Reading a Rate Table Carefully
A rate table can contain two kinds of information: values of a rate at particular times, and definite integrals of that rate over specified intervals. These entries are not interchangeable. A rate value describes what is happening at one time; an interval integral gives the exact accumulated change across an interval.
When a table supplies the interval integrals, you can use them to reconstruct changes in a quantity and, if one quantity value is known, find other values. The method does not require guessing what the rate did between listed times.
In this tutorial, an interval integral written in a table is treated as supplied information. A rate value such as \(r(2)\) is a point value; it does not, by itself, specify the integral over a time interval around \(2\).
How Interval Integrals Fit Together
Definite integrals are additive across adjacent intervals. If \(a\), \(c\), and \(b\) are in that order, then the accumulation from \(a\) to \(b\) is the sum of the accumulations from \(a\) to \(c\) and from \(c\) to \(b\). This lets us combine supplied entries or subtract known pieces from a larger interval.
Here is why this property works. If \(R\) is an antiderivative of \(r\) on the interval, the Fundamental Theorem of Calculus gives \(\int_x^y r(t)\,dt=R(y)-R(x)\). Therefore,
The intermediate terms cancel. This is the mathematical reason interval accumulations can be joined without knowing the detailed rate values between the endpoints. In a table problem, you can apply the same additivity to the definite integrals the table supplies.
A Reliable Table Method
Before calculating, identify what each entry represents. A column labelled \(r(t)\) gives rate values. An entry labelled \(\int_a^b r(t)\,dt\) gives accumulation over the entire stated interval. Keep the interval endpoints attached to each integral; two numerical entries cannot be added or subtracted correctly unless their intervals line up.
Write the exact lower and upper bounds for the change you need.
Combine adjacent integrals, or subtract known subintervals from a larger supplied integral.
Add the signed integral to the quantity at the lower endpoint to obtain the quantity at the upper endpoint.
Confirm that the interval pieces reconstruct the target interval and that the units and sign make sense.
Worked Example: Reconstructing Quantities from Adjacent Intervals
Worked Example: Reconstructing Quantities from Adjacent Intervals
A tank contains \(40\) liters at time \(t=0\). The table lists the filling rate \(r(t)\), in liters per minute, and the exact integral over each interval ending at the listed time. Find the amount in the tank at \(t=2\), \(t=5\), and \(t=8\) minutes.
| Time \(t\) (min) | Rate \(r(t)\) (L/min) | Supplied interval integral (L) |
|---|---|---|
| 0 | 3 | — |
| 2 | 4 | \(\int_0^2 r(t)\,dt=7\) |
| 5 | 2 | \(\int_2^5 r(t)\,dt=12\) |
| 8 | 5 | \(\int_5^8 r(t)\,dt=9\) |
Start with the known quantity \(Q(0)=40\) liters. From \(0\) to \(2\) minutes, the change is \(7\) liters:
From \(2\) to \(5\) minutes, the change is \(12\) liters. Thus,
From \(5\) to \(8\) minutes, the change is \(9\) liters. Therefore,
Check: The change from \(0\) to \(8\) minutes is \(7+12+9=28\) liters. Adding it to the initial \(40\) liters gives \(40+28=68\) liters, agreeing with \(Q(8)\). Each rate integral has units \((\text{L/min})(\text{min})=\text{L}\). The rate values at the listed times are not needed for these exact calculations because the interval integrals are already supplied.
Worked Example: Finding a Missing Interval Integral
Worked Example: Finding a Missing Interval Integral
A quantity \(Q\), measured in milligrams, changes at rate \(r(t)\), measured in milligrams per hour. The table supplies \(\int_1^5 r(t)\,dt=14\), \(\int_1^2 r(t)\,dt=3\), and \(\int_4^5 r(t)\,dt=2\). Find \(\int_2^4 r(t)\,dt\). If \(Q(1)=50\) mg, also find \(Q(2)\), \(Q(4)\), and \(Q(5)\).
Break the large interval \([1,5]\) into the three adjacent pieces \([1,2]\), \([2,4]\), and \([4,5]\). Additivity gives
Substitute the known values and solve for the missing integral:
Now use the known starting quantity. From \(t=1\) to \(t=2\), the change is \(3\) mg, so
The change from \(2\) to \(4\) is \(9\) mg. Hence
Finally, the change from \(4\) to \(5\) is \(2\) mg:
Check: The three changes sum to \(3+9+2=14\) mg, exactly the supplied change over \([1,5]\). Also, \(Q(5)-Q(1)=64-50=14\) mg. The units are milligrams because each integral combines a rate in mg/hr with time in hours.
Worked Example: Reversing the Bounds
Worked Example: Reversing the Bounds
A quantity \(Q\), measured in units, satisfies \(Q(2)=20\). A table gives \(\int_2^6r(t)\,dt=11\) and \(\int_4^6r(t)\,dt=5\). Find \(Q(4)\) and the change in \(Q\) from \(t=6\) back to \(t=2\).
First, split the supplied integral at \(t=4\):
Substitute the table values and isolate the missing piece:
Thus the quantity at \(t=4\) is
The full change from \(2\) to \(6\) is \(11\), so \(Q(6)=20+11=31\). Reversing the bounds reverses the sign:
Check: The change from the value at \(t=6\) back to the value at \(t=2\) is \(Q(2)-Q(6)=20-31=-11\), matching the reversed integral. The sign records the direction of the change between those endpoints.
What Rate Values Alone Do Not Tell You
A rate table might list only \(r(a)\) and \(r(b)\). Those two point values do not determine the exact integral \(\int_a^b r(t)\,dt\). Many different functions can have the same endpoint values while having different behavior between them, and therefore different integrals.
For example, suppose a rate has value \(4\) at both endpoints of the interval \([0,3]\). One possible rate is \(r_1(t)=4\), whose integral is
Another possible rate with the same endpoint values is \(r_2(t)=4+t(3-t)\), since \(r_2(0)=r_2(3)=4\). Its integral is
The same endpoint rate values are consistent with integrals of \(12\) and \(16.5\). Therefore, neither endpoint rates nor other isolated rate samples can substitute for an interval integral when an exact accumulation is required. An estimation method, if requested, produces an estimate under its stated method; it does not turn unprovided information into an exact integral.
Why the Table’s Interval Labels Matter
Every definite integral belongs to a particular interval. For example, \(\int_1^4 r(t)\,dt\) and \(\int_4^7 r(t)\,dt\) can be added to obtain \(\int_1^7 r(t)\,dt\), because the intervals meet at \(4\). But an entry for \(\int_1^4 r(t)\,dt\) cannot be treated as the change from \(1\) to \(5\) unless the interval from \(4\) to \(5\) is also accounted for.
Before combining entries, write the interval decomposition explicitly. This simple habit prevents both missing pieces and double-counting. It also makes a final check straightforward: the pieces should cover exactly the requested interval, with no gaps and no overlaps that count a segment twice.
In a contextual problem, report the quantity with its units and identify the relevant time. The integral gives the change over the interval; the quantity at the endpoint also depends on the value at the starting time. Without such a starting value, interval integrals can still determine changes, but they do not by themselves specify an absolute quantity value.
Check Your Understanding
For each question, keep the interval endpoints attached to the supplied information and distinguish a rate value from an interval integral.
- If \(\int_0^2r(t)\,dt=5\) and \(\int_2^6r(t)\,dt=9\), what is \(\int_0^6r(t)\,dt\)?
- If \(\int_1^7r(t)\,dt=18\) and \(\int_1^3r(t)\,dt=6\), what is \(\int_3^7r(t)\,dt\)?
- If \(Q(4)=30\) and \(\int_4^9r(t)\,dt=-8\), what is \(Q(9)\)?
- If \(\int_2^5r(t)\,dt=7\), what is \(\int_5^2r(t)\,dt\)?
- Why do the entries \(r(0)=3\) and \(r(4)=3\) fail to determine the exact value of \(\int_0^4r(t)\,dt\)?
