Averages and integrals

As any student of calculus knows, the integral of a function over a given range is closely related to the average value of the function over that range. One very useful average in studies of energy use is the daily average temperature. Building energy use is often a strong function of outdoor air temperature because about half of the energy required is for heating and air conditioning. Consequently, daily energy use in a building usually correlates well with daily average temperature.

Weather stations report temperature at regular intervals, and the best way to find the daily average is of course to take the average of all readings over a 24-hour period (assuming there are no gaps in the data). In historical studies however, sometimes the only information available is the daily high and low. But believe it or not, averaging the daily high and low temperatures gives a pretty good estimate of the daily average temperature.

Students in calculus classes are taught various methods of approximating an integral. The simplest is the rectangular rule, which in the case of daily temperature is equivalent to taking the average of all of the readings over the day (the integral would then be equal to the average multiplied by the total length of the interval, which is 24 hours).

Averaging the maximum and minimum of the function over the range gives about the same answer as the rectangular rule, but to my knowledge this method of approximating an integral is never taught. It would be interesting to know how “smooth” a curve has to be for this hold. It’s trivially true for a straight line, for example. What about second order curves?

Another interesting method of finding daily average temperature is to measure the temperature at 5:04:18 AM and 6:55:42 PM, and average the two readings. What is special about those two times? The answer is left as an exercise for the reader. Hint: it’s closely related to another method of approximating an integral, one that is occasionally still taught in classes on numerical analysis.

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2 Responses

  1. I suppose averaging the max and min temperatures works because temperatures are approximately sinusoidal. The method would be exact if temperatures were a constant plus a sine with a 24 hour period.

  2. I think it’s more complicated than that. As I said, it’s trivially true for any line: the average height equals the average of the max and the min. It’s almost true for a lot of second-order curves.

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