Using Accumulation Functions | AP Calc BC Unit 8 Study Guide

The Core Idea: Using Accumulation Functions and Definite Integrals in Applied Contexts

This topic establishes a fundamental connection between the rate at which a quantity changes and the net change in that quantity over a period of time. The core idea is that the definite integral acts as an "accumulator." By integrating a rate of change function over a specific interval, we can determine the total net accumulation, or net change, of the original quantity across that interval.

This concept is powerful because it allows us to move from information about a rate (e.g., velocity in meters per second, flow rate in gallons per minute) to information about an amount (e.g., net displacement in meters, net change in volume in gallons). This principle is applicable in a wide variety of real-world contexts. Furthermore, the value of this definite integral can often be determined by interpreting it as the geometric area between the rate function's graph and the horizontal axis.

Key Formulas & Concepts

The primary concept in this topic is the relationship between a function representing a rate of change and the net change in the related quantity.

The Net Change Theorem

If a quantity changes at a rate given by the function $f (t)$ , then the net change in that quantity from time $t = a$ to $t = b$ is given by the definite integral of the rate function.

$Net Change = \int_{a}^{b} f (t) d t$

If $F (t)$ is the original quantity, and $F^{'} (t) = f (t)$ , this can be expressed as:

$Net Change = F (b) - F (a) = \int_{a}^{b} F^{'} (t) d t$

The Accumulation Function

To find the total amount of a quantity at a specific time $b$ , we must know the amount at a starting time $a$ (an initial condition) and add the net change that occurred from $a$ to $b$ .

$Final Amount = Initial Amount + Net Change$

$F (b) = F (a) + \int_{a}^{b} F^{'} (t) d t$

Definite Integral as Net Area

The value of a definite integral $\int_{a}^{b} f (t) d t$ can be interpreted as the net area between the curve of $f (t)$ and the t-axis from $t = a$ to $t = b$ .

Area above the t-axis is counted as positive.
Area below the t-axis is counted as negative.

Understanding the Difference: Net Change vs. Final Amount

A critical distinction in applied problems is between finding the net change and finding the final amount.

Net Change: The definite integral $\int_{a}^{b} f (t) d t$ by itself only provides the net change. It answers the question, "By how much did the quantity increase or decrease between time $a$ and time $b$ ?" For example, if $\int_{1}^{5} v (t) d t = - 10$ meters, it means the particle's net displacement was 10 meters in the negative direction from $t = 1$ to $t = 5$ . It does not tell us the particle's final position.
Final Amount (or Final Position): To find the final amount, you must have a starting point. The formula $F (b) = F (a) + \int_{a}^{b} F^{'} (t) d t$ is used. It answers the question, "What is the value/position/amount of the quantity at time $b$ ?" Using the previous example, if the particle's position at $t = 1$ was $s (1) = 25$ , then its final position at $t = 5$ would be $s (5) = s (1) + \int_{1}^{5} v (t) d t = 25 + (- 10) = 15$ meters.

Core Concepts & Rules

The definite integral of a rate of change gives the net change of the quantity.
This principle can be applied to any context involving rates, such as velocity, population growth rates, water flow rates, etc.
To find the value of a quantity at a specific time, one must add the net change (the definite integral) to a known initial value of the quantity.
When a rate function is represented graphically, its definite integral can be calculated by finding the geometric net area between the graph and the horizontal axis over the desired interval.

Step-by-Step Example 1: Net Change in Volume

Problem: Water is pumped into a tank at a rate of $r (t) = 200 - 4 t$ liters per minute, where $t$ is the time in minutes, for $0 \leq t \leq 30$ . What is the net change in the volume of water in the tank from $t = 5$ to $t = 15$ ?

Step 1: Identify the Rate and the Interval

The rate of change is given by the function $r (t) = 200 - 4 t$ .

The interval is from $t = 5$ to $t = 15$ .

Step 2: Set up the Definite Integral

According to the Net Change Theorem, the net change in volume is the definite integral of the rate function over the given interval.

$Net Change = \int_{5}^{15} (200 - 4 t) d t$

Step 3: Evaluate the Integral

Find the antiderivative of $r (t)$ .

$\int (200 - 4 t) d t = 200 t - 4 \frac{t ^{2}}{2} + C = 200 t - 2 t^{2} + C$

Now, apply the Fundamental Theorem of Calculus.

$Net Change = [200 t - 2 t^{2}]_{5}^{15}$

$= (200 (15) - 2 (15)^{2}) - (200 (5) - 2 (5)^{2})$

$= (3000 - 2 (225)) - (1000 - 2 (25))$

$= (3000 - 450) - (1000 - 50)$

$= 2550 - 950$

$= 1600$

Step 4: State the Final Answer with Units

The net change in the volume of water in the tank from $t = 5$ to $t = 15$ minutes is 1600 liters.

Step-by-Step Example 2: Finding Position from a Velocity Graph

Problem: The graph below shows the velocity $v (t)$ , in meters per second, of a particle moving along the x-axis for $0 \leq t \leq 10$ . The graph consists of a line segment and a semicircle. The particle's position at $t = 0$ is $x (0) = 8$ . Find the position of the particle at $t = 10$ .

Using Accumulation Functions and Definite Integrals in Applied Contexts - AP Calculus BC Study Guide