Why does an antiderivating process give me an area under a curve? I know an area under a such curve $f(x)$ lying at $(a,b)$ can be get as:
$A=\displaystyle\int_{a}^{b}f(x)dx=F(b)-F(a)$
Where $F$ is the antiderivative of $f$ that is:
$F(x)=\displaystyle\int{f(x)dx}$
We also have:
$A=\displaystyle\sum_{i=1}^{\infty}f(x_i)\Delta x_i$
This summation also evaluates what the area is since we're creating an enough large partition on $(a,b)$ and summating the area of infinite retangles lying at this interval. That makes sense because given a curve $f$  it's easy imagine its area on $(a,b)$ being approximated by the area of a few retangles, for example:

So in order to have got the exact area we go $n$ to $\infty$, then we have that using $\displaystyle\sum_{i=1}^{\infty}f(x_i)\Delta x_i$ makes pretty much sense to area evaluating, but we all also know that:
$\displaystyle\int_{a}^{b}f(x)dx=\displaystyle\sum_{i=1}^{\infty}f(x_i)\Delta x_i=F(b)-F(a)$
Where, as I already have told, $F$ is the antiderivative of $f$.
So finally the question is:
Why does define $\displaystyle\int_{a}^{b}f(x)dx$ as $\displaystyle\sum_{i=1}^{\infty}f(x_i)\Delta x_i$ make sense, I mean what does $\displaystyle\int$ have to do with area evaluating? What does that have to do with the infinite summation of the retangles area?
PS: Assume $f$ to be continous and a finite area.
 A: This is an excellent question which I myself have asked quite a while ago. The answer here lies in the following proof:
Let $f(x)$ be a function continuous on the interval $[a,b]$.
Let the integral functional $\int_a^b f(x) \textrm{d}x$ denote the following expression:
$$\lim_{n\to\infty} \sum_{i=0}^{n-1} f\left(a+\dfrac{b-a}{n}i\right) \dfrac{b-a}{n}$$
Let $\Delta x=\dfrac{b-a}{n}$ and $x_i = a+i\Delta x$. This reduces the expression to
$$\lim_{n\to\infty} \sum_{i=0}^{n-1} f(x_i) \Delta x$$
This is the area under $f(x)$ from $a$ to $b$ because it's an infinite sum of rectangles of arbitrarily small length. With this in mind, let us now define $F(x)$ to be $\int_a^x f(t)\textrm{d}t$ for $x\in[a,b]$. We may now begin our proof.
Observe:
$$\dfrac{\textrm{d}F}{\textrm{d}x} = F'(x) = \lim_{h \to 0} \dfrac{F(x+h)-F(x)}{h}$$
$$\dfrac{\textrm{d}F}{\textrm{d}x} = F'(x) = \lim_{h \to 0} \dfrac{\int_a^{x+h} f(t)\textrm{d}t-\int_a^{x} f(t)\textrm{d}t}{h}$$
Based on what we know about areas and how they work, we can say that the area under $f(x)$ between $b$ and $c$ is equal to the area under $f(x)$ between $a$ to $c$ minus the area under $f(x)$ between $a$ to $b$, or $\int_b^c f(x) \textrm{d}x = \int_a^c f(x) \textrm{d}x - \int_a^b f(x) \textrm{d}x$.
This is also true because $\int_b^a f(x) \textrm{d}x=-\int_a^b f(x) \textrm{d}x$ and $\int_a^b f(x) \textrm{d}x + \int_b^c f(x) \textrm{d}x = \int_a^c f(x) \textrm{d}x$.
In any case, our original derivative equation reduces to
$$ F'(x) = \lim_{h \to 0} \dfrac{1}{h} \int_x^{x+h} f(t) \textrm{d}t $$
We may now apply the mean value theorem, which states the following:
$$\exists c \in [a,b] \textrm{ s.t. } f(c) = \dfrac{1}{b-a} \int_a^{b} f(t) \textrm{d}t$$
If we rewrite $F'(x)$ as being $\dfrac{1}{(x+h)-x} \int_x^{x+h} f(t) \textrm{d}t$, we will easily see that we can apply MVT:
$$\exists c \in [x,x+h] \textrm{ s.t. } f(c) = \dfrac{1}{(x+h)-x} \int_x^{x+h} f(t) \textrm{d}t$$
From this, we observe that $ f(c) = F'(x) $ for some $c$ between $x$ and $x+h$.
But as $h \to 0$, $x\leq c\leq x+h$ becomes the inequality $x\leq c\leq x$. And if we are to apply the squeeze theorem, we can say that the only way for this inequality to be true is if $c=x$. Therefore, $f(x) = F'(x)$. $\square$
A: It seems there are a few things that may be causing some confusion, so let us clarify some concepts here.

*

*Area: Technically, the notion of area enclosed by the curve and the $x$-axis is not well-defined a priori. However, there are special types of planar regions for which area is indeed well-defined. Examples of these are rectangles and triangles. For any convex polygon, we know how to compute the are enclosed. Thus, if we can approximate a planar region as a collection of rectangles, then it makes sense to talk about a concept of generalized area for that planar region, which is approximated by the sum of the well-defined areas of those rectangles. This approximation is made mathematically rigorous by defining what is called the integral of a function.

*Integral: The symbol $\int_a^bf(x)\,\mathrm{d}x$ denotes the integral of f, and it is defined as $\lim_{n\to\infty}\sum_{i=1}^nf(t_i)\Delta{x}_i
$. Since this is a definition, there is actually nothing to be proven here. This is because the integral is what we are using to define our notion of generalized area in the first place, as explained in the bullet point above, and the definition rigorously captures the idea that we are approximating the shape of the region enclosed by the function using rectangles.

*Antiderivatives: The symbol $\int_a^bf(x)\,\mathrm{d}x$ denotes the integral of $f$, not an antiderivative of $f$. An antiderivative of $f$ is simply a function $F$ such that $F'=f$. Antiderivatives, by themselves, have very little to do with the concept of generalized area. However, a relationship between antiderivatives and the integral of a function does exist. This relationship is given by the fundamental theorem of calculus.

Fundamental Theorem of Calculus
The fundamental theorem of calculus states two things. The first thing it states is that if $f$ is Riemann integrable on $(a,b)$, then $$\frac{\mathrm{d}}{\mathrm{d}x}\int_a^xf(t)\,\mathrm{d}t=f(x).$$ The second thing it states is that if $F'=f$ and $f$ is Riemann integrable, then $$\int_a^bf(t)\,\mathrm{d}t=F(b)-F(a).$$ This latter statement connects antiderivatives of a function to its integral.
