AP Calculus AB · Free-response skill

Accumulation from a table of rates

The move. From a table of an invented rate, build a Riemann sum with units, say whether it over- or underestimates and why, and use the Fundamental Theorem to find an amount and a net change.

How to do it

  1. Read the widths from the table. The listed times are often unevenly spaced, so each width is the gap between two neighboring times.
  2. Write the sum term by term, width times height, before giving the total, and give units: the rate’s units times the time units.
  3. For too high or too low, name the property that decides it: increasing or decreasing for left and right sums, concave up or down for trapezoidal and midpoint sums.
  4. For an amount at a later time, start from the known amount and add the integral of the rate in and subtract the integral of the rate out. Integrate a rate given by a formula with an antiderivative and the Fundamental Theorem.
  5. The integral of a derivative, ∫abQ′(t) dt\int_a^b Q'(t)\,dt, is Q(b)−Q(a)Q(b) - Q(a): the change in Q, with Q’s units.

Practice prompt Try it

During a holiday rush, packages arrive at a warehouse sorting floor at a rate of R(t)R(t) packages per hour, where tt is hours after 6 a.m. and 0≤t≤80 \leq t \leq 8. The function RR is increasing and has a continuous derivative. Selected values are in the table. At t=0t = 0 there are 400 packages on the floor.

Arrival rate R(t)R(t)

tt (hours)02368
R(t)R(t) (packages per hour)3044506170

Use the table and the information given in each part.

  1. Use a left Riemann sum with the four subintervals in the table to estimate ∫08R(t) dt\int_0^8 R(t)\,dt. Show the computation and give units.
  2. Is your estimate in part (a) more or less than the true value of the integral? Give a reason.
  3. Packages also leave the floor on trucks at a rate of D(t)=2t+4D(t) = 2t + 4 packages per hour. Using your estimate from part (a), estimate how many packages are on the floor at t=8t = 8.
  4. Find the value of ∫26R′(t) dt\int_2^6 R'(t)\,dt, and tell what it says about the arrival rate.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 6 points

Model answer, annotated

(a) The widths are 2, 1, 3 and 2 hours. Left sum: 2(30)+1(44)+3(50)+2(61)=60+44+150+122=3762(30) + 1(44) + 3(50) + 2(61) = 60 + 44 + 150 + 122 = 376 packages.

  • Point 1 Pairs each width with the rate at the left end of its subinterval and gives the total in packages.

(b) The estimate is less than the true value of the integral. R is increasing, so on each subinterval the left-endpoint value is the smallest value of R, and each rectangle lies below the graph of R.

  • Point 2 States too low and ties it to R increasing, the property that decides a left sum.

(c) Packages that leave: ∫08(2t+4) dt=[t2+4t]08=64+32=96\int_0^8 (2t + 4)\,dt = [t^2 + 4t]_0^8 = 64 + 32 = 96. Packages on the floor at t=8t = 8: about 400+376−96=680400 + 376 - 96 = 680.

  • Point 3 Uses an antiderivative and the Fundamental Theorem for the rate at which packages leave.
  • Point 4 Starts from 400, adds the estimated arrivals from part (a) and subtracts the ones that leave.

(d) By the Fundamental Theorem, ∫26R′(t) dt=R(6)−R(2)=61−44=17\int_2^6 R'(t)\,dt = R(6) - R(2) = 61 - 44 = 17. From t=2t = 2 to t=6t = 6 hours, the arrival rate increased by 17 packages per hour.

  • Point 5 Evaluates the integral of a derivative as a difference of table values.
  • Point 6 Says the change is in the rate, with rate units and the time span.