AP Calculus AB · Lesson 1

Average value, motion and accumulation

Know Know it

The average value of a continuous function f on [a,b][a, b] is 1b−a∫abf(x) dx\frac{1}{b - a}\int_a^b f(x)\,dx. It is the height of the rectangle on [a,b][a, b] with the same area as the region under f. If f is continuous, f takes that value at least once on the interval.

For a particle moving on a line with velocity v(t)v(t): the displacement from t=at = a to t=bt = b is ∫abv(t) dt\int_a^b v(t)\,dt, and the total distance traveled is ∫ab∣v(t)∣ dt\int_a^b |v(t)|\,dt. The position at time b is the starting position plus the displacement: x(b)=x(a)+∫abv(t) dtx(b) = x(a) + \int_a^b v(t)\,dt.

The same idea works for any rate. If R(t)R(t) is the rate at which an amount changes, then amount at b = amount at a + ∫abR(t) dt\int_a^b R(t)\,dt. With water flowing in at E(t)E(t) and out at L(t)L(t), the net rate is E(t)−L(t)E(t) - L(t).

Calculator On the exam

Each section of the exam has a part with a graphing calculator and a part without. You need a graphing calculator for the second multiple-choice part (13 questions) and the first free-response part (2 questions), and the testing app has one built in for those parts. The first multiple-choice part (29 questions) and the second free-response part (4 questions) allow no calculator.

In the calculator parts, use it to evaluate a definite integral numerically and to find where two curves meet or where a velocity is zero, then write the integral you evaluated. In the other parts, every integral in this unit must be done by hand, so keep practicing antiderivatives.

Trap Two averages that are not the average value

The average value is not the mean of the endpoint values. For f(x)=x3f(x) = x^3 on [0,2][0, 2], f(0)+f(2)2=4\frac{f(0) + f(2)}{2} = 4, but the average value is 12∫02x3 dx=2\frac{1}{2}\int_0^2 x^3\,dx = 2.

It is also not the average rate of change, f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}, which is the average of f′f', not of f. Ask which function is being averaged.

Worked example Easy

What is the average value of f(x)=3x2f(x) = 3x^2 on the interval [1,3][1, 3]?

  1. 2626
  2. 1313Answer
  3. 1515
  4. 2727

How to solve it

  1. ∫133x2 dx=[x3]13=27−1=26\int_1^3 3x^2\,dx = \left[x^3\right]_1^3 = 27 - 1 = 26.
  2. Divide by b−a=2b - a = 2: the average value is 13.

Why each choice is right or wrong

  • A. Incorrect. 26 is the integral ∫133x2 dx\int_1^3 3x^2\,dx; the average value divides it by the length of the interval, 2.
  • B. Correct. ∫133x2 dx=27−1=26\int_1^3 3x^2\,dx = 27 - 1 = 26, and dividing by the length 2 gives 13.
  • C. Incorrect. 15 is the mean of the endpoint values f(1)=3f(1) = 3 and f(3)=27f(3) = 27, which is not the average value of a curved function.
  • D. Incorrect. 27 is f(3)f(3), the largest value of f on the interval, not its average value.

Worked example Medium

Velocity v(t), in meters per second, of a cart on a track, for 0 \le t \le 5 seconds123456−3−2−1123450t (seconds)v(t)(0, 4)(2, 0)(5, −2)
Velocity v(t)v(t), in meters per second, of a cart on a track, for 0≤t≤50 \le t \le 5 seconds Two straight pieces. The graph starts at (0, 4) and falls in a straight line to (2, 0). It then keeps falling in a straight line, more gently, from (2, 0) to (5, -2). It is above the t-axis for 0 < t < 2 and below it for 2 < t < 5.

The graph shows the velocity of a cart. What is the total distance the cart travels from t=0t = 0 to t=5t = 5 seconds?

  1. 1 meter
  2. 4 meters
  3. −1-1 meter
  4. 7 metersAnswer

How to solve it

  1. On [0,2][0, 2] the area is a triangle above the axis: 12(2)(4)=4\frac{1}{2}(2)(4) = 4.
  2. On [2,5][2, 5] the area is a triangle below the axis: 12(3)(2)=3\frac{1}{2}(3)(2) = 3.
  3. Total distance adds the sizes: 4+3=74 + 3 = 7 meters. Displacement would be 4−3=14 - 3 = 1 meter.

Why each choice is right or wrong

  • A. Incorrect. 1 meter is the displacement, 4−34 - 3, where the backward motion cancels part of the forward motion.
  • B. Incorrect. 4 meters counts only the forward motion on [0,2][0, 2] and leaves out the 3 meters moved backward.
  • C. Incorrect. −1-1 meter subtracts the forward area from the backward area, 3−43 - 4, and a distance is never negative.
  • D. Correct. The cart moves 4 meters forward on [0,2][0, 2] and 3 meters back on [2,5][2, 5], so it travels 4+3=74 + 3 = 7 meters in all.

Explain Explain it

Task verbs: calculate, interpret

To interpret a definite integral of a rate, say what accumulates, how much, with units, and over what interval. The units are the rate's units times the input's units: liters per minute times minutes gives liters.

To find when an amount is greatest or least, set the net rate equal to 0 to find candidates, then compare the amount at those times and at the endpoints. This is the Candidates Test applied to an accumulation function.

Worked example Hard

A tank holds 120 liters of water at t=0t = 0. For 0≤t≤50 \le t \le 5 minutes, water flows in or out at the net rate r(t)=6t−12r(t) = 6t - 12 liters per minute. What is the least amount of water in the tank during this time?

  1. 108 litersAnswer
  2. 120 liters
  3. 96 liters
  4. 135 liters

How to solve it

  1. The net rate is 0 when 6t−12=06t - 12 = 0, so at t=2t = 2. It is negative before and positive after.
  2. Amount at each candidate: A(0)=120A(0) = 120, A(2)=120+∫02(6t−12) dt=120−12=108A(2) = 120 + \int_0^2 (6t - 12)\,dt = 120 - 12 = 108, A(5)=120+15=135A(5) = 120 + 15 = 135.
  3. The least amount is 108 liters, at t=2t = 2 minutes.

Why each choice is right or wrong

  • A. Correct. The amount is 120+∫0t(6s−12) ds=120+3t2−12t120 + \int_0^t (6s - 12)\,ds = 120 + 3t^2 - 12t, which is least at t=2t = 2, where it is 108 liters.
  • B. Incorrect. 120 liters is the starting amount, but the rate is negative for 0<t<20 < t < 2, so the tank loses water first.
  • C. Incorrect. 96 liters subtracts 12 liters per minute for 2 minutes and ignores the 6t6t part of the rate, which cuts the loss.
  • D. Incorrect. 135 liters is the amount at t=5t = 5, the greatest amount on the interval, not the least.

Interpret Your turn

Rain falls into a barrel at a rate of R(t)R(t) liters per hour, where t is hours after noon. A forecaster finds that ∫25R(t) dt=18\int_2^5 R(t)\,dt = 18 and 13∫25R(t) dt=6\frac{1}{3}\int_2^5 R(t)\,dt = 6. Tell what each of these two numbers means for the barrel, with units.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Says that 18 liters of rain fall into the barrel from 2 p.m. to 5 p.m.
  • Says that 6 liters per hour is the average rate at which rain falls into the barrel from 2 p.m. to 5 p.m.

Model answer

The integral ∫25R(t) dt=18\int_2^5 R(t)\,dt = 18 means that 18 liters of rain fall into the barrel from 2 p.m. to 5 p.m. The value 6 is the average value of R on that interval: from 2 p.m. to 5 p.m., rain falls into the barrel at an average rate of 6 liters per hour.

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