Net Change: What an Integral of a Rate Means
- State the Net Change Theorem and identify when a problem calls for it
- Compute a final amount from an initial amount and a rate
- Set up an in-minus-out problem and find when the amount is greatest
The theorem, and the sentence it licenses
The Net Change Theorem is FTC Part 1 read in applied language: ∫ₐᵇ f′(t)dt = f(b) − f(a). In words, the integral of a rate of change over an interval gives the net change in the quantity over that interval — not the amount, and not the rate. So if W′(t) is the rate at which water enters a tank in gallons per minute, then ∫₀¹⁰ W′(t)dt is the number of gallons by which the contents changed over those ten minutes. To get the amount at time 10, add the initial amount: W(10) = W(0) + ∫₀¹⁰ W′(t)dt. That single equation answers most applied integration questions on the exam, and writing it before substituting anything keeps the initial condition from being forgotten.
In minus out, and where the maximum is
A frequent setup gives two rates — water entering at R(t) and leaving at S(t) — and asks for the amount in the tank. The net rate is R(t) − S(t), and the amount at time b is the initial amount plus ∫ₐᵇ [R(t) − S(t)]dt. The follow-up question is when the amount is greatest, and the method is Unit 5 applied to this new function: the amount is maximized where its derivative, the net rate, changes from positive to negative — that is, where R(t) = S(t) with inflow exceeding outflow before and outflow exceeding inflow after. Then evaluate the amount there and compare with the endpoints, exactly as the Candidates Test requires. Students often find where R = S and stop; the justification and the endpoint comparison are separate points.
Average value, and the distinction from average rate
The average value of f on [a, b] is (1/(b − a))·∫ₐᵇ f(x)dx — the integral divided by the interval width, which is the height of a rectangle with the same area as the region. Two things to keep separate. The average value of a rate function has the units of that rate, so the average of a velocity in feet per second is a velocity in feet per second, not a distance. And average value is not the same as average rate of change, which is [f(b) − f(a)]/(b − a) and involves no integral at all — the first averages a function, the second is the slope of a secant line. Reading which one a question wants is usually a matter of noticing whether you are given f or f′.
Water enters a tank at R(t) = 12 gallons per hour and leaves at S(t) = 3t gallons per hour, for 0 ≤ t ≤ 6. The tank starts with 20 gallons. Find the amount at t = 6 and the time when the amount is greatest.
- 1.Net rate is R(t) − S(t) = 12 − 3t gallons per hour.
- 2.Amount at t = 6 is 20 + ∫₀⁶ (12 − 3t)dt. Antiderivative: 12t − 1.5t².
- 3.At t = 6: 12(6) − 1.5(36) = 72 − 54 = 18. At t = 0 it is 0. So the net change is 18 gallons.
- 4.Amount at t = 6 is 20 + 18 = 38 gallons.
- 5.For the maximum, set the net rate to zero: 12 − 3t = 0 gives t = 4.
- 6.The net rate is positive for t < 4 and negative for t > 4, so it changes from positive to negative at t = 4 and the amount is greatest there.
- 7.Its value: 20 + ∫₀⁴ (12 − 3t)dt = 20 + [12(4) − 1.5(16)] = 20 + (48 − 24) = 44 gallons. Compare with 20 at t = 0 and 38 at t = 6 — 44 is the largest.
The integral of a rate is a change, never an amount. Reporting 18 gallons as the contents of that tank instead of 38 is the single most common error on applied integration questions, and it happens because the initial condition is given in a different sentence from the rate.
A tank holds 50 liters at t = 0, and liquid flows in at r(t) liters per minute with ∫₀¹⁵ r(t)dt = 32. The amount in the tank at t = 15 is
Water enters at R(t) and leaves at S(t). The amount in the tank is greatest at a time when
Answer the 2 checkpoints as you read.
Sign in to save your progress