From Accumulation to an Average
A definite integral describes signed accumulation over an interval. An average value answers a different question: what single number summarizes the function’s values across that interval?
The definition combines these ideas. It uses the definite integral to represent the function’s accumulated values, then divides by the length of the interval. The result is a value with the same units as the function.
The Definition
Let \(f\) be continuous on the closed interval \([a,b]\), where \(a<b\). The average value of \(f\) on \([a,b]\), denoted \(f_{\mathrm{avg}}\), is the definite integral of \(f\) over the interval divided by the interval’s length.
Each part of the formula has a specific role:
- \(f(x)\) is the function whose values are being summarized.
- \([a,b]\) is the interval over which the average is defined.
- \(\int_a^b f(x)\,dx\) is the signed accumulation of the function over that interval.
- \(b-a\) is the interval’s length, which must be positive.
- \(f_{\mathrm{avg}}\) is the resulting average value; it has the same units as \(f(x)\).
The conditions matter. Continuity on \([a,b]\) guarantees that the definite integral exists, and \(a<b\) ensures that the denominator is not zero. An interval of zero length cannot be used in this definition.
Why Divide by the Interval Length?
One way to understand the formula is to approximate the function’s values at equally spaced points. The ordinary arithmetic mean of \(n\) sampled values is the sum of those values divided by \(n\). As the samples become more numerous and evenly spaced across the interval, that mean approaches the average value defined by the integral.
To see the connection, divide \([a,b]\) into \(n\) equal subintervals, each of width \(\Delta x=(b-a)/n\). Choose the right endpoint of each subinterval as a sample point, \(x_i=a+i\Delta x\), for \(i=1,\ldots,n\). The arithmetic mean of the sampled function values is
Since \(n\Delta x=b-a\), rewrite this mean by multiplying each summand by \(\Delta x\) and adjusting the factor outside the sum:
The sum on the right is a right-endpoint Riemann sum for the definite integral. Because \(f\) is continuous, these sums approach \(\int_a^b f(x)\,dx\) as \(n\) increases and the subinterval width approaches zero. Taking the limit therefore gives
Thus, the formula gives the limiting arithmetic mean of equally spaced function samples. The factor \(1/(b-a)\) is essential: it converts accumulated values over the whole interval into a value per unit of interval length.
Worked Example: Identifying Each Part
Worked Example: Identifying Each Part
Suppose \(f\) is continuous on \([1,5]\), and its definite integral on that interval is given as \(\int_1^5 f(x)\,dx=18\). Find the average value and identify the quantities in the definition.
The interval endpoints are \(a=1\) and \(b=5\), so its length is
The given signed accumulation is \(18\). Substitute the interval length and integral into the definition:
Therefore, the average value is \(4.5\). The denominator is \(4\), not the upper endpoint \(5\); the integral supplies the accumulated value, and division by the interval length produces the average.
As a check, multiplying the reported average by the interval length gives \(4.5(4)=18\), the given integral.
Sign and Units
The definite integral in the definition is signed. If positive and negative function values occur, their contributions can cancel. Therefore, the average value can be positive, zero, or negative. It is not defined by adding up only positive contributions.
Units provide another useful check. If \(x\) is measured in seconds and \(f(x)\) in meters per second, then the integral has units of meters: \((\text{meters per second})(\text{seconds})=\text{meters}\). Dividing by an interval length measured in seconds gives meters per second, the same units as the function.
Worked Example: A Function with Negative Average
Suppose a continuous function \(g\) is defined on \([-2,4]\), and the signed integral is given as \(\int_{-2}^{4}g(x)\,dx=-12\). Find its average value.
The interval length is
Apply the definition using the given integral:
The average value is \(-2\). Its negative sign is consistent with the negative signed accumulation. Checking by multiplication, \((-2)(6)=-12\), which recovers the given integral.
Worked Example: Including Units
Worked Example: Including Units
A continuous velocity function \(v(t)\), measured in meters per second, is defined from \(t=0\) to \(t=6\) seconds. Suppose the signed integral over that interval is \(\int_0^6 v(t)\,dt=30\) meters. Find the average value of the velocity function over the interval.
Here \(a=0\) seconds and \(b=6\) seconds. The interval length is \(6-0=6\) seconds. The integral is \(30\) meters, so
The average value is \(5\) meters per second. The units check: the integral’s units of meters are divided by an interval length in seconds, leaving meters per second. Multiplying \(5\text{ meters per second}\) by \(6\text{ seconds}\) gives \(30\) meters, the stated integral.
A Common Mistake: Dividing by an Endpoint
A frequent error is to divide the integral by the upper endpoint instead of by the interval’s length. The denominator is always \(b-a\), even when the interval does not start at zero.
Worked Example: Correcting the Denominator
Return to the given information \(\int_1^5 f(x)\,dx=18\). A mistaken calculation might divide by the upper endpoint:
This is incorrect because the interval runs from \(1\) to \(5\), so its length is \(5-1=4\), not \(5\). The definition gives
The check confirms the correction: \(4.5(5-1)=4.5(4)=18\). Using \(3.6\) instead would give \(3.6(4)=14.4\), not the given integral.
What the Definition Does—and Does Not—Say
The average value is a single number summarizing the function across a specified interval. It depends on both the function and the chosen interval, so changing either can change the average. The integral is not itself the average: it represents signed accumulation, while dividing by \(b-a\) accounts for the interval’s length.
The definition also does not assert, by itself, that the function takes the average value at a particular input. It defines a number through an integral and a length. Whether the function equals that number somewhere is a separate question.
Before applying the formula, check that the function is continuous on the stated closed interval, identify both endpoints, compute the length as upper endpoint minus lower endpoint, and keep the sign and units of the integral. These checks help make the meaning of the result clear, not just its numerical value.
Check Your Understanding
Use the definition to answer each question. Treat any stated integral value as given.
- State the formula for the average value of a continuous function \(f\) on \([a,b]\), and identify what \(b-a\) represents.
- If \(\int_2^8 f(x)\,dx=24\), what is the average value of \(f\) on \([2,8]\)?
- If the signed integral on an interval is negative, must the average value be negative? Explain using the definition.
- A function has units of liters per minute, and its input is time measured in minutes. What units does its average value have?
- Why is dividing by \(b\) incorrect when finding the average value on \([a,b]\) with \(a\ne0\)?
