Part 2 of the Fundamental Theorem says that integrating a rate of change gives the total change. Applied to velocity from Position, Velocity and Acceleration, it measures how far something moved.
Key idea
Net change theorem: $\displaystyle\int_a^b F'(t)\,dt = F(b) - F(a)$. For velocity $v(t)$: displacement $= \displaystyle\int_a^b v(t)\,dt$, but total distance $= \displaystyle\int_a^b \lvert v(t) \rvert\,dt$. To compute the second, split the interval where $v$ changes sign.
Why it works
A rate times a short time is a small change: $v(t)\,\Delta t$ is how far you move in time $\Delta t$. Adding those small changes and taking the limit gives the total change. When $v < 0$ you move backward, and the integral subtracts that movement, which is right for displacement (where you end up) but wrong for distance (how far the odometer turns). Using $\lvert v \rvert$ counts every bit of motion as positive.
Worked example: Displacement and distance
A particle moves with $v(t) = t^2 - 4t + 3$ m/s for $0 \le t \le 4$. Find its displacement and total distance.
- Antiderivative: $F(t) = \frac{t^3}{3} - 2t^2 + 3t$. Displacement $= F(4) - F(0) = \frac{64}{3} - 32 + 12 = \frac{4}{3}$.
- $v = (t - 1)(t - 3)$ is positive on $[0, 1)$, negative on $(1, 3)$, positive on $(3, 4]$.
- Pieces: $F(1) - F(0) = \frac{4}{3}$; $F(3) - F(1) = 0 - \frac{4}{3} = -\frac{4}{3}$; $F(4) - F(3) = \frac{4}{3} - 0 = \frac{4}{3}$.
- Distance: $\frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$.
Answer: Displacement $\frac{4}{3}$ m; total distance 4 m.
Worked example: A flow rate
Water flows into a tank at $r(t) = 2t + 5$ liters per minute. The tank holds 40 liters at $t = 0$. How much does it hold at $t = 6$?
- Change $$= \int_0^6 (2t + 5)\,dt = \big[t^2 + 5t\big]_0^6 = 36 + 30 = 66$$ liters.
- Amount $= 40 + 66 = 106$.
Answer: 106 liters
- Population: if $P'(t)$ is a growth rate in people per year, $\int_a^b P'(t)\,dt$ is the change in population.
- Cost: integrating the marginal cost from 100 to 200 items gives the extra cost of those 100 items.
- Water: integrating a flow rate in liters per minute gives liters added (negative rates count as water removed).
- The units always work out: (units per time) times time = units.
Common mistake
Reporting displacement when total distance is asked: answering $\frac{4}{3}$ m in the first example. The particle went forward, came back, and went forward again; the odometer counts all of it. Check: if the velocity ever changes sign on the interval, distance and displacement are different.