Does that mean that ∫f(x)dx is f.(dx, dx, …) = f(x_0)·dx + f(x_1)·dx + f(x_2)·dx + … for all x in the domain?No, though it is true that the integral with respect to x, as an operation, is the limit of that summation as dx approaches 0. But the point wasn't about any particular numerical or symbolic manipulation we could do (this was a graduate-level calculus class, after all; we all knew how to actually integrate things).
Generally, when you learn inner products of functions, you learn the definition
f·g = ∫(f(x)g(x))dx where the concatenation there of f(x) and g(x) represents scalar multiplication. We all know how integration works, so this becomes how we define inner products.
My professor's point was to reverse the primacy there. We have a sense from vector operations of what inner products are; that can inform our intuition of what an integral is. That is, rather than saying "I know how to do an integral so I can now take an inner product of two functions", say "I have an intuition of what an inner product is, that is, the projection of one vector onto another to form a scalar, and that should inform my intuition of what an integral is".
The larger motivation for the whole talk was introducing Clifford algebras and the symmetry between dot product generalized to inner product opposed by outer product on the one hand, and cross product generalized to wedge product opposed by interior product on the other hand.
And, just to finish the mathjerking, the whole point of the course was to get to:
(δΩ)∫ω = (Ω)∫dω
ie, the most general case of Stokes' theorem. But that takes a lot of sussing out of what the differential operator d actually is.