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Home›Calculus›Integrals

Net change and average value

A cart with velocity t² − 4t + 3 drives 4 meters in total over four seconds, yet ends only 4/3 of a meter from where it started.

Integrating a rate

Read the Fundamental Theorem the other way round. If F is any quantity and F′ is its rate of change, then ∫ab F′(x) dx = F(b) − F(a). OpenStax calls this the net change theorem: integrating a rate of change gives the total change. Equivalently, F(b) = F(a) + ∫ab F′(x) dx, the new value is the old value plus the accumulated change. If water flows into a tank at a steady 5 liters per minute for 12 minutes, the integral is 5 × 12 = 60 liters; with a varying rate the integral does the adding for you.

Displacement and distance

For motion, the rate is velocity. ∫ab v(t) dt is the displacement, the change in position, which counts backward motion as negative. The distance traveled is ∫ab |v(t)| dt, which counts every meter. Take v(t) = t² − 4t + 3 = (t − 1)(t − 3) meters per second on [0, 4], with antiderivative t³/3 − 2t² + 3t. The displacement is 64/3 − 32 + 12 − 0 = 4/3 meters. But v is negative between t = 1 and t = 3, when the cart rolls backward.

Split the interval at the sign changes. From 0 to 1 the cart moves forward 4/3 meters. From 1 to 3 the integral is −4/3, so it moves back 4/3 meters. From 3 to 4 it moves forward 4/3 meters again. Net: 4/3 − 4/3 + 4/3 = 4/3. Total distance: 4/3 + 4/3 + 4/3 = 4 meters. The two answers differ because backward motion cancels forward motion in one and adds to it in the other. Questions about where something ends up want displacement; questions about fuel, wear or a trip meter want distance.

v(t) = t² − 4t + 3 on [0, 4]. Three pieces of area 4/3: displacement 4/3, distance 4.

The same split works for any rate that changes sign, such as money flowing in and out of an account or water pumped into and out of a tank.

Average value of a function

An average of n numbers is their sum divided by n. For a function on [a, b], the integral plays the role of the sum and the length b − a the role of n: favg = (1/(b − a)) ∫ab f(x) dx. Geometrically, it is the height of the rectangle on [a, b] with the same area as the region under the curve. For x² on [0, 2], the average is (8/3)/2 = 4/3. For sin x on [0, π], it is 2/π ≈ 0.637. The mean value theorem for integrals in OpenStax 5.3 adds that a continuous function actually reaches its average value at some point of the interval.

The cart here is a made-up example chosen so that every piece comes out to 4/3. Real velocity data come as a table of readings, and then a Riemann sum of the readings plays the role of the integral.
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Velocity t² − 4t + 3 on [0, 4]. The red rectangles are backward motion. The exact integral, 4/3, is the displacement.
  1. Water enters at 5 L/min for 12 min. Total liters?
  2. v(t) = t² − 4t + 3 on [0, 4]. Displacement in meters (3 decimals)?
  3. Same motion: total distance traveled in meters?
  4. During which time is that cart moving backward?
    • 1 < t < 3
    • 0 < t < 1
    • 3 < t < 4
    • never
  5. The dropped rock has v(t) = −32t. Displacement from t = 0 to 2 (ft)?
  6. Average value of x² on [0, 2] (3 decimals)?
  7. Average value of sin x on [0, π] (3 decimals)?
  8. A tank holds 100 L. Water drains at r(t) = 2t L/min for 5 min. How many liters remain?
  9. A trip meter measures:
    • ∫ |v(t)| dt, the distance
    • ∫ v(t) dt, the displacement
    • v(t) at the end
    • the average velocity
  10. favg on [a, b] equals:
    • (1/(b − a)) ∫ab f(x) dx
    • (f(a) + f(b))/2
    • ∫ab f(x) dx
    • f((a + b)/2)
Key points
  • Net change theorem: ∫ab F′(x) dx = F(b) − F(a).
  • Displacement ∫ v dt can cancel; distance ∫ |v| dt cannot.
  • Split at sign changes to get distance.
  • Average value: (1/(b − a)) ∫ab f(x) dx.
Sources for this course (6)
  • OpenStax, Calculus Volume 1 , chapter 5 Integration: 5.1 Approximating Areas (Archimedes and the method of exhaustion; sigma notation with Σ 3^i for i = 1 to 5 = 363; left and right endpoint approximations L4 = 1.75 and R4 = 3.75 for x² on [0, 2], L6 = 3.4375 and R6 = 5.6875 for x²/2 on [0, 3]; (x − 1)³ + 4 on [0, 2] approaching 8; Riemann sums named for the 19th-century mathematician Bernhard Riemann; upper and lower sums: openstax.org/books/calculus-volume-1/pages/5-1-approximating-areas)
  • 5.2 The Definite Integral (limit of Riemann sums, integrability of continuous functions, net signed area, properties: .../pages/5-2-the-definite-integral)
  • 5.3 The Fundamental Theorem of Calculus (Newton and Leibniz, among others, late 1600s and early 1700s; mean value theorem for integrals; parts 1 and 2: .../pages/5-3-the-fundamental-theorem-of-calculus)
  • 5.4 Integration Formulas and the Net Change Theorem (.../pages/5-4-integration-formulas-and-the-net-change-theorem)
  • 4.10 Antiderivatives (.../pages/4-10-antiderivatives)
  • MacTutor, University of St Andrews: Archimedes (287 BC to 212 BC; area of a parabolic segment; π bounded with 96-sided inscribed and circumscribed polygons: mathshistory.st-andrews.ac.uk/Biographies/Archimedes) and Leibniz (first used the ∫ f(x) dx notation in a manuscript of 21 November 1675: mathshistory.st-andrews.ac.uk/Biographies/Leibniz).

Checked: 2026-10-17