8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts

4 min read•june 18, 2024

AP Calculus AB/BC ♾️

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🎥Watch: AP Calculus AB/BC - Interpreting the Meaning of a Derivative/Integral

Accumulation problems are word problems where the rate of change of a quantity is given and we are asked to calculate the value of the quantity accumulated over time. These problems are solved using definite integrals.

What should you know about this topic? 🧠

For this topic, you’ll need to know how to do two things: interpret the meaning of a definite integral and determine the net change of an accumulation problem.

In order to interpret the meaning of a definite integral, you must know two important things. Firstly, a function defined as an integral represents an accumulation of a rate of change. Second, the definite integral of the rate of change of a quantity over an interval gives the net change of that quantity over that interval.

What will this look like on the AP Exam? ✍

In order to understand this concept, let’s look at an example from an actual AP exam. The following question is from the 2000 AP Calculus AB Exam posted on the College Board website. All credits to College Board.

2000 AB FRQ #4: Water is pumped into a tank at a constant rate of 8 gallons per minute. Water leaks out of the tank at the rate of √(t + 1) gallons per minute, for 0 ≤ t ≤ 120. At time t = 0, the tank contains 30 gallons of water. 🥛

(a) How many gallons of water leak out of the tank from time t = 0 to t = 3 minutes?

(b) How many gallons of water are in the tank at time t =3 minutes?

(d) At what time t, for 0 ≤ t ≤ 120, is the amount of water in the tank a maximum? Justify your answer.

Before you look at the explanations, try to answer yourself!

Answers to FRQ #4

The first part of the question asked for the amount of water that leaked out of the tank in the first 3 minutes. To solve this, you must integrate the leak function from 0 to 3. When we integrate, we get the following:

The second part asked for the amount of water in the tank after t = 3 minutes. Our amount at t = 0 was given as 30. So, we start with 30 gallons and then add the amount that accumulated over three minutes. Water is pumped in at a constant rate of 8 gallons/minute, so we add 24 gallons, for a running total of 54 gallons. Finally, we just determined in part (a) that 14/3 gallons leaked out in the first 3 minutes, so from 54 we subtract 14/3 gallons to get 148/3 gallons, or about 49.3 gallons.
The third part asked for an expression for A(t), the amount of water at any time t. We know we have an initial value of 30, so that should be added in the equation. We also know that we have a constant flow of water into the tank of 8 gallons per minute. Since t is in minutes, we can represent this as 8t. Finally, we were able to find the amount leaking out by integrating the function √(t + 1). So, our function A(t) should look something like one of the following:

Alternatively, you could've said that dA/dt = 8 - √(t + 1). Then, taking the integral of that function, you would find that A(t) = 8t - 2/3(t + 1)^(3/2) + C. After that, you would set A(0) = 30 (due to our starting value) to find that C = 92/3. Thus, if you chose this strategy, your equation would be:

The final part asks you to find an extreme point, more specifically a maximum. To do this, we want to take the derivative of A(t) and find where it is equal to zero. This indicates a point of inflection. Then, we want to make sure that A'(t) goes from positive to negative at that point, which indicates that A(t) was increasing before that and is now decreasing (aka, a maximum). The derivative of A(t) is 8 - √(t + 1). Setting that equal to 0, we can see that √(t + 1) should equal 8. 64 is the square of 8, so t + 1 should equal 64. Therefore, t = 63. Plugging in the point before that (62), we would have a positive number. Plugging in the point after (64), we would have a negative number. Therefore, t = 63 is the time at which the tank is fullest.