Why Interval Lengths Matter
Suppose a quantity varies over time, and you know its average rate during several consecutive time intervals. To find its average rate over the entire period, you need to combine those interval averages.
The key point is that an average over a long interval represents more of the overall period than an average over a short interval. The appropriate combination therefore weights each interval average by the length of the interval. The result is a length-weighted average.
For example, if one rate applies on an interval lasting one hour and another rate applies on an interval lasting three hours, the second rate contributes for three times as long. Taking the ordinary mean of the two rates would treat those intervals as if they had equal lengths.
Definition and Setup
For an integrable function \(f\) on an interval \([a,b]\), where \(a<b\), its average value on that interval is
The integral gives the accumulated signed quantity, and division by the interval length gives the average value per unit of the input variable. If \(f\) is a rate, for instance, its integral over time is a total amount, and its average value has the same units as the rate.
Now divide \([a,b]\) into adjacent intervals with endpoints \(x_0,x_1,\ldots,x_n\), where \(a=x_0<x_1<\cdots<x_n=b\). Write \(\Delta x_i=x_i-x_{i-1}\) for the positive length of the \(i\)th interval, and let \(A_i\) be the average of \(f\) on that interval.
Each interval has a positive length, and all the intervals together cover \([a,b]\) without gaps or overlaps except at their shared endpoints. Those conditions let us add their accumulated quantities to obtain the total accumulated quantity.
The Length-Weighted Formula
The whole-interval average can be found directly from the subinterval averages and lengths:
The formula has a useful interpretation. Each product \(\Delta x_iA_i\) is the accumulated quantity on its subinterval. Add those quantities to get the total, then divide by the total interval length.
Why the Formula Works
The derivation follows from the definition of average and the additivity of the definite integral across adjacent intervals. For each subinterval, rearranging the definition of \(A_i\) gives
The integral over the whole interval is the sum of the integrals over its subintervals. Also, the total length is the sum of the subinterval lengths. Substituting these facts into the definition of the whole-interval average gives a proof:
For each \(i\), the accumulated quantity is \(\Delta x_iA_i\), so additivity gives \(\int_a^b f(x)\,dx=\sum_{i=1}^n\Delta x_iA_i\).
The adjacent intervals partition \([a,b]\), so \(b-a=\sum_{i=1}^n\Delta x_i\).
By the definition of average, \(f_{\mathrm{avg},[a,b]}=\dfrac{\int_a^b f(x)\,dx}{b-a}=\dfrac{\sum_{i=1}^n\Delta x_iA_i}{\sum_{i=1}^n\Delta x_i}\).
This proves the result. It also explains why the weights are interval lengths: multiplying an average by the length recovers the accumulated quantity that belongs to that interval.
Worked Examples
Worked Example: Combining Average Production Rates
A factory produces items at an average rate of \(4\) items per hour during the first \(2\) hours, \(7\) items per hour during the next \(3\) hours, and \(10\) items per hour during the final hour. Find the average production rate over the full \(6\)-hour period.
First calculate the accumulated production during each interval by multiplying its average rate by its duration:
The total production is \(8+21+10=39\) items. The total duration is \(2+3+1=6\) hours. Therefore,
Check by recovering the total production: \(6\text{ h}\cdot 6.5\text{ items/h}=39\text{ items}\), which agrees with the sum of the three interval totals. The combined average is \(6.5\) items per hour.
Worked Example: Unequal Intervals and an Incorrect Unweighted Mean
A variable has average value \(3\) on an interval of length \(1\), and average value \(9\) on the adjacent interval of length \(3\). Find its average over the combined interval.
The two subinterval totals are \(1\cdot 3=3\) and \(3\cdot 9=27\). Thus the total accumulated quantity is \(3+27=30\), and the combined length is \(1+3=4\). The combined average is
The check is \(4\cdot 7.5=30\), matching the accumulated quantity. By contrast, the unweighted mean \((3+9)/2=6\) would imply a total of \(4\cdot 6=24\), not \(30\). It is therefore incorrect for this data.
Worked Example: Finding an Unknown Subinterval Average
A function has average value \(6\) over an interval of length \(8\). On the first \(3\) units of that interval, its average value is \(4\). Find its average value on the remaining \(5\) units.
The total accumulated quantity over all \(8\) units is \(8\cdot 6=48\). The first subinterval contributes \(3\cdot 4=12\). Therefore the remaining interval must contribute \(48-12=36\). If its average is \(A\), its length-weighted total is \(5A\), so
Check by recombining: \((3\cdot 4+5\cdot 7.2)/8=(12+36)/8=48/8=6\), the stated average over the full interval. The unknown subinterval average is \(7.2\).
When the Ordinary Mean Works
If all \(n\) subintervals have the same positive length \(L\), the weighted formula becomes
Thus, equal lengths guarantee that the ordinary arithmetic mean of the subinterval averages gives the combined average. If lengths are unequal, the ordinary mean is not guaranteed to give the combined average for arbitrary subinterval averages. It can still agree in special cases—for example, if each subinterval average is \(3\), both the weighted and unweighted means are \(3\), regardless of the interval lengths. In general, use the length-weighted formula.
Units and a Practical Check
The interval lengths and averages must be interpreted consistently. If a function is a rate measured in items per hour and interval lengths are measured in hours, multiplying them gives items. Dividing the sum of those products by the total number of hours gives items per hour again.
A reliable check is to multiply the proposed combined average by the total interval length. The result should equal the sum of the subinterval totals, where each total is its interval length times its average. If those quantities do not agree, check the interval lengths, the arithmetic, and the units.
The formula applies when the intervals are adjacent and together make up the whole interval, and when the function is integrable there. It combines averages over intervals; it does not replace the need to know each interval’s average or length.
Check Your Understanding
Use interval lengths as weights in each calculation, and check your result by recovering the total accumulated quantity.
- A function has average \(5\) on an interval of length \(2\), and average \(11\) on an adjacent interval of length \(1\). What is the average on their union?
- Why does multiplying a subinterval average by its length give the accumulated quantity on that interval?
- Two adjacent intervals have equal length and averages \(4\) and \(10\). Find their combined average.
- Can the ordinary mean of two averages on unequal-length intervals ever equal the combined average? Explain briefly.
- A total interval of length \(6\) has average \(8\). Its first \(2\) units have average \(5\). What is the average on the remaining \(4\) units?
