Replace a varying function by an equal-total constant

Try

A changing temperature is recorded over three hours. Imagine replacing it by a constant temperature that gives the same area under its time graph. The rectangle’s width is three hours. What height makes its area match the original? This height defines a time average, not necessarily the midpoint of two endpoint readings.

Predict

For f(x)=x² on [0,3], predict whether the average is 9/2, the mean of the endpoint values, or something smaller. The convex curve lies below its endpoint secant over the interval, so the endpoint-based rectangle suggests an overestimate. Use integration to settle the value.

Experiment

The integral is 9 and the interval length is 3, giving average value 3. Draw the parabola and a rectangle of width 3 and height 3; their areas match even though their shapes differ.

Function on its interval	Integral	Interval length	Integral average
x² on [0,3]	9	3	3
2x+1 on [1,4]	18	3	6
Constant 7 on [2,5]	21	3	7

For the linear example, endpoint averaging also gives 6. Explain geometrically why a trapezoid formula is exact for a line, and why that special case should not be generalized to every continuous curve.

Understand

For an integrable function on [a,b] with a<b, the average value is (1/(b−a))∫ₐᵇf(x)dx. The factor divides by interval length, not by the number of measurements. If f is temperature and x is hours, the integral has degree-hour units; dividing by hours returns degrees. The average has the same units as the original function.

If f is continuous, it attains this average somewhere in the interval. The extreme value theorem provides minimum m and maximum M. Integral comparison bounds the average between m and M. The intermediate value theorem then gives at least one c with f(c) equal to the average. The point need not be the interval midpoint and need not be unique. For x² on [0,3], it is c=√3, not 1.5.

Continuity is a sufficient hypothesis for that attainment conclusion. A step function taking value 0 on [0,1/2) and 1 on [1/2,1] has average 1/2 but never takes that value. It is integrable, so its average exists; the continuous mean-value theorem simply does not apply. Distinguish existence of an average from existence of an input attaining it.

For T(t)=20+6t−t² on [0,3], integration gives 78 degree-hours and time average 26 degrees. Endpoint readings are 20 and 29, averaging 24.5. The difference is caused by the nonlinear variation over time, not by an arithmetic inconsistency. A finite set of readings would require an interpolation or sampling model before either method could claim a physical time average.

Changing the weighting changes the question. Equal-time averaging differs from equal-distance averaging on a trip with changing speed. Similarly, an average over space need not equal an average over sampled locations if sampling is uneven. This lesson uses uniform interval weighting; weighted integrals require the weighting function to be stated and normalized separately.

Apply

Show the integral, interval length and resulting units in each calculation. In the existence proof, name both EVT and IVT and their roles. For the discontinuous example, demonstrate that the average lies outside the function’s range rather than claiming that a discontinuous function has no average.

Continue

The next lessons expand the range of integrands you can handle symbolically. Integration by parts reverses the product rule, while substitution reversed the chain rule.