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

148 Views Asked by At
  • " 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?

1

There are 1 best solutions below

0
On BEST 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 $$\int_{[a,b]} f(t)\>{\rm d}t=\int_a^b f(t)\>dt\qquad(a<b)\ .$$