Rate-In/Rate-Out: Mastering Accumulation Word Problems (AP Style)

One of the most common FRQ types. It involves a system where something is being added and removed simultaneously at given rates.

Core Theorem

The Net Change Theorem: The amount of a quantity at time

t

is given by

A(t) = A(0) + \int_{0}^{t} [R_{in}(x) - R_{out}(x)] dx

Step-by-Step SOP

1
Step 1: Define the Accumulation Function
Establish the total amount formula: $A(t) = A(0) + \int_{0}^{t} [R_{in}(x) - R_{out}(x)] dx$ . This ensures you don't forget the initial condition.
2
Step 2: Identify Rates and Limits
Determine the 'Rate In' and 'Rate Out' functions. Set the integral limits based on the time interval provided in the question (e.g., from $t=0$ to $t=10$ ).
3
Step 3: Integrate the Net Rate
Find the antiderivative of $[R_{in}(t) - R_{out}(t)]$ . Remember integration rules, such as $\int e^{kt} dt = \frac{1}{k}e^{kt}$ .
4
Step 4: Evaluate and Add Initial Value
Use the Fundamental Theorem of Calculus to calculate the definite integral. Finally, add the initial amount $A(0)$ to get the total quantity.

Practice Exercises

Example 01Medium

Water is pumped into a tank at

R(t) = 20e^{0.02t}

gal/min. Water drains out at

D(t) = 12

gal/min. If the tank has 50 gallons at

t=0

, how much is in the tank at

t=10

NEED A HINT?

Apply the Net Change Theorem: Total = Initial +

\int_{a}^{b} (R(t) - D(t)) dt

. Be careful when evaluating

e^0

; it is not zero!

SHOW DETAILED EXPLANATION

Step 1: Set up the Accumulation Function

The amount of water at

t=10

is given by the initial amount plus the integral of the net rate of change:

A(10) = 50 + \int_{0}^{10} (20e^{0.02t} - 12) dt

Step 2: Find the Antiderivative

Integrate the net rate:

\int (20e^{0.02t} - 12) dt = \frac{20}{0.02}e^{0.02t} - 12t = 1000e^{0.02t} - 12t

Step 3: Evaluate the Definite Integral

Apply the Fundamental Theorem of Calculus:

[1000e^{0.02t} - 12t]_{0}^{10} = (1000e^{0.2} - 120) - (1000e^0 - 0) \approx 1101.4 - 1000 = 101.4

Step 4: Add Initial Amount

Add the net change to the starting 50 gallons:

A(10) = 50 + 101.4 = 151.4

gallons.

Common Pitfalls

⚠
The 'Zero' TrapPlugging in $t=0$ doesn't always result in 0. In this case, $1000e^{0} = 1000$ . Always subtract the lower limit carefully.
⚠
Missing Initial ValueThe integral only calculates the **Net Change**. Don't forget to add the initial 50 gallons to your final result.
⚠
Rate vs. Amount $R(t)$ and $D(t)$ are rates. To find the total gallons, you must integrate. Never just calculate $R(10) - D(10)$ for the total amount.

Ready to test it?

Practice Integration & Accumulation multiple-choice questions, timed and fully explained.

MCQ Practice

On the AP exam

See how this shows up on the free-response section: step-by-step method, scoring tips and practice.

FRQ Practice

Past AP Exams

Real AP exam questions on this topic

Official AP Calculus AB free-response questions, each with a full step-by-step solution.

Gary Chang

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5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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Step 1: Define the Accumulation Function

Step 2: Identify Rates and Limits

Step 3: Integrate the Net Rate

Step 4: Evaluate and Add Initial Value