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? 
 A: 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):


*

*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
$$\int_a^bf(x) dx := F(b)-F(a).$$

*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.
