Does equating definitionally “ indefinite integral” with “ primitive” turn the fundamental theorem of calculus into a tautology?

• " Indefinite integral" is sometimes equated with " primitive" ( https://en.wikipedia.org/wiki/Antiderivative).

• The fundamental theorem of calculus establishes a link between differentiation and integration, saying, informally, that one is the inverse process of the other.

• So, roughly, the FTC states that every indefinite integral of a function f is also a primitive of f.

• But, if I first definte " indefinite integral of f " as " primitive of f", the FTC appears as a tautology : " every primitive of f is a primitive of f".

My question : (1) should one say that " indefinite integral" and " primitive " actually denote the same function ( or the same set of function) but , in fact, differ conceptually ( I mean, differ as to their definitions); and that (2) the interest of FTC lies in the fact that it shows the extentional identity of these two expressions, in spite of their intensional / conceptual difference?

• Which Fundamental theorem of calculus are you speaking of? There are two of them, for me. – Bernard Feb 11 '20 at 11:47
• The FTC (you should say what form you have in mind) involves definite integrals. What is your definition of the definite integral? That will show whether your form of FTC is a tautology. There is real depth to the connection between differentiation and (definite) integration, so if you choose a formulation of FTC that makes it a tautology then you have merely moved the subtlety of it somewhere else. – KCd Feb 11 '20 at 12:06

1 Answer

Write $$\int_a^b f(t)\>dt:=F(b)-F(a)\ ,$$ where $$F$$ is a primitive ("antiderivative") of $$f$$, and $$\int_{[a,b]} f(t)\>{\rm d}t:=\lim_\ldots\sum_{k=1}^N f(\xi_k)\>|x_k-x_{k-1}|\ ,$$ where the RHS is some limit of Riemann sums. The FTC then says that $$\int_{[a,b]} f(t)\>{\rm d}t=\int_a^b f(t)\>dt\qquad(a