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

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

This topic explores one of the most powerful applications of the definite integral: calculating the total accumulation or net change of a quantity when its rate of change is known. If we have a function representing a rate, such as the velocity of a particle (rate of change of position) or the rate at which water flows into a tank (rate of change of volume), the definite integral of that rate function over an interval gives the total net change in the original quantity over that same interval. This fundamental concept, often called the Net Change Theorem, allows us to solve a wide variety of real-world problems involving accumulation.

Furthermore, this topic introduces a specific application of the definite integral to find the "average value" of a function over a closed interval. Just as we can find the average of a discrete set of numbers, the definite integral provides a way to find the average value of a function that has infinitely many values over an interval. This is achieved by integrating the function over the interval and then dividing by the length of the interval, effectively "smoothing out" the function's values to a single representative value.

Key Formulas

The principles in this topic are captured by two key formulas derived from the properties of the definite integral.

1. The Net Change Theorem

If $f^{\prime }(x)$ represents the rate of change of a quantity $f(x)$, then the definite integral of $f^{\prime }(x)$ from $a$ to $b$ gives the net change in $f(x)$ over the interval $[a,b]$.

$$\text{Net Change}={\int }_a^b{f}^{\prime }(x)dx=f(b)-f(a)$$

2. The Formula for the Average Value of a Function

The average value of a continuous function $f(x)$ on the closed interval $[a,b]$ is given by:

$$\text{Average Value}=\frac{1}{b-a}{\int }_a^{b}f(x)dx$$

Understanding Net Change vs. Final Amount

A critical distinction in applied problems is between "net change" and the "final amount" of a quantity. The definite integral of a rate function, ${\int }_a^{b}R(t)dt$, directly calculates the net change from time $a$ to time $b$. It tells you how much the quantity has increased or decreased over that period.

However, many problems ask for the final amount of the quantity at a specific time, say $t=b$. To find this, you must know the initial amount at the start of the interval, $t=a$, and add the net change that occurred. This relationship is a rearrangement of the Net Change Theorem:

$$\text{Final Amount}=\text{Initial Amount}+\text{Net Change}$$

$$f(b)=f(a)+{\int }_a^b{f}^{\prime }(t)dt$$

This formula is essential for solving accumulation problems where an initial condition is provided. For example, if you know the amount of water in a tank at $t=0$ and the rate at which water is being added, you can find the amount at any later time $t$.

Core Concepts & Rules

• The definite integral of a rate of change gives the net change of the original quantity. If $R(t)$ is the rate of change of $Q(t)$, then ${\int }_a^{b}R(t)dt=Q(b)-Q(a)$.

• Definite integrals can be used to model accumulation in various contexts, such as particle motion (integrating velocity to find displacement), resource consumption (integrating rate of consumption to find total consumed), or population growth (integrating rate of growth to find change in population).

• To find the total amount of a quantity at a specific time, you must add the net change (calculated by the definite integral) to the initial amount.

• The average value of a function $f(x)$ over an interval $[a,b]$ is calculated by finding the definite integral of $f(x)$ over that interval and dividing the result by the length of the interval, $b-a$.

Step-by-Step Example 1: Basic Application

Problem: A company produces widgets at a rate given by $r(t)=100+4t$ widgets per hour, where $t$ is the number of hours since the factory opened at 8:00 AM. Find the net change in the number of widgets produced between 9:00 AM and 11:00 AM.

Solution:

1. Identify the Rate Function and Interval:

   The rate of production is $r(t)=100+4t$.

   The time interval is from 9:00 AM ($t=1$) to 11:00 AM ($t=3$).

2. Set up the Definite Integral:

   The net change in the number of widgets is the definite integral of the rate function over the time interval $[1,3]$.

   $$\text{Net Change}={\int }_1^3(100+4t)dt$$

3. Find the Antiderivative:

   The antiderivative of $100+4t$ is $100t+4\cdot \frac{t^2}{2}=100t+2t^2$.

4. Evaluate the Definite Integral using the Fundamental Theorem of Calculus:

   $${\left[100t+2t^2\right]}_1^3$$

   $$=(100(3)+2(3{)}^2)-(100(1)+2(1{)}^2)$$

   $$=(300+2(9))-(100+2)$$

   $$=(300+18)-(102)$$

   $$=318-102=216$$

5. State the Final Answer with Units:

   The net change in the number of widgets produced between 9:00 AM and 11:00 AM is 216 widgets.

Step-by-Step Example 2: Exam-Style Application

Problem: At time $t=0$ hours, a water tank holds 500 gallons of water. For $0\le t\le 8$, water is pumped into the tank at a rate modeled by $W(t)=20\sin (0.2t^2)$ gallons per hour.

(a) To the nearest whole number, how many gallons of water are in the tank at time $t=8$ hours?

(b) What is the average rate at which water is pumped into the tank from $t=0$ to $t=8$ hours?

Solution:

(a) Find the amount of water at $t=8$ hours.

1. Identify the Goal: We need the total amount of water at $t=8$, not just the net change. This means we need the initial amount plus the accumulated amount.

   $$\text{Amount at }t=8=\text{Initial Amount}+\text{Net Change from }t=0\text{ to }t=8$$

2. Set up the Expression:

   The initial amount is 500 gallons. The net change is the integral of the rate function $W(t)$ from $t=0$ to $t=8$.

   $$\text{Amount}=500+{\int }_0^{8}20\sin (0.2t^2)dt$$

3. Evaluate using a Calculator:

   This integral is difficult to compute by hand. We use a calculator's numerical integration feature.

   $${\int }_0^{8}20\sin (0.2t^2)dt\approx 111.916$$

   Now, add the initial amount:

   $$\text{Amount}\approx 500+111.916=611.916$$

4. State the Final Answer with Units:

   To the nearest whole number, there are 612 gallons of water in the tank at time $t=8$ hours.

(b) Find the average rate of water flow from $t=0$ to $t=8$ hours.

1. Identify the Goal: We need the average value of the rate function $W(t)$ on the interval $[0,8]$.

2. Set up the Average Value Formula:

   $$\text{Average Rate}=\frac{1}{8-0}{\int }_0^{8}W(t)dt$$

   $$\text{Average Rate}=\frac{1}{8}{\int }_0^{8}20\sin (0.2t^2)dt$$

3. Evaluate using a Calculator:

   From part (a), we already know the value of the integral.

   $$\text{Average Rate}\approx \frac{1}{8}(111.916)\approx 13.9895$$

4. State the Final Answer with Units:

   The average rate at which water is pumped into the tank from $t=0$ to $t=8$ is approximately 13.990 gallons per hour.

Using Your Calculator

This topic frequently appears on the calculator-active section of the AP Exam. The primary tool you will use is the numerical integration function, often labeled fnInt.

To find the Net Change or Final Amount:

Suppose you need to find the amount of a pollutant in a lake at $t=10$ days, given an initial amount of 50 kg and a rate of pollution $P(t)=0.5e^{0.1t}$ kg/day.

1. The expression is $50+{\int }_0^{10}0.5e^{0.1t}dt$.

2. On a TI-84 style calculator, you would enter this as:

   50 + fnInt(0.5*e^(0.1*X), X, 0, 10)

3. The calculator will evaluate this expression to find the final amount.

To find the Average Value:

Suppose you need to find the average temperature, given by $T(t)=75-10\cos (\frac{\pi t}{12})$ degrees Fahrenheit, from $t=0$ to $t=24$ hours.

1. The expression is $\frac{1}{24-0}{\int }_0^{24}(75-10\cos (\frac{\pi t}{12}))dt$.

2. On a TI-84 style calculator, you would enter this as:

   (1/24) * fnInt(75 - 10*cos(π*X/12), X, 0, 24)

3. The calculator will evaluate this to find the average temperature.

AP Exam Quick Hit

Common Question Types

• Rate In/Rate Out: A scenario where a quantity is both entering and leaving a system. You might be given a rate in, $R_{in}(t)$, and a rate out, $R_{out}(t)$. The net rate of change is $R_{in}(t)-R_{out}(t)$. To find the total amount at time $T$, you would calculate: $InitialAmount$ + ${\int }_0^T(R_{in}(t)-R_{out}(t))dt$.

• Particle Motion: Given a velocity function $v(t)$ and an initial position $x(a)$, you are asked to find the position at a later time $x(b)$. This is a direct application of the accumulation formula: $x(b)=x(a)+{\int }_a^{b}v(t)dt$.

• Average Value in Context: You are given a function representing a quantity (e.g., temperature, height, velocity) and asked to find its average value over a specified interval. Example: "The temperature of a wire is given by $T(x)$ for $0\le x\le 10$. Find the average temperature of the wire."

Common Mistakes

• Forgetting the Initial Condition: When asked for a final amount (e.g., "position at $t=5$" or "gallons in tank at $t=10$"), students often calculate only the definite integral (the net change) and forget to add the given initial amount.

• Omitting the $\frac{1}{b-a}$ for Average Value: A very common error is to correctly set up and evaluate the integral for an average value problem but forget to multiply the result by $\frac{1}{b-a}$.

• Confusing Average Value and Average Rate of Change: The average value of a function $f(x)$ is $\frac{1}{b-a}{\int }_a^{b}f(x)dx$. The average rate of change of $f(x)$ is $\frac{f(b)-f(a)}{b-a}$. These are different concepts and are calculated differently.

• Incorrect Units: Forgetting to provide units or using the wrong ones. The integral of a rate (e.g., meters/sec) gives an amount (meters). The average value of a function has the same units as the function itself (e.g., the average value of a velocity in m/s is also in m/s).