a to t = b.

Recap

  • Differentiating goes x → v → a. Integrating goes back, a → v → x, but gives only the change, so you need a starting value.
  • Displacement = ∫ (a to b) v(t) dt = x(b) − x(a). Signed: backward motion cancels forward motion.
  • Total distance = ∫ (a to b) |v(t)| dt. Never negative.
  • By hand: split at the times where v changes sign, integrate v on each piece, take absolute values, add. With a calculator: integrate |v| directly.
  • On a velocity graph, area above the t-axis is forward motion and area below is backward motion.

Key relationships

Want Use Units (m, s)
Position at time b x(b) = x(a) + ∫ (a to b) v(t) dt m
Velocity at time b v(b) = v(a) + ∫ (a to b) a(t) dt m/s
Displacement ∫ (a to b) v(t) dt m
Total distance ∫ (a to b) |v(t)| dt m
Average velocity (1/(b − a)) ∫ (a to b) v(t) dt m/s
Average speed (total distance)/(b − a) m/s

Assumptions behind the method

  • Motion is along a straight line, with a chosen positive direction.
  • v and a are continuous on the interval, so the Fundamental Theorem applies.
  • Direction changes happen only where v changes sign.

Mistakes to avoid

  1. Calling ∫ v dt the distance. It is displacement.
  2. Writing |∫ v dt| for distance. The absolute value goes inside.
  3. Forgetting x(a) when asked for a position.
  4. Splitting where a = 0 instead of where v changes sign.
  5. Average speed from displacement instead of distance.
  6. Wrong units: ∫ a dt is a velocity, not a distance.

Quick self-check

  1. v(t) = 4 − 2t on [0, 3]. Find the displacement and the total distance. (Displacement 3; distance 5: 4 forward on [0, 2], then 1 back on [2, 3])
  2. x(0) = 5 and ∫ (0 to 2) v(t) dt = −7. Find x(2). (−2)
  3. v(0) = 3 m/s and a(t) = 2 m/s² for all t. Find v(5). (13 m/s = 3 + 2 × 5)

Next: practice questions.

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