Difficulty: Intermediate | Prerequisites: Sections 4.1–4.3 (antiderivatives, definite integrals, Fundamental Theorem of Calculus)
The net change theorem says that integrating a rate function over an interval gives you the total accumulated change. When the rate is a velocity, the integral gives displacement (net change in position, which can be negative). To get total distance travelled, you integrate the absolute value of velocity instead.
Net change theorem
If F'(t) = f(t), then the integral of f(t) from a to b equals F(b) - F(a). In plain terms, integrating a rate of change over an interval recovers the total change in the original quantity.
Displacement
The net change in position of an object over a time interval, found by integrating velocity. Think of it as where you end up minus where you started, which can be zero or negative if you double back.
Total distance travelled
The sum of all movement regardless of direction, found by integrating the absolute value of velocity. In simple terms, this is your odometer reading: it only goes up, even when you reverse.
Rate function
A function r(t) describing how quickly some quantity changes per unit time. Think of it as the speedometer needle for whatever quantity you are tracking.
Position function
The antiderivative of velocity plus an initial condition. It tells you where the object is on the number line at any time t.
When you know a rate of change r(t) and want the total quantity that has flowed, integrate r(t) over the relevant interval.
Example: sand pours from a bag at r(t) = 100 - 3t mL/min, and the bag empties at t = 20 min.
Total sand = integral from 0 to 20 of (100 - 3t) dt
= [100t - (3/2)t^2] from 0 to 20
= (2000 - 600) - 0 = 1400 mL
The key insight: the rate function tells you the flow per minute, and the definite integral adds all those tiny flows together.
Position change (displacement) = integral of v(t) dt over the interval.
Given v(t) = t(6 - t) = 6t - t^2 and initial position s(0) = -2:
Displacement over [0, 16] = integral from 0 to 16 of (6t - t^2) dt
= [3t^2 - t^3/3] from 0 to 16
= (768 - 4096/3) - 0 = (2304/3 - 4096/3) = -1792/3
Position at t = 16: s(16) = -2 + (-1792/3) = -6/3 - 1792/3 = -1798/3
A negative displacement means the object ended up to the left of where it started.
Total distance = integral of |v(t)| dt. You must split the interval wherever v(t) changes sign.
v(t) = t(6 - t) = 0 when t = 0 or t = 6. So v(t) > 0 on (0, 6) and v(t) < 0 on (6, 16).
Total distance = integral from 0 to 6 of (6t - t^2) dt + integral from 6 to 16 of (t^2 - 6t) dt
First piece: [3t^2 - t^3/3] from 0 to 6 = 108 - 72 = 36
Second piece: [t^3/3 - 3t^2] from 6 to 16 = (4096/3 - 768) - (72 - 108) = (4096/3 - 768) + 36 = 4096/3 - 732 = (4096 - 2196)/3 = 1900/3
Total distance = 36 + 1900/3 = 108/3 + 1900/3 = 2008/3
The critical step most students skip: find where velocity is zero and split the integral there. Without the split, you get displacement, not distance.
\text{Net Change Theorem: } \int_a^b f(t)\,dt = F(b) - F(a)\text{Displacement: } s(b) - s(a) = \int_a^b v(t)\,dt\text{Total Distance: } \int_a^b |v(t)|\,dtTo evaluate the total distance integral, find all zeros of v(t) in [a, b], split the interval at those zeros, and integrate the positive and negative pieces separately (flipping the sign on intervals where v < 0).
Engineers use net change integrals constantly: if you know the flow rate of water through a pipe (litres per second), integrating gives you total volume delivered. The displacement-vs-distance distinction is exactly the difference between your GPS showing "you are 3 km from home" and your fitness tracker showing "you walked 8 km."
Students often think integrating velocity always gives total distance. It does not. The plain integral of v(t) gives displacement, which can be negative or zero even when the object has moved.
Students sometimes forget the initial condition. The displacement integral gives the change in position, not the final position. You must add s(0) to get s(t).
A common error is setting up the distance integral without finding where v(t) = 0. If you skip the sign analysis, the negative and positive parts cancel, and you get displacement instead of distance.
Some students think distance can be negative. It cannot. Total distance is always non-negative.
Displacement vs total distance is a near-certain exam topic. Expect at least one problem that requires you to split an integral at a sign change.
Problems that give a rate function and ask for a total (like the sand problem) test whether you can recognise the net change theorem in a non-velocity context.
Watch for problems that give initial position: you need it for final position but not for displacement or distance.
True or false: integrating velocity from t = a to t = b always gives total distance. (False, it gives displacement.)
Fill in the blank: to convert a displacement integral into a distance integral, you integrate the ______ of v(t). (absolute value)
True or false: if an object's velocity is negative over the entire interval, the displacement is negative. (True.)
Fill in the blank: the net change theorem states that the integral of f(t) from a to b equals ______ . (F(b) - F(a), where F' = f)
True or false: total distance travelled can be zero even if the object moved. (False. Distance is zero only if the object never moved.)
Q: Water leaks from a tank at a rate of r(t) = 50 - 2t litres per minute for 0 <= t <= 25. How many litres leaked out in total?
A: Integrate r(t) from 0 to 25: [50t - t^2] from 0 to 25 = 1250 - 625 = 625 litres.
Q: A particle has velocity v(t) = 4 - t m/s. Find its displacement over [0, 8].
A: Integral from 0 to 8 of (4 - t) dt = [4t - t^2/2] from 0 to 8 = 32 - 32 = 0. The particle returns to its starting position.
Q: For the same particle with v(t) = 4 - t, find the total distance travelled over [0, 8].
A: v(t) = 0 at t = 4. Distance = integral from 0 to 4 of (4 - t) dt + integral from 4 to 8 of (t - 4) dt = [4t - t^2/2] from 0 to 4 + [t^2/2 - 4t] from 4 to 8 = 8 + 8 = 16 m.
Q: An object starts at position s(0) = 5 and has velocity v(t) = 3t^2 - 12. Where is it at t = 3?
A: Displacement = integral from 0 to 3 of (3t^2 - 12) dt = [t^3 - 12t] from 0 to 3 = (27 - 36) - 0 = -9. So s(3) = 5 + (-9) = -4.
This material connects directly to the Fundamental Theorem of Calculus (Sec. 4.3), which provides the theoretical foundation for evaluating definite integrals via antiderivatives. It also sets up Section 5.1 (area between curves), because the same logic of integrating to find a total carries over: instead of rate times time, you are adding up vertical slices of area. If you are comfortable with net change, Riemann sums (Sec. 4.2) become easier to interpret as approximations of the same accumulated total.
net change theorem, displacement vs distance, total distance travelled, integrating velocity, integrating rate functions, position from velocity, absolute value integral, antiderivative applications, MTH 132, Calculus I, Michigan State University, Section 4.4, recitation 14, signed area, accumulation function