From a Rate to a Change in the Quantity
A rate describes how a quantity changes. If \(Q(t)\) is a quantity at time \(t\), its instantaneous rate of change is \(Q'(t)\). For example, if \(Q\) measures the amount of water in a tank in liters and \(t\) is measured in minutes, then \(Q'(t)\) has units of liters per minute.
Over a time interval, the quantity has a net change: its final value minus its initial value. The rate may vary during that interval, so the rate at a single instant does not by itself describe the whole change. The average value of the rate gives a single rate that represents the quantity’s net change over the entire interval.
Definition and Conditions
For an integrable rate function \(r\) on an interval \([a,b]\), where \(a<b\), its average value is defined as the definite integral of the rate divided by the length of the interval.
To interpret this average as net change divided by elapsed time, the rate must describe the derivative of the quantity. In particular, suppose \(Q\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(Q'(t)=r(t)\) there, with \(r\) continuous on \([a,b]\). The Fundamental Theorem of Calculus then gives \(\int_a^b r(t)\,dt=Q(b)-Q(a)\), so the identity applies.
The units make the relationship clear. The integral of a rate over time has units of quantity, and dividing by elapsed time gives units of quantity per time. For example, liters divided by minutes gives liters per minute.
Why the Identity Works
The definite integral of a rate over an interval is the net change in the quantity when that rate is its derivative. Dividing that net change by the interval length produces the average rate over the interval. The argument is a direct application of the Fundamental Theorem of Calculus.
By definition, average rate is \(\frac{1}{b-a}\int_a^b r(t)\,dt\).
Since \(Q'(t)=r(t)\), the integral of the rate is \(\int_a^b Q'(t)\,dt\).
\(\int_a^b Q'(t)\,dt=Q(b)-Q(a)\).
The result is \(\frac{Q(b)-Q(a)}{b-a}\), net change per unit of time.
Combining those steps gives the complete derivation:
This identity links two ways to describe the same average: averaging the rate across the interval, or dividing the quantity’s net change by the time elapsed. It does not say that the rate is constant during the interval.
Worked Example: Average Water-Inflow Rate
Worked Example: Average Water-Inflow Rate
Water flows into a tank at a rate \(r(t)=4+2t\) liters per minute, where \(t\) is in minutes after the observation begins. Find the average inflow rate during the first \(3\) minutes, and verify the result using the change in the amount of water.
Step 1: Average the rate over the interval. Here \(a=0\) and \(b=3\), so the interval length is \(3-0=3\) minutes.
Step 2: Check using the quantity’s net change. An antiderivative of \(r(t)=4+2t\) is \(4t+t^2\). Thus the amount added by time \(t\), measured relative to its initial amount, is \(Q(t)=4t+t^2\). Its change over the interval is
Divide the net change by the elapsed time: \(21\text{ liters}/3\text{ minutes}=7\text{ liters per minute}\), which agrees with the integral calculation. A quantity already in the tank at the start would add the same amount to \(Q(0)\) and \(Q(3)\), so it would not alter their difference.
Worked Example: Finding an Average Rate from Measurements
Worked Example: Finding an Average Rate from Measurements
A reservoir contains \(350\) liters of water at 2:00 p.m. and \(428\) liters at 5:00 p.m. Suppose the reservoir’s net rate of change is continuous during this period. Find the average net rate of change of its water amount.
Step 1: Find the net change. The final amount minus the initial amount is
Step 2: Find the elapsed time. From 2:00 p.m. to 5:00 p.m. is \(5-2=3\) hours.
Step 3: Divide the net change by the elapsed time. If \(Q(t)\) is the amount of water and \(r(t)=Q'(t)\) is its net rate of change, the identity gives
Check: An average increase of \(26\) liters per hour for \(3\) hours corresponds to a net increase of \(26\cdot3=78\) liters, matching the measured change. This result is an average over the three hours; it does not mean the reservoir gained exactly \(26\) liters in each hour.
Worked Example: Using an Average Rate to Find Net Change
Worked Example: Using an Average Rate to Find Net Change
During a \(15\)-minute interval, a substance’s temperature has an average rate of change of \(-1.2\) degrees Celsius per minute. If its temperature at the start is \(80^\circ\text{C}\), find its net temperature change and its temperature at the end of the interval.
Step 1: Use average rate times interval length. Rearranging the average-rate identity gives net change as average rate multiplied by elapsed time.
The negative sign indicates a decrease. Add the net change to the starting temperature:
Check: The change from the stated starting and ending temperatures is \(62-80=-18^\circ\text{C}\). Dividing by \(15\) minutes gives \(-18/15=-1.2\) degrees Celsius per minute, the given average rate.
Common Pitfall: Confusing Net Change with Average Rate
The integral of a rate gives net change, not average rate by itself. To obtain the average rate, divide the net change by the interval length.
For instance, in the water-inflow example, the integral over the first \(3\) minutes is \(21\) liters. Calling \(21\) liters per minute the average rate would be incorrect: \(21\) is the net amount added, with units of liters. The correct calculation is
It is also not generally correct to average only the rate’s values at the two endpoints. For example, let \(r(t)=t^2\) on \([0,3]\). The arithmetic mean of the endpoint rates would be \(\frac{0^2+3^2}{2}=\frac{9}{2}\). But the average value over the interval is
The endpoint calculation gives \(9/2\), not \(3\). The interval average depends on the rate across the whole interval, through its integral, rather than just on its two endpoint values.
Using the Identity in Context
Before calculating, identify the quantity and the rate that describes its change. If the quantity is \(Q(t)\), write \(r(t)=Q'(t)\). Then identify the interval endpoints, determine the net change \(Q(b)-Q(a)\), and divide by \(b-a\) to find the average rate. Alternatively, integrate \(r(t)\) over the interval and divide by its length.
Check the units and signs in the final result. A positive net change over a positive time interval gives a positive average rate; a negative net change gives a negative average rate. The rate’s units should be quantity per unit time, while the integral’s units should be quantity.
This relationship is useful whenever a varying rate describes a measured quantity. It lets you translate between a rate accumulated across time and an observed difference between the quantity’s final and initial values. The average summarizes the interval as a whole; it need not match the rate at either endpoint or remain constant at any time.
Check Your Understanding
Use the relationship between average rate and net change to answer the following questions.
- A quantity satisfies \(Q'(t)=r(t)\) on \([a,b]\). Write the formula for the average value of \(r\) using \(Q(a)\), \(Q(b)\), \(a\), and \(b\).
- A container gains \(45\) liters over \(5\) hours. What is its average net rate of change, including units?
- A rate is measured in meters per second and is integrated over a time interval measured in seconds. What units does the integral have?
- For \(r(t)=3t\) on \([0,4]\), compute the average value of \(r\) using the integral definition.
- Why can the arithmetic mean of the endpoint rates fail to equal the average rate over the full interval?
