Displacement, Total Distance and Net Change
- Distinguish displacement from total distance and choose the right integral for each
- Apply the Net Change Theorem to recover an amount from a rate and an initial value
- Interpret an integral of a rate in context, with units
Two integrals, two questions
Displacement over [a, b] is ∫ₐᵇ v(t) dt — signed, so backward motion cancels forward motion. Total distance is ∫ₐᵇ |v(t)| dt, which counts every bit of travel as positive. They agree only when v never changes sign. To evaluate the distance integral by hand, find where v = 0, split the interval there, integrate each piece, and add the absolute values of the results. Answering the displacement integral when the question said "total distance" is the single most repeated error in this unit.
The Net Change Theorem
The integral of a rate of change is the net change in the quantity: ∫ₐᵇ F′(t) dt = F(b) − F(a). Rearranged for the form the exam actually asks, F(b) = F(a) + ∫ₐᵇ F′(t) dt — final amount equals initial amount plus accumulated change. Whenever a problem hands you a rate and one known value, this is the tool. The initial value is not optional: an integral of a rate alone gives the change, never the amount.
Interpretation carries points of its own
If R(t) is in gallons per minute and t in minutes, ∫₀¹⁰ R(t) dt is in gallons — the units of the integral are the product of the integrand's and the variable's. A complete interpretation names the quantity, the value, the units and the time interval: "the tank gained 47 gallons during the first 10 minutes." On free response, that sentence is worth as much as the computation that produced the number.
A particle moves along a line with velocity v(t) = t² − 4t (in m/s) for 0 ≤ t ≤ 5. Its position at t = 0 is x = 3. Find the position at t = 5, the displacement, and the total distance traveled.
- 1.Displacement is ∫₀⁵ (t² − 4t) dt = [t³/3 − 2t²] from 0 to 5 = (125/3 − 50) = −25/3 ≈ −8.33 m.
- 2.Position at t = 5 is the initial position plus the displacement: 3 + (−25/3) = −16/3 ≈ −5.33 m.
- 3.For total distance, find where v = 0: t² − 4t = t(t − 4) = 0 at t = 0 and t = 4. The sign of v changes at t = 4.
- 4.On [0, 4]: ∫₀⁴ (t² − 4t) dt = 64/3 − 32 = −32/3, so that leg covers 32/3 meters. On [4, 5]: ∫₄⁵ (t² − 4t) dt = (125/3 − 50) − (64/3 − 32) = 7/3 meters.
- 5.Total distance = 32/3 + 7/3 = 39/3 = 13 meters.
On the calculator section you can evaluate ∫|v(t)| dt directly and skip the splitting. On the no-calculator section, finding where v = 0 is unavoidable — and it is also where the reasoning point lives.
A particle has v(t) > 0 on (0, 3) and v(t) < 0 on (3, 7). Which statement is true for the interval [0, 7]?
Water enters a tank at rate R(t) gallons per hour. The tank holds 40 gallons at t = 0 and ∫₀⁵ R(t) dt = 18. How much is in the tank at t = 5?
Read the verb. "How far did it travel" is total distance; "what is its position" and "how far from its starting point" are displacement questions. The integrals differ by absolute-value bars and the answers differ by a lot.
Answer the 2 checkpoints as you read.
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