# why the fundamental theorem of calculus is deep? can we think it as a definition?

I failed to understand the deepness of the fundamental theorem of calculus. For the second part, can we think it as a definition to make it just a tautology?

If $f$ is a continuous function on an open interval $I$ and $a$ is any point in $I$, and if $F$ is defined by $$F(x)=\int_a^xf(t)\,dt,$$ then $F'(x)=f(x)$ at each point in I.

I read that it is deep since it relates the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral. But can we just start from this definite integral first to define this as a baseline for anti derivative?

• Then your problem will be to prove that your new definition of the integral gives the same results as "area under the curve" definition (Riemann sums or whatever) you now don't have any connection to anymore. Which will be the FTC in disguise ... – hmakholm left over Monica Sep 2 '16 at 13:21

The deepness of the result comes from the following:

We can define two "operations" on (a certain class of) functions (namely, the functions where both definitions make sense, i.e. Riemann integrable functions):

1. Given a function $f(x)$, we can find a function $F(x)$ (in fact: infinitely many, all of them differing by a constant) such that $F'(x) = f(x)$. We usually denote $$F(x) = \int f(x) dx,$$ and given $a\le b$ two real numbers, define $$\int_a^bf(x) dx := F(b)-F(a).$$
2. Given a function $f(x)$ and two real numbers $a\le b$, we can define the Riemann integral $$\int_a^bf(x) dx$$ as a limit of certain sums on subdivisions of the interval $[a,b]$. I won't bother with the details, as I believe that you (should) know them already.

Now notice that, even though I have denoted them by the same symbols, these two definitions are different a priori. The fundamental theorem of calculus now states:

Theorem: These two definitions of $$\int_a^bf(x)dx$$ are equivalent.

Therefore, it allows us to compute the Riemann integral of a function without having to take a limit over partitions etc., but simply (well, you get what I mean) finding an antiderivative $F(x)$ and computing the difference of its evaluation at the extremal points.

• thanks. I guess i was confused that why do we need the first kind of definition at all if it not related to the second definition? – ahala Sep 2 '16 at 13:35
• -1: The fundamental theorem of the calculus for the Riemann integral does not say these two definitions are equivalent: the existence of an antiderivative does not imply Riemann integrability. The Lebesgue integral suffers from the same defect. You need the Henstock-Kurzweil gauge integral to get this equivalence. – Rob Arthan Sep 2 '16 at 20:39
• @RobArthan Notice that I didn't specify the class of functions we are working on. Of course for the theorem to apply you need the two definitions to make sense... – Daniel Robert-Nicoud Sep 2 '16 at 21:28
• By not saying that the "certain class" of functions has to comprise the Riemann integrable functions, you are marginalising the most important issue, which is that definition (1) is only equivalent to definition (2) if you assume that the process of definition of (2) is applicable in the first place. You don't need to assume Riemann integrability to "make sense" of the first definition, so FTC is profoundly unsatisfying if you come clean about what the "certain class" is. – Rob Arthan Sep 2 '16 at 21:58
• @RobArthan Allright, I'll modify my answer. – Daniel Robert-Nicoud Sep 2 '16 at 22:20