From a Rate to a Net Change
A rate describes how a quantity changes over time. If \(Q(t)\) is the amount of a quantity at time \(t\), and \(r(t)\) is its instantaneous rate of change, then \(r(t)=Q'(t)\). Integrating that rate over an interval adds up the changes during the interval.
The result is the net change in \(Q\) between the beginning and end of that interval. It is not, by itself, the amount of \(Q\) at the endpoint. A complete contextual response should identify the quantity, its rate, the time interval, the integral, and what the evaluated result means.
Building the Integral from the Context
Before evaluating anything, determine what each part of the prompt represents. Identify the quantity being tracked, the function giving its rate, and the interval over which that rate applies. The bounds are the beginning and ending times stated in the question; they are not chosen from the formula or adjusted to make the integral positive.
If a prompt says that \(r(t)\) is measured in liters per minute and asks about the interval from \(t=2\) to \(t=6\) minutes, then the net change is represented by \(\int_2^6 r(t)\,dt\). The variable \(t\) in the integral is the time variable. The product of rate units and time units is liters, the units of the quantity’s change.
Decide what amount or measurement is changing, and use a clear symbol for it.
Confirm that the given function is an instantaneous rate of change, then read the interval endpoints from the prompt.
Integrate the rate from the starting time to the ending time, keeping the bounds in the stated order.
Report the result with units and describe it as a net increase or net decrease in the quantity over the interval.
For the integral to have this interpretation, the rate function must represent the quantity’s instantaneous rate over the interval. In the usual AP Calculus setting, a continuous rate function is sufficient. If a rate is negative on part of the interval, its contribution to the definite integral is negative: the quantity is decreasing during that part.
Why the Integral Gives Net Change
The connection follows from the Fundamental Theorem of Calculus. Suppose \(Q\) is differentiable on \([a,b]\), and its derivative is the rate \(r\). Since \(Q\) is an antiderivative of \(r\), evaluating the definite integral gives the difference between the endpoint values of \(Q\).
This result also explains why the integral alone does not tell you the endpoint amount. The integral gives the difference \(Q(b)-Q(a)\). It does not supply \(Q(a)\), the starting amount. Therefore, describing the integral as the ending quantity would confuse a change with a quantity.
By assumption, \(Q'(t)=r(t)\) on \([a,b]\).
Because \(Q\) is an antiderivative of \(r\), \(\int_a^b r(t)\,dt=Q(b)-Q(a)\).
\(Q(b)-Q(a)\) compares the quantity at the end with the quantity at the start. It is a change in \(Q\), not \(Q(b)\) alone.
Worked Examples
Worked Example: Net Change in a Liquid Volume
A tank’s volume changes at the rate \(r(t)=3t-2\) liters per minute, where \(t\) is measured in minutes. Find the net change in volume from \(t=0\) to \(t=2\), and interpret the result.
The rate is given in liters per minute and the interval is \([0,2]\), so the net change is:
The rate is negative from \(t=0\) to \(t=\frac{2}{3}\) and positive from \(t=\frac{2}{3}\) to \(t=2\). These contributions have opposite signs, but the evaluated definite integral is \(2\) liters. Thus the tank’s volume has a net increase of \(2\) liters over the interval. This answer is a change in volume, not the volume at \(t=2\).
Check: An antiderivative is \(F(t)=\frac{3}{2}t^2-2t\), and \(F'(t)=3t-2\), the given rate. The endpoint calculation gives \(F(2)-F(0)=2-0=2\) liters, as required.
Worked Example: Net Change in the Number of Visitors
A museum’s visitor count changes at the rate \(v(t)=12-2t\) visitors per minute, where \(t\) is measured in minutes. Find the net change in visitor count from minute \(1\) to minute \(5\).
The rate function describes the change in visitor count, so the requested net change is:
The units are visitors, because \((\text{visitors per minute})(\text{minutes})=\text{visitors}\). The net change in visitor count from minute \(1\) to minute \(5\) is an increase of \(24\) visitors.
Check: \(F(t)=12t-t^2\) is an antiderivative because \(F'(t)=12-2t=v(t)\). Evaluating at the bounds gives \(F(5)-F(1)=35-11=24\), which confirms the calculation.
Worked Example: A Rate That Changes Sign
The volume of water in a reservoir changes at the rate \(r(t)=t^2-4t+3\) cubic meters per minute, with \(t\) in minutes. Find the net change in volume from \(t=0\) to \(t=4\).
Use the given endpoints as the integration bounds:
The net change is \(\frac{4}{3}\) cubic meters, so the reservoir’s volume has a net increase of \(\frac{4}{3}\) cubic meters over the four-minute interval. The rate is zero at \(t=1\) and \(t=3\), and its sign changes across those times. The negative contribution during the middle part of the interval offsets some of the positive contributions elsewhere; the definite integral gives the resulting signed change.
Check: For \(F(t)=\frac{t^3}{3}-2t^2+3t\), differentiation gives \(F'(t)=t^2-4t+3=r(t)\). Also, \(F(4)=\frac{4}{3}\) and \(F(0)=0\), so \(F(4)-F(0)=\frac{4}{3}\), matching the integral.
Writing a Complete Contextual Response
A numerical result by itself may not communicate what was calculated. A strong response states the integral, evaluates it, includes units, and identifies the result as a net change over the specified interval. If the evaluated integral is positive, the net change is an increase; if it is negative, it is a decrease. If it is zero, the quantity has no net change over the interval.
For instance, writing “the answer is \(24\)” leaves the reader to infer both the units and the meaning. Writing “The net change in the visitor count from minute \(1\) to minute \(5\) is an increase of \(24\) visitors” explicitly connects the calculation to the context.
| Part of response | What it communicates |
|---|---|
| Definite integral with bounds | Which rate is accumulated and over what interval |
| Evaluated value | The numerical net change |
| Units | The units of the quantity’s change |
| Contextual sentence | Which quantity changed and whether it increased or decreased overall |
Common Mistake: Calling the Integral the Ending Amount
Suppose a rate integral evaluates to \(21\) liters over an interval. The following conclusion is incorrect: “The volume at the end of the interval is \(21\) liters.” The integral gives the net change, so that conclusion assigns the result the wrong meaning.
The correction is: “The volume has a net increase of \(21\) liters over the interval.” This statement is justified by the integral and its units. The ending volume is a different quantity: the integral alone does not give it, because the integral represents the difference between the ending and starting volumes.
Check Your Understanding
For each question, explain what the integral represents in the context, not only how to evaluate it.
- A quantity \(Q(t)\) has rate \(r(t)\) units per hour. What are the units of \(\int_2^7 r(t)\,dt\), and what does it represent?
- A rate function is negative throughout \([a,b]\). What does the sign of \(\int_a^b r(t)\,dt\) say about the net change?
- In one sentence, explain why \(\int_a^b r(t)\,dt\) is not generally the value of the quantity at time \(b\).
- For \(r(t)=5-t\) units per minute on \([1,4]\), write the definite integral that represents net change. Do not evaluate it.
- What calculation can you use to check that an antiderivative used to evaluate a rate integral is correct?
