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antiderivatives, integrals, and units
Posted:
Nov 8, 1997 8:05 AM


I know we had a rather lengthy exchange on antiderivative versus indefinite integral terminology recently, but in working on some simple examples I have discovered something I did not realize before, and I thought others might find it interesting.
Here is such an example:
Suppose the integrand function is given by 30t and the units are miles per hour. Problemfind the net change of position over the one hour time interval [1,2].
By a simple use of the second form of the Fundamental Theorem of Calculus the answer can be computed by choosing an antiderivative expression such as the convenient 15t^2, evaluating it at the endpoints and subtracting the results in the proper order.
All very routine and basic. However, thing about the units. It seems to me that the limits of integration should retain their designation as "hours" as the computation with the antiderivative is carried out. We certainly want the computed difference to have units "miles." Therefore, it seems to me that the antiderivative expression, 15t^2 should have the same units as the integrand expression (miles per hour) and therefore a different unit designation than the integral.
Does this not give another reason for reserving the word "integral" exclusively for the number computed from the Fundamental Theorem (or approximated in any number of ways)? Since in applied context, antiderivatives would retain the unit designation of the rate of change function, the use of the word "integral" in a phrase to describe an antiderivative or a family of antiderivatives could cause confusion.
Sincerely,
Richard Sisley keckcalc@earthlink.net



