Why the Bounds Matter
A rate tells us how quickly a quantity changes at a particular time. To find the change over a time interval, we accumulate that rate across the interval. The bounds of the integral must therefore match the interval asked about—not simply begin at time zero.
If a question asks what happens from hour 3 to hour 7, the relevant accumulation is from \(t=3\) to \(t=7\). Integrating from \(t=0\) to \(t=7\) would include an earlier period that the question did not ask about. This remains true even when the quantity itself has a known value at time zero.
From Context to Bounds
Before writing an integral, identify what the time variable means. It might represent hours after midnight, days since an experiment began, or minutes since a machine was switched on. The bounds must use that same time scale.
For instance, if \(t\) is measured in hours after midnight, an interval from 9:30 a.m. to noon has bounds \(t=9.5\) and \(t=12\). If instead \(s\) is measured in hours since 9:30 a.m., the same interval has bounds \(s=0\) and \(s=2.5\). Both choices can describe the event correctly, but the rate formula must be expressed in the corresponding variable.
Determine what is changing and what the rate measures, including its units.
Read whether time is measured from a reference point such as midnight or the start of an event.
Express its beginning and ending times using the stated time variable.
Place the start time in the lower bound and the end time in the upper bound, then interpret the signed result.
The Accumulation Rule
Suppose \(Q(t)\) is a quantity and \(r(t)\) is its rate of change, so that \(Q'(t)=r(t)\) throughout the interval from \(a\) to \(b\). If \(r\) is continuous on that interval, the Fundamental Theorem of Calculus gives the change in \(Q\):
The expression on the left is the ending value minus the starting value. It is not the ending value itself. If the starting amount is known, add the integral’s signed result to that starting amount to obtain the ending amount.
Why the Rule Works
Let \(F\) be an antiderivative of \(r\) on \([a,b]\), so \(F'(t)=r(t)\). Since \(Q'(t)=r(t)\) as well, the two functions \(Q\) and \(F\) have the same derivative on the interval. Their difference is therefore constant. Applying the Fundamental Theorem of Calculus gives:
The last equality holds because \(Q-F\) is constant, so its value at \(b\) equals its value at \(a\). Subtracting those equal constant differences leaves \(Q(b)-Q(a)=F(b)-F(a)\). This explains why evaluating an antiderivative at the correct endpoints returns the change in the original quantity.
Worked Example: A Quantity Begins Accumulating After Time Zero
Worked Example: Change from Day 2 to Day 5
A tank’s water amount \(W(t)\), measured in liters, changes at the rate \(r(t)=4t-6\) liters per day, where \(t\) is time in days since monitoring began. Find the change in the amount from day 2 to day 5.
The interval starts at \(t=2\) and ends at \(t=5\), so the change is:
An antiderivative is \(2t^2-6t\). Evaluate it at both bounds and subtract the lower-bound value:
The water amount has a net increase of \(24\) liters from day 2 to day 5. The units follow from multiplying liters per day by days. As a check, differentiating \(2t^2-6t\) gives \(4t-6\), the stated rate. The bounds also match the requested interval: this calculation does not include the period from day 0 to day 2.
Intervals After Time Zero
An interval beginning after time zero is not unusual. It may describe a monitoring period, a machine’s operating window, or a later part of a process. The integral’s lower bound records the beginning of the period being studied; it does not have to be zero.
If \(r(t)\) is positive throughout the interval, the integral is positive and the quantity increases. If \(r(t)\) is negative throughout, the integral is negative and the quantity decreases. If the rate changes sign, positive and negative contributions both affect the signed change. The sign of the final result tells the direction of the net change over the full interval.
Worked Example: A Decreasing Amount over a Later Interval
Worked Example: Volume from Minute 4 to Minute 8
A reservoir’s volume \(V(t)\), measured in liters, changes at the rate \(r(t)=-0.4t+1.2\) liters per minute, where \(t\) is time in minutes since observation began. The volume is \(50\) liters at \(t=4\). Find its change from minute 4 to minute 8, and then find the volume at minute 8.
Use the interval’s endpoints as the bounds. An antiderivative of the rate is \(-0.2t^2+1.2t\):
The volume changes by \(-4.8\) liters, meaning it decreases by \(4.8\) liters. Starting from \(50\) liters at minute 4, the volume at minute 8 is \(50+(-4.8)=45.2\) liters.
Check the rate’s sign on the interval: \(r(4)=-0.4\) and \(r(8)=-2.0\), and the linear rate stays negative between those times. A negative change is consistent with a decreasing volume. Differentiating the antiderivative gives \(-0.4t+1.2\), confirming the integration.
Clock Time and Elapsed Time
A common source of incorrect bounds is mixing a time coordinate with the duration of an interval. A clock time such as 12:00 noon is not the same number as a duration of 12 hours unless the chosen time variable makes it so. Always translate the context according to the variable used in the rate formula.
The next example calculates one interval in two consistent ways. It shows that changing the time origin changes the formula and bounds together, but not the change being measured.
Worked Example: The Same Event with Two Time Variables
Worked Example: Accumulation from 9:30 a.m. to Noon
Water flows into a container at \(r(t)=0.4t+2\) gallons per hour, where \(t\) is hours after midnight. Find the amount added from 9:30 a.m. to noon.
In hours after midnight, 9:30 a.m. is \(9.5\), and noon is \(12\). An antiderivative is \(0.2t^2+2t\):
Thus \(15.75\) gallons are added during the event. Now let \(s\) be the number of hours since 9:30 a.m. Then \(t=s+9.5\), so the rate becomes \(0.4(s+9.5)+2=0.4s+5.8\). The bounds in this variable are \(s=0\) and \(s=2.5\):
The two calculations agree. In the second calculation, the formula was rewritten for elapsed time; using the original formula \(0.4t+2\) with bounds \(0\) and \(2.5\) would mix two different time coordinates. Differentiating \(0.2s^2+5.8s\) gives \(0.4s+5.8\), confirming the second antiderivative.
A Common Bounds Mistake
Suppose the question asks for the change from day 2 to day 5 using the rate \(r(t)=4t-6\). A tempting but incorrect setup is:
This answer includes the interval from day 0 to day 2, so it does not answer the question. The correct bounds are 2 and 5, giving:
The correction is not a change in integration technique; it is a correction to what period the integral represents. Before calculating, state in words what the lower and upper bounds mean.
Interpreting the Result in Context
After evaluating a rate integral, express the result as a signed change and relate it to the quantity. A result of \(+12\) meters means the quantity increased by 12 meters over the interval; a result of \(-12\) meters means it decreased by 12 meters. The ending value is found only if the starting value on that same interval is known: ending value equals starting value plus signed change.
Check these details before reporting an answer:
- Interval: Do the bounds correspond to the requested start and end times?
- Time scale: Are the bounds measured from the same origin as the rate’s variable?
- Sign: Does the sign make sense given the rate’s behavior over the interval?
- Units: Does rate times time give units of the accumulated quantity?
- Meaning: Have you identified the result as a change, rather than automatically calling it the ending amount?
Check Your Understanding
For each question, identify the bounds before doing any calculation.
- A rate \(r(t)\) is measured in liters per hour, with \(t\) in hours after midnight. Write an integral for the change from 1:00 p.m. to 4:00 p.m.
- A quantity has rate \(r(t)=3-t\), with \(t\) in days since observation began. Write and evaluate the signed change from day 1 to day 4.
- A process begins at 10:00 a.m. and ends at 1:30 p.m. If \(s\) measures hours since the process began, what are the bounds? What are they if \(t\) measures hours after midnight?
- In one sentence, explain why integrating a rate from time zero can give the wrong answer when the requested interval starts later.
- If the integral of a rate from \(a\) to \(b\) equals \(-7\) kilograms, what does the sign tell you about the quantity’s change over that interval?
