Accumulation in Context — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

A bathtub fills from the tap while the drain is open. Whether the water rises or falls at any moment depends on one comparison: is more coming in than going out? How much water is in the tub at 7:10 depends on how much was there at 7:00, plus everything that came in, minus everything that drained, in between. That bookkeeping is one integral.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $W'(t)$ is continuous, $\int_2^5 W'(t)\,dt$ $=$

$W(5) - W(2)$
$W(5)$
$W'(5) - W'(2)$
$W(2) - W(5)$

Show the answer

The integral of a rate of change is the net change.

Correct: $W(5) - W(2)$ :
$W(5)$ :
$W'(5) - W'(2)$ :
$W(2) - W(5)$ :

2. A trapezoidal sum on a subinterval of width 3 with endpoint values 4 and 10 adds

21
42
30
14

Show the answer

$3 \cdot \frac{4 + 10}{2} = 21$ .

Correct: 21:
42:
30:
14:

Part 4 · See it

See it first

Two rates on 0 ≤ t ≤ 12 minutes: water flows in at R(t) = 10 + 3 sin(πt/6) liters per minute (blue) and out at D(t) = 4 + t liters per minute (crimson). From t = 0 to 6 the rate in is larger and the gap is shaded green: the amount rises. The curves cross at t = 6, where both rates are 10. After that the rate out is larger and the gap is shaded vermillion: the amount falls. The amount is greatest at t = 6. — Water flows in at R(t) and out at D(t). The amount rises while R > D, falls while D > R, and peaks where the rates cross at t = 6. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A quantity gains at a rate in and loses at a rate outits net rate of change is $A'(t) = R(t) - D(t)$
Integrate the net rate over an intervalthe net change in the amount: $\int_a^b (R(t) - D(t))\,dt$
Add the amount at the start $A(t)$ $= A(a)$ $+ \int_a^t (R(s) - D(s))\,ds$
$R - D$ changes signthe amount has a relative extremum there; compare it with the endpoints to find the absolute one

Part 6 · Key ideas

Key ideas

Amount now = amount at the start + $\int$ (rate in − rate out). The initial amount is part of the answer.
$A'(t) = R(t) - D(t)$ : the amount increases when the rate in is larger and decreases when the rate out is larger.
An integral of a rate has the accumulated unit: gallons per hour integrated over hours gives gallons.
An interpretation names the quantity, its units and the time interval.
For the greatest or least amount on a closed interval: find where R − D changes sign, then compare with the endpoints (candidates test).

Part 7 · Misconception

A common mistake

The wrong idea: The tank holds 50 L at $t = 0$ , so at $t = 9$ it holds $\int_0^9 (R(t) - D(t))\,dt$ liters.

What actually happens: That integral is the change in the amount. The amount itself is $50$ $+ \int_0^9 (R(t) - D(t))\,dt$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Visitors entering a museum

The rate E(t) at which visitors enter a museum, in people per hour, t hours after opening
$t$ (hours)	0	2	3	5	8
$E(t)$ (people per hour)	60	110	150	90	30

1. Use a right Riemann sum with the table's subintervals to estimate $\int_0^8 E(t)\,dt$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine), in people.

Show the answer

$2(110)$ $+ 1(150)$ $+ 2(90)$ $+ 3(30)$ $= 220$ $+ 150$ $+ 180$ $+ 90$ $= 640$ people.

Answer: 640 people

2. Use a trapezoidal sum with the table's subintervals to estimate $\int_0^8 E(t)\,dt$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine), in people.

Show the answer

$2 \cdot \frac{60 + 110}{2}$ $+ 1 \cdot \frac{110 + 150}{2}$ $+ 2 \cdot \frac{150 + 90}{2}$ $+ 3 \cdot \frac{90 + 30}{2}$ $= 170$ $+ 130$ $+ 240$ $+ 180$ $= 720$ people.

Answer: 720 people

3. What does $\int_0^8 E(t)\,dt$ mean in this context?

The number of visitors inside the museum at t = 8 hours
The number of visitors who enter the museum in the first 8 hours after opening
The average rate at which visitors enter during the 8 hours
The rate at which visitors enter at t = 8 hours

Show the answer

$E$ is a rate in people per hour, so $\int_0^8 E(t)\,dt$ is the number of people who enter from $t = 0$ to $t = 8$ .

The number of visitors inside the museum at t = 8 hours: Visitors also leave, and some may have been there before.
Correct: The number of visitors who enter the museum in the first 8 hours after opening: The integral of the rate in is the number who came in over the interval.
The average rate at which visitors enter during the 8 hours: That would be the integral divided by 8.
The rate at which visitors enter at t = 8 hours: That is $E(8) = 30$ people per hour.

4. A pond holds $P_0$ gallons at $t = 0$ . Water enters at $R(t)$ and leaves at $D(t)$ gallons per day. Which gives the amount at time $t$ ?

$\int_0^t (R(s) - D(s))\,ds$
$P_0 + R(t) - D(t)$
$P_0$ $+ \int_0^t (R(s) + D(s))\,ds$
$P_0$ $+ \int_0^t (R(s) - D(s))\,ds$

Show the answer

$P(t)$ $= P(0)$ $+ \int_0^t P'(s)\,ds$ with $P' = R - D$ .

$\int_0^t (R(s) - D(s))\,ds$ : The starting amount is missing.
$P_0 + R(t) - D(t)$ : Rates are added to an amount without integrating; the units do not match.
$P_0$ $+ \int_0^t (R(s) + D(s))\,ds$ : Water leaving is subtracted.
Correct: $P_0$ $+ \int_0^t (R(s) - D(s))\,ds$ : Start amount plus net change.

5. Snow falls on a driveway at $2t$ cm per hour while melting removes it at 3 cm per hour, for $0 \le t \le 5$ . At $t = 0$ the snow is 10 cm deep. Find the depth $H(t)$ .

Type an expression for $H(t) =$ .

Show the answer

$H(t)$ $= 10$ $+ \int_0^t (2s - 3)\,ds$ $= 10$ $+ t^2 - 3t$ .

Answer: $H(t) =$ $t^{2} - 3t + 10$

6. Water enters a tank at $R(t)$ and leaves at $D(t)$ . When is the amount of water increasing at $t = 4$ ?

When $R'(4) > 0$
When $R(4) > 0$
When $\int_0^4 R(t)\,dt > \int_0^4 D(t)\,dt$
When $R(4) > D(4)$

Show the answer

The amount's rate of change is $R(t) - D(t)$ ; it is increasing at $t = 4$ when $R(4) - D(4) > 0$ .

When $R'(4) > 0$ : The rate in growing does not mean it beats the rate out.
When $R(4) > 0$ : The rate out might be larger than $R(4)$ .
When $\int_0^4 R(t)\,dt > \int_0^4 D(t)\,dt$ : That says the amount is larger than at the start, not that it is rising at t = 4.
Correct: When $R(4) > D(4)$ : $A'(4) = R(4) - D(4) > 0$ .

Part 9 · Summary

Summary

With a rate in

R

and a rate out

D

, the amount satisfies

A'(t) = R(t) - D(t)

and

A(t)

= A(a)

+ \int_a^t (R - D)

. Integrals of rates carry accumulated units and need an interpretation with quantity, units and interval. The amount peaks or bottoms out where

R - D

changes sign or at an endpoint.

Part 10 · Up next

What comes next

Part 11 · Connections