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.
Best Answer
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, $f(x) = F'(x)$. $\square$