From a Rate to a Change
A rate describes how a quantity changes at an instant. For example, a population’s growth rate describes how quickly its size is changing at each time. To find the population’s change across an entire time interval, we need to combine the rate over that interval.
The definite integral performs this accumulation. The connection is not a separate rule to memorize without explanation: it follows directly from the Fundamental Theorem of Calculus. If we know the rate of a quantity and the interval over which it acts, integration gives the quantity’s net change on that interval.
Definition and Conditions
Let \(Q(t)\) represent a quantity at time \(t\). Its instantaneous rate of change is \(Q'(t)\). Suppose that \(Q\) is differentiable on an interval containing \([a,b]\), and that \(Q'\) is continuous on \([a,b]\). The continuity condition allows us to apply the Fundamental Theorem of Calculus to the rate function.
The order of the endpoint values matters: the change from \(a\) to \(b\) is the later value minus the earlier value, \(Q(b)-Q(a)\). If the quantity’s value at \(a\) is known, the theorem can be rearranged to find the value at \(b\): add the net change to the starting value.
Why the Theorem Works
The key observation is that \(Q\) is an antiderivative of its rate \(r\), because \(Q'(t)=r(t)\). The Fundamental Theorem of Calculus says that integrating a continuous function over \([a,b]\) gives the difference in values of any antiderivative at the endpoints. Here, that antiderivative is \(Q\) itself.
Write \(r(t)=Q'(t)\), where \(Q\) is the quantity being tracked.
By the Fundamental Theorem of Calculus, \(\int_a^b r(t)\,dt\) equals the endpoint difference of an antiderivative of \(r\).
Since \(Q'(t)=r(t)\), the endpoint difference is \(Q(b)-Q(a)\), the quantity’s net change.
In equation form, the full justification is:
This reasoning also explains why the rate must describe the derivative of the quantity in question. If a function is not the rate of change of \(Q\), integrating it does not, by this theorem, give the change in \(Q\).
Worked Examples
Worked Example: Finding a Population Change
A population’s growth rate is \(r(t)=3t^2+2t\) individuals per year, where \(t\) is measured in years. Find the population’s change from \(t=0\) to \(t=2\). If the population is 18 individuals at \(t=0\), find its value at \(t=2\).
Because \(r(t)\) is the population’s rate, the Net Change Theorem gives:
An antiderivative is \(t^3+t^2\), since \(\frac{d}{dt}(t^3+t^2)=3t^2+2t\). Evaluate at the endpoints:
The population changes by 12 individuals over the interval. Starting from 18 individuals, its value at \(t=2\) is \(18+12=30\) individuals.
Check: Differentiating \(t^3+t^2\) returns the given rate, \(3t^2+2t\). Evaluating the antiderivative at 2 gives 12, and at 0 gives 0, so the endpoint difference is 12 as calculated.
Worked Example: Change in an Inventory
An inventory level \(Q(t)\) changes at the rate \(r(t)=4-t\) items per hour from \(t=0\) to \(t=3\). If the inventory starts at 20 items, find its net change and its level after 3 hours.
Integrate the rate over the interval:
At the upper endpoint, \(4\cdot 3=12\) and \(\tfrac{1}{2}\cdot 3^2=\tfrac{9}{2}\), so the upper-end value is \(12-\tfrac{9}{2}=\tfrac{24}{2}-\tfrac{9}{2}=\tfrac{15}{2}\). The lower-end value is 0. Therefore:
Add this change to the initial inventory to obtain the final level:
Check: The derivative of \(4t-\tfrac{1}{2}t^2\) is \(4-t\), the stated rate. Its values at 3 and 0 are \(\tfrac{15}{2}\) and 0, respectively, confirming the calculated change. Adding that change to 20 gives \(\tfrac{55}{2}\).
Worked Example: Accumulated Inflow
Water flows into a container at the rate \(r(t)=6\cos(t)\) liters per minute during the interval \(0\leq t\leq \frac{\pi}{2}\). Find the change in the amount of water during that interval. If the container starts with 10 liters, how much water is in it at the end?
Integrate the rate from the beginning to the end of the interval. An antiderivative of \(6\cos(t)\) is \(6\sin(t)\):
The amount of water changes by 6 liters. The final amount is the initial 10 liters plus that change:
Check: Differentiating \(6\sin(t)\) gives \(6\cos(t)\). Its value at \(\frac{\pi}{2}\) is 6 and its value at 0 is 0, so the integral is \(6-0=6\) liters, as found.
A Common Mistake: Subtracting Rates
The theorem says to subtract the quantity’s endpoint values, not the rate’s endpoint values. A rate tells how quickly the quantity is changing at a particular time. The difference between two rate values describes a change in the rate; it is not the accumulated change in the quantity.
For example, suppose the rate is \(r(t)=2t+1\) on \([1,4]\). A mistaken calculation might subtract the endpoint rates:
That calculation gives the change in the rate’s value, not the change in the quantity. To find the quantity’s change, integrate the rate:
Correction: The quantity’s net change is 18, not 6. Checking, the derivative of \(t^2+t\) is \(2t+1\), and its endpoint values are 20 and 2; their difference is 18.
When the Theorem Is Useful
The Net Change Theorem applies whenever a quantity is tracked by a rate that is its derivative and the rate is continuous on the interval. The quantity might represent a population, an inventory, an amount of water, or another changing measurement. The theorem turns information about instantaneous change into information about change across a whole interval.
Keep the two related results distinct:
The first equation gives a change, not an endpoint value on its own. The second uses the starting value to recover the endpoint quantity. This distinction helps ensure that an answer responds to the actual question: whether it asks for a change over an interval or a quantity at a particular time.
Check Your Understanding
For each question, identify what quantity the rate describes and use the Net Change Theorem where appropriate.
- If \(Q'(t)=5t\) on \([0,2]\), what is the net change in \(Q\) over the interval?
- A quantity begins at 12 and changes at rate \(r(t)=3\) for \(0\leq t\leq 4\). What is its value at \(t=4\)?
- State the continuity condition used in the Net Change Theorem.
- Why does \(\int_a^b Q'(t)\,dt\) equal \(Q(b)-Q(a)\)?
- For a rate \(r(t)=t+2\) on \([1,3]\), explain why \(r(3)-r(1)\) is not the quantity’s net change.
