I put a lot of effort into figuring out how to state my ideas in a way that would make sense to beginning calculus students. From the reactions I'm getting in the newsgroup, though, I now realize that it would be easier for mathematicians if I stated them in a more conventional mathematical language.
So here's the main thing I claim:
Theorem: Let Appl(f,I) be a mapping that associates a real number to each ordered pair (f,I) consisting of a Riemann-integrable function f defined on some reasonable subset of the real line and a compact subinterval I of the domain of f. Suppose further that Appl is additive over disjoint intervals (and over intervals which intersect only in a single point). Let G(x,y) be a continuous function of two variables. Suppose now that Appl(f,I) equals the integral over I of G(x,f(x)) whenever f is a constant function. Suppose also that the mapping Appl is monotonic, i.e. making f larger [resp. smaller] will always result in a larger [smaller] value for Appl(f,I). THEN: Appl(f,I) will equal the integral over I of G(x,f(x)) for all Riemann-integrable functions f.
Proof: Since both sides of the asserted equation are additive over disjoint intervals, the equation is valid not only for constant functions but for step functions. Now f can be obtained as a limit of step functions, and in fact there exist step functions s(x) and S(x) with s(x) <= f(x) <= S(x), and where s and S can be chosen so that the difference between Appl(s,I) and Appl(S,I) (as well as the difference between the corresponding integrals) is arbitrarily small.* The result now follows from the fact that both the mapping Appl and the integral are monotonic. (*I think that this follows from the hypotheses given. To say that s is a (uniformly) small function is to say that there exists a small real number c such that s(x) <= c for all x. It now follows from the stated hypotheses that Appl(s,I) will be small. However I need to deal with the case where there are two step functions s_1 and s_2 such that s_1 - s_2 is small. Unfortunately, I can't assume that Appl respects subtraction. So there might be a glitch here, as far as proving that Appl(s_1,I) - Appl(s_2,I) is small. Actually, it suffices to consider the case where s_1 and s_2 are constant functions. But maybe I need an additional hypothesis on the function G(x,y). )
For a more general theorem, replace "monotonic" by "continuous," where this means that by making f_1 and f_2 sufficiently close to each other (measured in the uniform topology with respect to the interval I), the numbers Appl(f_1,I) and Appl(f_2,I) can be made arbitrarily close to each other.
This theorem will enable one to derive formulas for the most common applications of integration without talking about limits of sums and going through detailed calculations. However there do exist applications (arc length, surface area) which cannot be handled this way. I believe (although without certainty) that somewhat similar ideas can be used in these cases.
-- Trying to understand learning by studying schooling is rather like trying to understand sexuality by studying bordellos. -- Mary Catherine Bateson, Peripheral Visions lady@Hawaii.Edu