From an Accumulated Total to an Average Level
A definite integral records an accumulated total across an interval. But sometimes the total alone does not describe the function’s typical level: a total of 12 accumulated over three units of input has a different scale from the same total accumulated over six units.
To define an average level, compare the function with a constant function that accumulates the same total over the same interval. The constant’s height is the function’s average value. This comparison leads directly to the average-value formula.
The integral is a signed accumulated total: portions where \(f\) is below zero contribute negatively. Accordingly, this definition uses the net signed total, not necessarily the sum of all positive and negative contributions taken as positive quantities.
Deriving the Formula by Matching Totals
For a constant function with value \(c\), its graph is a horizontal line. On an interval of length \(b-a\), its accumulated total is the signed area of a rectangle: height \(c\) times width \(b-a\). Thus its integral is \(c(b-a)\).
By definition, the average value \(f_{\mathrm{avg}}\) is the constant that has the same integral as \(f\). Therefore it must satisfy the equation below. Since \(b-a>0\), we can divide both sides by the interval length.
This also establishes that the matching constant exists and is unique: the displayed formula gives a value for it, and any constant with the same total must satisfy the same equation. The condition \(a<b\) matters because it makes the interval length positive and the division valid.
Calculate \(\int_a^b f(x)\,dx\), using the signed integral.
The length of \([a,b]\) is \(b-a\).
A constant height \(c\) accumulates \(c(b-a)\) over the interval.
Set \(c(b-a)=\int_a^b f(x)\,dx\) and divide by \(b-a\).
Why the Formula Is an Average
There is another way to see the same relationship. Divide \([a,b]\) into \(n\) equal subintervals, each of width \(\Delta x=(b-a)/n\). Choose a sample point \(x_i^*\) in the \(i\)th subinterval. The corresponding rectangle has approximate accumulated total \(f(x_i^*)\Delta x\). Adding these rectangle totals gives a Riemann sum for the integral.
Now divide that approximate total by the full interval length \(b-a\). Because \(\Delta x=(b-a)/n\), the result is the arithmetic mean of the \(n\) sampled function heights. In the limit, the Riemann sums approach the integral, so these sampled means approach the integral divided by the interval length.
Since \(f\) is integrable, the tagged Riemann sums converge to \(\int_a^b f(x)\,dx\) as the equal subintervals become arbitrarily narrow. Dividing the sums by the fixed positive number \(b-a\) preserves their limit. Therefore,
This explains the connection to an ordinary arithmetic mean: the integral average is the limit of the means of sampled heights on increasingly fine, equal-width partitions. The constant-function argument gives the exact formula; the Riemann-sum argument explains why dividing by interval length captures an average level.
Worked Examples
Worked Example: A Linear Function on an Interval
Find the average value of \(f(x)=2x+1\) on \([0,3]\), then verify it by comparing accumulated totals.
First find an antiderivative. Since the derivative of \(x^2+x\) is \(2x+1\), an antiderivative is \(F(x)=x^2+x\). Evaluate the definite integral:
The interval length is \(3-0=3\). Divide the accumulated total by this length:
Check by using the constant function with value \(4\). Its integral on this interval is
The original function and the constant function both have accumulated total \(12\) on \([0,3]\), so the computed average value \(4\) satisfies the definition.
Worked Example: A Function with a Negative Contribution
Suppose \(f(x)=3\) on \([0,2]\) and \(f(x)=-1\) on \((2,5]\). Find its average value on \([0,5]\). The value at the single dividing point does not change the integral.
The first piece contributes \(3\) across an interval of length \(2\), giving \(3\cdot2=6\). The second piece contributes \(-1\) across an interval of length \(5-2=3\), giving \((-1)\cdot3=-3\). Thus the net signed total is
The full interval has length \(5-0=5\), so
Check the matching constant: a constant value of \(3/5\) on an interval of length \(5\) has total
The negative contribution reduces the net total, and the matching constant is positive because the total remains positive. Treating the contribution \(-3\) as \(+3\) would instead give the wrong net total.
Worked Example: A Quadratic Function with Physical Units
Let \(t\) measure time in hours and let \(f(t)=t^2\) represent a temperature, in degrees Celsius, on \(1\leq t\leq3\). Find its average value over that time interval.
An antiderivative of \(t^2\) is \(t^3/3\), since \(\frac{d}{dt}(t^3/3)=t^2\). The accumulated integral is
The interval length is \(3-1=2\) hours. Hence
For a check, the constant temperature \(13/3\) degrees Celsius lasting \(2\) hours accumulates
This matches the original integral. The units also check: dividing degree-Celsius hours by hours leaves degrees Celsius, the same unit as the original function’s values.
Signed Totals and Interval Length
The formula applies whether the total is positive, zero, or negative. For a negative total, the average is also negative. For example, a total of \(-8\) over an interval of length \(4\) gives an average of \(-8/4=-2\). If the same total is spread over an interval of length \(8\), the average is \(-8/8=-1\). The latter average is larger, though its magnitude is smaller.
In general, if a fixed nonzero total is divided by a longer positive interval length, the magnitude of the resulting average decreases. For a positive total, the average gets smaller; for a negative total, the average gets larger, closer to zero. Describing the average as simply “smaller” when the interval grows would therefore be incorrect for negative totals.
If the signed total is zero and the interval has positive length, the average is zero. This does not require \(f\) to be zero everywhere: positive and negative contributions can cancel in the integral.
Common Mistake: Forgetting to Divide by Length
Suppose an integral over an interval of length \(3\) equals \(12\). A common error is to report \(12\) as the average value. But \(12\) is the accumulated total, not the constant height that would produce that total over the interval.
The correction can be checked directly: the constant function with value \(12\) accumulates \(12\cdot3=36\), not \(12\). The constant function with value \(4\) accumulates \(4\cdot3=12\), as required.
Why the Average-Value Formula Is Useful
An integral gives a total, while the average-value formula converts that total into a representative constant level over a specified interval. This is useful when comparing functions measured over intervals of different lengths: their totals alone may not be comparable, but dividing each total by its interval length puts the averages on the scale of the function values.
The formula also provides a consistency check. The average must have the same units as \(f\), because the integral’s units include one factor of input units and division by interval length removes that factor. And multiplying the proposed average by the interval length must recover the original signed total.
Check Your Understanding
Use the matching-total definition and the integral-over-length formula to answer these questions.
- If an integrable function has integral \(18\) on an interval of length \(6\), what is its average value?
- Why must the interval length be positive to use the formula as stated?
- A function has signed integral \(-10\) on an interval of length \(5\). What is its average, and what constant function has the same total?
- If the integral has units of meters per second times seconds, what units does the average value have?
- Explain why a zero integral over an interval of positive length gives average value zero, without implying that the function is zero at every point.
