From a Rate to a Net Change
A rate describes how a quantity changes as its input changes. For example, a rate might describe the change in a population per year, the change in a position per second, or the change in a stored amount per minute. Knowing the rate throughout an interval lets us determine the quantity’s net change over that interval.
The key operation is a definite integral. If \(Q(t)\) is a quantity and \(r(t)\) is its rate, then \(r(t)=Q'(t)\). The integral of that rate from \(t=a\) to \(t=b\) gives the change from the quantity’s value at \(a\) to its value at \(b\). If the question asks for the final quantity itself, we also need the initial value.
This distinction matters: the integral gives a change, not automatically the quantity’s endpoint value. To find the endpoint value, add the change to the initial value.
Why the Definite Integral Gives the Change
The relationship follows from the Fundamental Theorem of Calculus. Since \(Q'(t)=r(t)\), the function \(Q\) is an antiderivative of \(r\). The theorem says that integrating a continuous function over an interval is equivalent to evaluating an antiderivative at the endpoints and subtracting.
Rearranging this equation shows why the initial value is needed when we want the final value:
More generally, suppose \(F\) is any antiderivative of \(r\), so \(F'(t)=r(t)\). Then \(Q'(t)=F'(t)\), which means \((Q-F)'(t)=0\). By the Mean Value Theorem, a function with derivative zero throughout an interval is constant there. Thus \(Q(t)-F(t)\) has the same value at \(a\) and \(b\), giving \(Q(b)-Q(a)=F(b)-F(a)\). The Fundamental Theorem identifies this endpoint difference with \(\int_a^b r(t)\,dt\). This is why the accumulated rate records the quantity’s net change.
A Reliable Calculation Process
For an application, identify the rate and the interval before evaluating anything. Keep track of whether the question asks for a change or for the quantity’s value at the end.
Write the rate function as \(r(t)=Q'(t)\), where \(Q\) is the quantity being tracked.
Use the starting input \(a\) as the lower bound and the ending input \(b\) as the upper bound.
Find an antiderivative \(F\), then calculate \(F(b)-F(a)\). This is the net change.
If the question asks for \(Q(b)\), calculate \(Q(a)+\int_a^b r(t)\,dt\).
Differentiate the antiderivative to check the rate, and verify that the initial and final values differ by the computed integral.
Worked Examples
Worked Example: Net Change from a Linear Rate
A quantity changes at the rate \(r(t)=4t+1\) for \(0\leq t\leq 3\). Find its net change. If its initial value is \(40\), find its value at \(t=3\).
First, integrate the rate. An antiderivative is \(F(t)=2t^2+t\), because $$ F'(t)=4t+1. $$
Evaluate at the interval’s endpoints: $$ \int_0^3 (4t+1)\,dt =F(3)-F(0) =(2\cdot 3^2+3)-(2\cdot 0^2+0) =(18+3)-0 =21. $$ The net change is \(21\).
Add that change to the initial value: $$ Q(3)=Q(0)+21=40+21=61. $$
Check: A quantity function with this rate and initial value is \(Q(t)=2t^2+t+40\). Its derivative is \(Q'(t)=4t+1\), and \(Q(0)=40\). At the endpoint, \(Q(3)=18+3+40=61\). The endpoint values differ by \(61-40=21\), as required.
Worked Example: A Rate on an Interval That Does Not Start at Zero
Suppose \(Q'(t)=3t^2+2\) from \(t=1\) to \(t=3\), and \(Q(1)=75\). Find the net change and \(Q(3)\).
An antiderivative of the rate is \(F(t)=t^3+2t\), since $$ F'(t)=3t^2+2. $$ Evaluate the definite integral using \(1\) and \(3\), the actual interval endpoints: $$ \int_1^3 (3t^2+2)\,dt =F(3)-F(1) =(3^3+2\cdot3)-(1^3+2\cdot1) =(27+6)-(1+2) =33-3 =30. $$ The net change is \(30\).
Now use the initial value: $$ Q(3)=Q(1)+30=75+30=105. $$
Check: The function \(Q(t)=t^3+2t+72\) has derivative \(3t^2+2\). Also, \(Q(1)=1+2+72=75\), and \(Q(3)=27+6+72=105\). The difference is \(105-75=30\), matching the integral.
Worked Example: An Exponential Rate
A quantity has rate \(r(t)=e^t\) from \(t=0\) to \(t=\ln 4\). If \(Q(0)=12\), find the net change and the value at the end of the interval.
An antiderivative of \(e^t\) is \(F(t)=e^t\), because $$ F'(t)=e^t. $$ Therefore, $$ \int_0^{\ln 4} e^t\,dt =F(\ln 4)-F(0) =e^{\ln 4}-e^0 =4-1 =3. $$ The net change is \(3\).
Adding the initial value gives $$ Q(\ln 4)=12+3=15. $$
Check: A quantity function meeting the rate and initial condition is \(Q(t)=e^t+11\). Its derivative is \(e^t\), and \(Q(0)=1+11=12\). At the endpoint, \(Q(\ln 4)=4+11=15\), so the change is \(15-12=3\), exactly the computed integral.
Net Change Is Not the Same as the Endpoint Value
A common error is to calculate the definite integral correctly and then report it as the final quantity. The integral answers a specific question: how much did the quantity change over the interval? It does not, by itself, include the amount that was already present at the start.
The integral calculation is indeed \(21\), but that is the change, not the endpoint value. Use the initial value: $$ Q(3)=Q(0)+\int_0^3(4t+1)\,dt=40+21=61. $$ A check confirms the correction: \(Q(t)=2t^2+t+40\) has the specified derivative, starts at \(40\), and ends at \(61\).
There is also a useful way to distinguish the two kinds of questions. If asked “How much did \(Q\) change?”, report \(Q(b)-Q(a)\), or the integral. If asked “What is \(Q\) at \(b\)?”, include the initial value \(Q(a)\). Both questions use the same rate integral; the requested quantity determines whether the initial value is added.
Choosing the Bounds and Interpreting the Result
The bounds come from the interval over which the rate is being accumulated. If the rate is given from \(t=a\) to \(t=b\), the net change is \(\int_a^b r(t)\,dt\). The lower bound is the start and the upper bound is the end. Using different bounds would describe a different interval, so it would generally produce a different change.
The integral can be evaluated by any appropriate method already established for definite integrals. In the examples, each rate had a straightforward antiderivative. In every case, the endpoint substitution was \(F(b)-F(a)\), not the reverse. After the integral is evaluated, the same relationship applies whether or not an explicit formula for \(Q(t)\) is known.
That last point is useful in applications: a rate function and one initial value are enough to determine the quantity at an endpoint. You do not need to first construct a formula for the quantity at every input. The definite integral supplies its change over the specified interval, and the initial value anchors that change.
Check Your Understanding
For each question, distinguish the net change from the endpoint value when both are relevant.
- If \(r(t)=2t+3\) on \([0,2]\), what is the net change? Show the antiderivative and endpoint substitution.
- If the rate in question 1 applies to a quantity with initial value \(18\), what is the quantity at \(t=2\)?
- If \(Q'(t)=t^2\) on \([1,4]\), write the definite integral that gives the net change and evaluate it.
- For \(r(t)=e^t\) on \([0,\ln 3]\), find the net change. What additional information would be needed to determine the final quantity?
- A student calculates \(\int_a^b r(t)\,dt\) and calls the result \(Q(b)\). What condition on the initial value would make that conclusion valid?
