From a Rate to an Accumulated Change
A rate describes how quickly a quantity changes. A definite integral of that rate over a time interval represents the total change in the quantity during that interval. In a contextual response, calculating the integral is only part of the task: the conclusion should identify what changed, over which interval, by how much, and in what units.
For example, if \(r(t)\) measures the rate at which water enters a tank, then integrating \(r\) from time \(a\) to time \(b\) gives the change in the amount of water from time \(a\) to time \(b\). The integral is not itself the amount of water in the tank at time \(b\). It describes how much the amount changed.
The bounds specify the start and end of the interval. The variable \(t\) is the input to the rate, and \(dt\) indicates that the rate is accumulated with respect to time. When the rate is given in quantity-units per time-unit, the integral has quantity-units.
Why the Integral Represents Change
Over a very short time interval, a rate is approximately constant. If the rate is about \(r(t_i)\) during an interval of length \(\Delta t\), the quantity changes by approximately \(r(t_i)\Delta t\). Adding these small changes over the whole interval gives an approximation to the total change. As the intervals become narrower, this sum approaches the definite integral.
This also explains the units. Each term in the sum has units of rate multiplied by units of time. For instance, \((\text{liters per minute})(\text{minutes})=\text{liters}\). The integral therefore has the units of the quantity that changes, not the units of the rate.
There is also a direct calculus justification. If \(Q'(t)=r(t)\) and \(r\) is continuous, then \(Q\) is an antiderivative of \(r\). The Fundamental Theorem of Calculus states that evaluating an antiderivative at the endpoints gives the accumulated change:
The endpoint order matters: the result is the value at the end minus the value at the start. A positive result means the quantity ended higher than it began; a negative result means it ended lower.
Worked Example: Water Entering a Tank
Worked Example: Water Entering a Tank
Water enters a tank at a rate \(r(t)=4+2t\) liters per minute, where \(t\) is measured in minutes. How does the amount of water change from \(t=0\) to \(t=3\)?
The accumulated change is the integral of the rate over the given time interval:
An antiderivative of \(4+2t\) is \(4t+t^2\), since its derivative is \(4+2t\). Evaluate it at the bounds:
The rate has units of liters per minute and the interval has units of minutes, so the change is measured in liters. Because the rate is positive throughout the interval and the integral is positive, the amount of water increases by \(21\) liters from \(t=0\) to \(t=3\).
Check: Differentiating the antiderivative gives \(\frac{d}{dt}(4t+t^2)=4+2t\), the stated rate. Also, \(4+2t\) is positive for every \(t\) from \(0\) to \(3\), consistent with an increase.
Units and the Meaning of the Result
To write a complete interpretation, name both the quantity that changes and the interval. Include the units of the accumulated change, and use wording that agrees with the sign. For example, “the amount increases by \(21\) liters from \(t=0\) to \(t=3\) minutes” is more informative than “the integral is \(21\).”
| Integral result | Interpretation of the change |
|---|---|
| Positive | The quantity is greater at the ending time than at the starting time. |
| Negative | The quantity is less at the ending time than at the starting time. |
| Zero | The quantity has the same ending and starting values, even if it changed in between. |
If \(Q(a)\) is the initial quantity, then the endpoint relationship is \(Q(b)=Q(a)+\int_a^b r(t)\,dt\). This equation clarifies the role of the integral: it is the change added to the initial amount, not the ending amount on its own. The accumulated change can be interpreted even when no initial value is provided.
Worked Example: A Quantity with a Negative Net Change
Worked Example: A Quantity with a Negative Net Change
A stored material changes at the rate \(r(t)=2t-5\) kilograms per hour, where \(t\) is measured in hours. Determine how the amount changes from \(t=0\) to \(t=4\).
Integrate the rate from the starting time to the ending time:
The units are \((\text{kilograms per hour})(\text{hours})=\text{kilograms}\), so the integral represents a change of \(-4\) kilograms. Thus the amount decreases by \(4\) kilograms from \(t=0\) to \(t=4\). The negative sign gives the direction of the change; the decrease itself is described using the positive magnitude \(4\).
The rate is negative when \(0\leq t<2.5\) and positive when \(2.5<t\leq4\), because \(2t-5=0\) at \(t=2.5\). So the quantity first decreases and then increases. The integral accounts for both parts and gives the overall change across the full interval.
Check: The derivative of \(t^2-5t\) is \(2t-5\), as required. The rate is negative for a longer portion of the interval than it is positive, and its values are negative over much of that interval, which agrees with the negative accumulated change.
When the Rate Varies Continuously
A rate need not be constant. If it is given by a function, the definite integral accounts for its changing values throughout the interval. In a response, keep the rate’s units visible until after the integral is evaluated. This makes it easier to report the correct units for the resulting change.
Worked Example: A Decreasing Flow Rate That Remains Positive
A fluid enters a container at the rate \(r(t)=8e^{-t/4}\) gallons per hour, where \(t\) is measured in hours. Find the change in the amount of fluid from \(t=0\) to \(t=4\).
The rate is positive throughout this interval because \(8>0\) and \(e^{-t/4}>0\). Its value decreases as time increases, but the flow is still into the container. The accumulated change is
The exact change is \(32(1-e^{-1})\) gallons, approximately \(20.2\) gallons to three significant figures. It is a positive change, so the amount in the container increases by approximately \(20.2\) gallons over the four-hour interval.
Check: Differentiating the proposed antiderivative gives \(\frac{d}{dt}(-32e^{-t/4})=8e^{-t/4}\). The rate remains positive on \([0,4]\), so a positive accumulated change is consistent with fluid entering the container.
Common Pitfall: Reporting the Rate Instead of the Change
A rate and an accumulated change have different units. If a rate is measured in liters per minute, its integral over minutes is measured in liters. Reporting the integral in liters per minute confuses the rate with the change it produces.
Another error is to report a negative integral as a positive increase. For the stored material example, a student might write:
The calculation gives a negative change, so the conclusion contradicts the result. The correction is: “The amount decreases by \(4\) kilograms from \(t=0\) to \(t=4\).” If an initial amount \(Q(0)\) were known, the ending amount would be \(Q(0)-4\) kilograms, not \(Q(0)+4\) kilograms.
Bounds must also match the requested direction in time. Reversing the bounds changes the sign: \(\int_b^a r(t)\,dt=-\int_a^b r(t)\,dt\). When the prompt asks how the quantity changes from an earlier time to a later time, use the earlier time as the lower bound and the later time as the upper bound.
A Concise Interpretation Checklist
Before submitting a contextual explanation of accumulated change, check that it answers the question in context, rather than stopping at a bare numerical value.
State what is changing and what the given rate measures.
Place the starting time below the integral sign and the ending time above it.
Rate units multiplied by time units give units of the changing quantity.
Describe a positive result as an increase and a negative result as a decrease.
The integral gives the change; the ending amount also depends on the starting amount.
Check Your Understanding
For each question, include the meaning of the integral in context, not only a numerical answer.
- A machine produces material at \(5\) kilograms per hour for \(3\) hours. What quantity does the integral of this rate over the interval represent, and what are its units?
- If \(r(t)\) is measured in people per day, what units does \(\int_2^7 r(t)\,dt\) have, and which time interval does it describe?
- A rate integral over an interval equals \(-12\) liters. What does the sign say about the ending amount compared with the starting amount?
- Explain why the integral of a rate in dollars per month over an interval measured in months has units of dollars.
- If an integral of a rate gives a positive change of \(6\) meters, is the ending position necessarily \(6\) meters? Explain what additional quantity would be needed to determine the ending position.
