I can tell why the fundamental theorem of arithmetic and the fundamental theorem of algebra are fundamental, but, indeed, I cannot convince myself why the fundamental theorems of calculus are fundamental.

Any feedback is highly appreciated!

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    $\begingroup$ The word Fundamental is basically wrong. It was fundamental when it was proven. After that the theory evolved and it may become "not that much fundamental" as in the case of the fundamental theorem of Algebra. For Calculus it still stands with its importance. $\endgroup$ – Lolman Jan 1 '15 at 12:15
  • $\begingroup$ Chou: Why is the fundamental theorem of algebra fundamental? $\endgroup$ – Jonas Meyer Jan 2 '15 at 1:00
  • $\begingroup$ @JonasMeyer: Why is the fundamental theorem of arithmetic fundamental? $\endgroup$ – Megadeth Jan 2 '15 at 2:06
  • $\begingroup$ @Chou: Approximately half of your post is the claim, "I can tell why the fundamental theorem of arithmetic and the fundamental theorem of algebra are fundamental," which is why I'm asking you. It also might help inform what "fundamental" means in your question. $\endgroup$ – Jonas Meyer Jan 2 '15 at 2:29
  • $\begingroup$ Nope respectfully, I feel that any reply of mine would cause unnecessary side issues such as finding or inferring from merely a part of my words, which may in turn form a vicious circle. $\endgroup$ – Megadeth Jan 2 '15 at 2:52

I agree with Lolman: sometimes we believe that a theorem is fundamental only because we call it so for historical reasons. In Italy we call it "The Torricelli-Barrow Theorem", probably because we don't think it is so... fundamental!

Apart from jokes, this theorem is important because it establishes a link between to concepts that have little in common: the concept of antiderivative and the concept of (Riemann) integral. Basically, we tend to believe that antiderivatives are (indefinite) integrals only because the fundamental theorem of calculus holds. Just pick one of the few textbooks that write $D^{-1}f$ instead of $\int f$, and you'll realize that definite and indefinite integrals are rather distinct subjects.

But then comes the FToC, and we learn that (for continuous functions) $x \mapsto \int_a^x f$ is an antiderivative of $f$, and, as a consequence, that $\int_a^b f = F(b)-F(a)$ for any antiderivative $F$ of $f$. The FToC should be called The Very Important Theorem of Calculus nowadays, since it has been generalized by measure theory. However, mathematical analysis wouldn't go very far without it.

  • $\begingroup$ Thanks, but $x \to \int_{a}^{x}f$ is an indefinite integral. I also believe that it may be an unfortunate name due to history, like "random variables". $\endgroup$ – Megadeth Jan 1 '15 at 12:28
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    $\begingroup$ Not at all: it is a function of $x$. Please do not confuse a primitive function with the set of all primitive functions (which is the indefinite integral). $\endgroup$ – Siminore Jan 1 '15 at 12:31
  • $\begingroup$ Wow, you may check with, say Apostol's Calculus (2nd edition, p. 120). It is then clear that the function is an indefinite integral. A primitive is an indefinite integral plus a constant. $\endgroup$ – Megadeth Jan 1 '15 at 12:38
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    $\begingroup$ Apostol's terminology looks a bit obsolete. Usually the indefinite integral of $f$ is the set of all functions $F$ with $F'=f$. In other words, $\int f$ is the set of all primitives of $f$. Of course you can call indefinite integral any primitive, but I can tell you that it is not customary nowadays. $\endgroup$ – Siminore Jan 1 '15 at 13:08
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    $\begingroup$ Probably just because Torricelli was Italian.. $\endgroup$ – nbubis Jan 1 '15 at 19:30

The important reason is that it connects differentiation and integration, hence, as a corollary, difficult to calculate definite integral with much easier to calculate indefinite integral.

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    $\begingroup$ Numerical analysis is obsolete then! For all the integral where you don't have an simple expression for the solution... $\endgroup$ – Lolman Jan 1 '15 at 12:15
  • $\begingroup$ @Lolman Certainly, not so optimistic. But definite integral from $x$ is as easy to calculate as integral from $x^{2015}$, which would be horrible only from definition of Riemann integral. (certainly the second is compared with an easy first). $\endgroup$ – Przemysław Scherwentke Jan 1 '15 at 12:18

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