Average Rate as a Contextual Quantity
A rate describes how quickly a quantity changes. A tank’s volume might change in liters per hour, a population might change in people per day, or an object’s velocity might describe how its position changes in meters per second. When a rate varies during a time interval, a single rate can summarize its overall effect: the average rate over the interval.
The key is to connect the rate to the change it produces. If \(r(t)\) is the instantaneous rate of change of a quantity \(Q(t)\), then the integral of \(r\) over a time interval gives the net change in \(Q\). Dividing that net change by the interval’s duration gives the average rate.
The result has the same units as \(r\). For example, if \(r\) is measured in liters per hour, its average over an interval is also measured in liters per hour. The integral has units of liters, because it represents a change in volume; dividing by elapsed hours restores units of liters per hour.
Why the Formula Gives an Average Rate
Suppose \(Q(t)\) is an amount and \(r(t)=Q'(t)\) is its instantaneous rate of change. The Net Change Theorem states that the net change in \(Q\) from \(a\) to \(b\) is the integral of its rate. The average rate of change is, by definition, that net change divided by elapsed time.
Since \(Q'(t)=r(t)\), the Net Change Theorem gives \(Q(b)-Q(a)=\int_a^b r(t)\,dt\).
The interval lasts \(b-a\) time units, so the average rate of change is \(\frac{Q(b)-Q(a)}{b-a}\).
Replacing \(Q(b)-Q(a)\) with its integral gives \(r_{\text{avg}}=\frac{1}{b-a}\int_a^b r(t)\,dt\).
This also explains why an average rate is a useful summary. If the quantity had instead changed at one constant rate \(r_{\text{avg}}\) throughout the interval, its net change would be \(r_{\text{avg}}(b-a)\). The definition makes this equal to the actual net change:
Thus the average rate is the constant rate that would produce the same net change over the same amount of time. It does not claim that the actual rate stayed constant.
Worked Example: Filling a Tank
Worked Example: Finding the Average Filling Rate
A tank’s volume is increasing at \(r(t)=8+3t\) liters per hour, where \(t\) is measured in hours since monitoring began. Find the average rate of increase from \(t=2\) to \(t=6\), and interpret it.
The duration is \(6-2=4\) hours. The average rate is the integral of the rate divided by this duration:
An antiderivative is \(8t+\frac{3}{2}t^2\), since its derivative is \(8+3t\). Therefore, the net volume change is
Divide the \(80\)-liter increase by the \(4\)-hour duration:
Interpretation: From hour 2 to hour 6, the tank’s volume increased at an average rate of \(20\) liters per hour. As a check, \(20\text{ liters per hour}\times4\text{ hours}=80\text{ liters}\), the net change found by integration.
Signed Rates and Net Change
A contextual rate can be positive or negative. A positive rate means the quantity is increasing; a negative rate means it is decreasing. The integral therefore gives net change, with increases and decreases contributing opposite signs. The average rate also keeps that sign. A positive average does not mean the rate was positive at every instant; it means the net change over the interval was positive.
Worked Example: A Population That First Grows and Then Declines
A model gives a population’s rate of change as \(p'(t)=40-8t\) people per day for \(0\leq t\leq8\), with \(t\) in days. Find and interpret the average rate of population change over these eight days.
The average rate is the net population change divided by eight days:
An antiderivative is \(40t-4t^2\), whose derivative is \(40-8t\). Evaluate the net change:
Now divide by the duration:
Interpretation: The population had an average net increase of \(8\) people per day over the eight days. The instantaneous rate becomes negative when \(40-8t<0\), or \(t>5\), so the model includes a period of decline. That decline is already accounted for by the signed integral. To check, \(8\text{ people per day}\times8\text{ days}=64\text{ people}\), matching the net change.
Average Velocity and Net Displacement
Velocity is a signed rate of change of position. Its integral gives net displacement, not necessarily total distance traveled. Consequently, average velocity is net displacement divided by elapsed time. If velocity changes direction, the positive and negative parts can cancel in the integral.
Worked Example: Average Velocity with a Change of Direction
An object moves along a straight line with velocity \(v(t)=2t-6\) meters per second for \(1\leq t\leq5\), where \(t\) is in seconds. Find its average velocity over this interval.
The time interval lasts \(5-1=4\) seconds. Integrate velocity to obtain net displacement:
The antiderivative checks because \(\frac{d}{dt}(t^2-6t)=2t-6\). Divide displacement by elapsed time:
Interpretation: The object’s average velocity is \(0\) meters per second because its net displacement over the interval is zero. This does not mean it remained still. Its velocity is negative when \(t<3\) and positive when \(t>3\), so the signed displacement contributions cancel. Checking the result, \(0\text{ meters per second}\times4\text{ seconds}=0\text{ meters}\).
A Common Mistake: Forgetting to Divide by Time
The integral of a rate over time gives a change in the underlying quantity, not the average rate itself. In the tank example, the integral was \(80\) liters. Reporting \(80\) liters per hour as the average rate would incorrectly attach rate units to a net change and would skip division by the \(4\)-hour duration.
Another tempting shortcut is to average only the rates at the endpoints. That is not generally the average value of a changing rate. For instance, if a rate is \(r(t)=t^2\) units per minute on \(0\leq t\leq2\), the endpoint rates are \(0\) and \(4\). Their arithmetic mean is \(2\), but the average rate over the interval is
The endpoint average gives a different answer here. Use the integral over the full interval unless the rate has a special form that justifies another method.
Reading the Result in Context
Before presenting an average-rate answer, check three features:
- Interval: Use the requested start and end times, and divide by \(b-a\), not by \(b\) unless \(a=0\).
- Meaning of the integral: Identify the net change in the relevant quantity and keep its sign.
- Units and interpretation: Divide quantity units by time units, then state what the resulting rate means in the situation.
These checks help distinguish a rate from an accumulated amount. For example, liters and liters per hour describe different quantities: the first can represent a volume change, while the second represents how quickly volume changes. The average rate is the net change spread across the elapsed time, not the net change by itself.
Check Your Understanding
For each question, state the average rate with units and explain what it means in context.
- A reservoir’s volume changes at \(r(t)=12-2t\) cubic meters per hour for \(0\leq t\leq4\). Find its average rate of change over the interval.
- A quantity increases by a net \(90\) kilograms over \(6\) hours. What is its average rate of change, and what calculation checks your answer?
- A velocity function has a negative integral over a time interval. What does the sign of the average velocity say about net displacement?
- Explain why integrating a rate and stopping at that result does not usually answer a question asking for average rate.
- For \(r(t)=t^2\) units per minute on \(0\leq t\leq3\), find the average rate and compare it with the arithmetic mean of the endpoint rates.
