The average value of f on [a,b] is (1/(b−a))∫f. Why is that the right definition, when an average is normally a sum divided by a count and there are infinitely many values here?

D. Because the integral plays the sum's role and the length the count's.

Sample at n points and average; multiply and divide by the spacing and it becomes (1/(b−a))·Σf(xᵢ)Δx, a Riemann sum. So it is not an analogy, the integral does the summing and (b−a) does the counting. The Mean Value Theorem for Integrals adds that a continuous function actually attains its average somewhere, which is the derivative statement on the integral side.