Why Average Concentration Depends on Time
A concentration measured at one instant describes the mixture at that instant. If the concentration changes during an observation, a single value is often needed to summarize the whole interval. The time average provides that summary while giving equal weight to each moment in the interval.
For example, a concentration that is high for only a short time does not necessarily have the same time average as one that stays high throughout. The average depends on both the concentration values and how long the concentration spends at those values.
In this tutorial, \(c(t)\) denotes concentration at time \(t\), and the observation interval is from \(t=a\) to \(t=b\), where \(b>a\). The time variable and concentration units should be stated clearly before calculating.
Definition of Time-Average Concentration
The numerator accumulates concentration over time. Dividing by the duration converts that accumulated quantity into a representative concentration. The result has the same units as \(c(t)\), not the units of the integral.
For instance, if \(c(t)\) is measured in milligrams per liter and \(t\) in hours, then the integral has units of milligram-hours per liter. Dividing by a duration in hours leaves milligrams per liter.
The bounds and the divisor must refer to the same interval. If the observation begins at \(a\) and ends at \(b\), use \(\int_a^b c(t)\,dt\) and divide by \(b-a\), not by \(b\) unless \(a=0\).
Why the Formula Gives an Average
The integral can be understood as the limit of sums over short time intervals. Divide \([a,b]\) into \(n\) equal pieces, each of width \(\Delta t=(b-a)/n\), and choose a time \(\tau_i\) in the \(i\)th piece. The average of the \(n\) sampled concentration values is \(\frac{1}{n}\sum_{i=1}^n c(\tau_i)\). Since \(1/n=\Delta t/(b-a)\), this sampled average can be rewritten as
As the pieces become shorter, the sum \(\sum_{i=1}^n c(\tau_i)\Delta t\) approaches the definite integral \(\int_a^b c(t)\,dt\), by the definition of the Riemann integral for a continuous function. Therefore, the average of the sampled concentrations approaches
This explains why the interval duration is the divisor: the integral accumulates concentration across time, and dividing by the total time produces the average over that time.
There is also a useful interpretation in terms of a constant concentration. The average \(\overline{c}\) is the constant value whose concentration-time integral over the interval equals the integral for the changing concentration:
This equality is an interpretation of the average; it does not mean that the concentration actually stayed constant.
Worked Examples
Worked Example: A Linear Concentration Model
Suppose a concentration changes according to \(c(t)=2+\frac{t}{2}\) milligrams per liter, where \(t\) is measured in minutes. Find its time average from \(t=0\) to \(t=4\) minutes.
The observation duration is \(4-0=4\) minutes. Apply the definition:
An antiderivative of \(2+\frac{t}{2}\) is \(2t+\frac{t^2}{4}\). Evaluate at the bounds:
Now divide by the four-minute duration:
Check: A constant concentration of \(3\ \text{mg/L}\) over four minutes gives an integral of \(3\times4=12\ \text{mg}\cdot\text{min}/\text{L}\), matching the original integral. The average concentration is \(3\ \text{mg/L}\).
Worked Example: A Nonlinear Concentration Model
Suppose \(c(t)=1+t^2\) milligrams per liter, with \(t\) in hours. Find the average concentration over the interval from \(t=0\) to \(t=2\) hours.
The duration is \(2-0=2\) hours. Integrate the model over the interval:
Divide this concentration-time integral by the duration:
Check: Multiplying the proposed average by two hours gives \(\frac{7}{3}\times2=\frac{14}{3}\ \text{mg}\cdot\text{h}/\text{L}\), the integral calculated from the model. The average is therefore \(\frac{7}{3}\ \text{mg/L}\).
Worked Example: A Piecewise Concentration Model
Suppose concentration is measured in milligrams per liter and time in hours. On the interval from \(0\) to \(4\) hours, let
The interval has duration \(4-0=4\) hours. Integrate on each part and add the results:
Thus the integral over the whole observation interval is \(8+9=17\ \text{mg}\cdot\text{h}/\text{L}\). Divide by four hours:
Check: A constant concentration of \(\frac{17}{4}\ \text{mg/L}\) over four hours gives \(\frac{17}{4}\times4=17\ \text{mg}\cdot\text{h}/\text{L}\), the sum of the two integrals. The average is \(\frac{17}{4}\ \text{mg/L}\), or \(4.25\ \text{mg/L}\).
Interpreting the Result and Its Units
A time-average concentration is a summary of the concentration values across a specified observation interval. It is not necessarily the concentration at the start, the concentration at the end, or a value the model actually takes at a measured time.
For a continuous concentration model, the average lies between the smallest and largest concentrations on the interval. Indeed, if \(m\leq c(t)\leq M\) throughout \([a,b]\), integrating gives
Because \(b-a>0\), dividing by the duration shows \(m\leq\overline{c}\leq M\). This is a useful check: an average outside the model’s concentration range indicates an error in the calculation or in the interpretation of the model.
Keep the units visible during the calculation. If time is measured in minutes, the integral carries a factor of minutes; if time is measured in hours, it carries a factor of hours. In either case, dividing by the duration returns the original concentration units. Changing the time unit consistently does not change the physical average concentration.
The integral of concentration alone is a concentration-time quantity. It is not, by itself, a total amount of a substance. A total amount would require additional information, such as the volume associated with the concentration. The average concentration and the total amount answer different questions.
A Common Mistake: Averaging the Endpoints
A tempting shortcut is to take the arithmetic mean of the concentration at the beginning and the end. That gives the time average for a linear concentration model over an interval, but it does not generally work for a nonlinear model.
For the model \(c(t)=1+t^2\) on \(0\leq t\leq2\), the endpoint concentrations are \(c(0)=1\ \text{mg/L}\) and \(c(2)=5\ \text{mg/L}\). The endpoint shortcut gives
But the integral calculation gives the correct time average, \(\frac{7}{3}\ \text{mg/L}\), not \(3\ \text{mg/L}\). The mistake is treating the curved model as though its concentration changed linearly between its endpoints. Use the endpoint shortcut only when linearity on the interval is known; otherwise, calculate the definite integral and divide by the duration.
Practical Calculation Checklist
Write down \(c(t)\), the units of concentration, and the start and end times \(a\) and \(b\).
Calculate \(b-a\), using the same time units used by the model.
Evaluate \(\int_a^b c(t)\,dt\). For a piecewise model, calculate and add the integrals on its pieces.
Divide the integral by \(b-a\), confirm that the result has concentration units, and verify that it lies within the model’s range when appropriate.
Check Your Understanding
Use the time-average formula and show the integral and duration in each calculation.
- A concentration model is \(c(t)=4+t\) milligrams per liter, with \(t\) in minutes. What is its average from \(t=0\) to \(t=2\)?
- For \(c(t)=2+t^2\) milligrams per liter on \(0\leq t\leq1\) hour, write the units of the definite integral and of the time average.
- A concentration model is \(c(t)=6-t\) milligrams per liter on \(1\leq t\leq3\) hours. What duration belongs in the denominator?
- Why does the arithmetic mean of the endpoint concentrations not generally give the time-average concentration?
- If a computed average is greater than every concentration value in the model on the interval, what should you check?
