The Net Change Theorem
Integrating a rate of change gives the net change in the quantity: \(\int_a^b F'(x)\,dx = F(b) - F(a)\). For motion, that split personality matters: integrating velocity gives displacement, while integrating speed \(|v(t)|\) gives the total distance travelled.
By the end you'll be able to read a definite integral of a rate as a net change, and compute both the displacement \(\int_a^b v\,dt\) and the total distance \(\int_a^b |v|\,dt\) of a moving particle — splitting the integral where the velocity changes sign.
Predict: for this particle, is the total distance bigger than the size of the displacement? Decide, then toggle to |v(t)| and compare the two totals.
A particle moves with velocity \(v(t) = t^2 - 4\) m/s on \([0, 3]\). It drifts backward until \(t = 2\) (where \(v = 0\)), then pushes forward. The toggle switches what the shading measures: signed area of \(v\) (displacement) or area of \(|v|\) (total distance). Hover or focus a shaded region for its exact value.
Backward piece [0, 2]: −16/3 ≈ −5.33 m · forward piece [2, 3]: +7/3 ≈ +2.33 m · displacement = −3.00 m
The Net Change Theorem says the definite integral of a rate of change gives the net change of the quantity; for motion, integrating velocity gives displacement while integrating speed — the absolute value of velocity — gives total distance travelled.
If \(F'\) is the rate at which \(F\) changes, then \(\int_a^b F'(x)\,dx = F(b) - F(a)\). This is the Fundamental Theorem of Calculus read as a statement about accumulation: integrating a rate recovers the total change accumulated over \([a, b]\), so the new value is the old value plus the accumulated change, \(F(b) = F(a) + \int_a^b F'(x)\,dx\). Because the definite integral is a net signed area — the same signed area you built Riemann sums for — intervals where the rate is negative subtract. That's why \(\int_a^b v(t)\,dt\) is displacement: backward motion (\(v < 0\)) reduces net position. To make every bit of motion count positively, integrate the speed instead: total distance \(= \int_a^b |v(t)|\,dt\). Removing the absolute value means handling \(|v|\) piecewise — use additivity to split the integral at each zero of \(v\) and flip the sign on the negative pieces.
Your car's odometer and your GPS disagree on purpose. Drive 10 km to a store and 10 km back: the GPS says your position changed by 0 (displacement, \(\int v\,dt\)), but the odometer added 20 km (distance, \(\int |v|\,dt\)) — it can't count backward. The same net-change reading works for any rate: integrate litres per minute flowing in and out of a tank and you get the net change in volume; integrate a population's growth rate and you get the net change in population. Whenever the rate changes sign, "net" and "total" part ways.
A particle moves with \(v(t) = t^2 - 4\) m/s on \([0, 3]\) — the one in the diagram above. Displacement: \(\int_0^3 (t^2 - 4)\,dt = \left[\tfrac{t^3}{3} - 4t\right]_0^3 = (9 - 12) - 0 = \) \(-3\) m — it ends up 3 m behind its start. Total distance: \(v(t) = 0\) at \(t = 2\), with \(v < 0\) on \([0, 2]\) and \(v > 0\) on \([2, 3]\), so split there: \(\int_0^2 -(t^2 - 4)\,dt = \left[4t - \tfrac{t^3}{3}\right]_0^2 = 8 - \tfrac{8}{3} = \tfrac{16}{3}\) and \(\int_2^3 (t^2 - 4)\,dt = \left[\tfrac{t^3}{3} - 4t\right]_2^3 = -3 - \left(-\tfrac{16}{3}\right) = \tfrac{7}{3}\). Total distance \(= \tfrac{16}{3} + \tfrac{7}{3} = \) \(\tfrac{23}{3} \approx 7.67\) m.
A particle moves with \(v(t) = 2t - 6\) m/s on \([0, 4]\). Displacement: \(\int_0^4 (2t - 6)\,dt = \left[t^2 - 6t\right]_0^4 = 16 - 24 = -8\) m. Total distance: \(v = 0\) at \(t = 3\), with \(v < 0\) before and \(v > 0\) after. The backward piece gives \(\int_0^3 (6 - 2t)\,dt = \left[6t - t^2\right]_0^3 = 9\). Now finish it: the forward piece is \(\int_3^4 (2t - 6)\,dt = \) ____, so the total distance is \(9 + \) ____ \(= \) ____ m.
Reveal the answer
\(\int_3^4 (2t - 6)\,dt = \left[t^2 - 6t\right]_3^4 = (16 - 24) - (9 - 18) = -8 + 9 = 1\), so total distance \(= 9 + 1 = \) 10 m — while the displacement is only \(-8\) m. The 9 m of backtracking and the 1 m forward partially cancel in the displacement, but the distance keeps both.
More info — why this is just the Fundamental Theorem in disguise
Nothing new is being proved here: \(\int_a^b F'(x)\,dx = F(b) - F(a)\) is exactly the evaluation half of the Fundamental Theorem of Calculus you used in the last few lessons — the Net Change Theorem is that identity read in the other direction. Instead of "evaluate the integral of \(f\) by finding an antiderivative," it says "the integral of a rate \(F'\) measures how much \(F\) changed." Position is the antiderivative of velocity, volume is the antiderivative of flow rate, population is the antiderivative of growth rate. The only genuinely new move in this lesson is the absolute value: since \(|v|\) has no elementary antiderivative formula in one piece, you fall back on the additivity of the definite integral over subintervals and integrate each sign-piece separately. The OpenStax section in Dive deeper below works several more rate examples if you want extra reps.
Check your understanding
A cyclist's velocity is \(v(t) = 6 - 2t\) m/s on \([0, 5]\). What is the total distance travelled?
A particle's velocity satisfies \(\int_0^4 v(t)\,dt = 0\). What can you conclude?
Water flows in and out of a tank at a rate \(r(t)\) litres per minute, and \(\int_0^{10} r(t)\,dt = -20\). What does this tell you?
On \([0, 6]\) a velocity graph encloses area 5 between the curve and the \(t\)-axis where \(v < 0\), then area 8 where \(v > 0\). What are the displacement and the total distance?
Recap
- Net Change Theorem: \(\int_a^b F'(x)\,dx = F(b) - F(a)\) — the definite integral of a rate of change is the net change in the quantity, so \(F(b) = F(a) + \int_a^b F'(x)\,dx\).
- Displacement \(= \int_a^b v(t)\,dt\): a net signed area, where intervals with \(v < 0\) subtract.
- Total distance \(= \int_a^b |v(t)|\,dt\): integrating the speed makes every piece of motion count positively.
- To integrate \(|v|\), split the integral at the zeros of \(v\) and flip the sign of the pieces where \(v < 0\) — never integrate \(v\) and call it distance.
Dive deeper
- OpenStax Calculus Volume 1 — 5.4 Integration Formulas and the Net Change Theorem State the Net Change Theorem and contrast displacement with total distance
Sources
- The Net Change Theorem — Displacement vs. Total Distance