What is the proof that anti-derivative gives function = area under curve? For many years now I have thought about this but have not been able to get a clear answer. We all know that $\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ gives us a function we call as the derivative that gives the gradient of the function at $x$. To get the anti-derivative, we can use the $\int$ of the derivative and get back the original $f(x)$.
This part of $\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ has been explained to me many times since high school and is crystal clear. By using this I obtain that derivative of $\sin$ is $\cos$ and derivative of $x^n$ is $nx^{n-1}$. What has never been explained to me is the proof that $\int$ of a function in fact gives the area under $f(x)$ and that also using the Riemann sum. I have always wondered about this and have not been able to get the answer. What is the proof of this?
 A: For simplicity, assume that $f(x)\ge0$ for all $x\in[a,b]$.
Riemann sums give the sum of areas of certain pairwise disjoint rectangles (overlapping boundary lines don't count), which by any reasonable concept of area is the area of the union of these rectangle. And if we can give a meaning to "area under the curve" then this area is "obviously" between the Riemann lower sums (which describe areas of subsets of the area under the curve) and Riemann upper sums (same with supersets). If the function $f$ is Riemann integrable on $[1,b]$ (which just says that lower and upper Riemann sums converge to a fixed number $c$ as the corresponding partitions become finer and finer (the rectangles become thinner and thinner), then we say $\int_a^bf(x)\,\mathrm dx=c$ and as this integral is between all lower and upper Riemann sums, the same number $c$ must be what we call area under the curve.
If you consider $F(x):=\int_0^xf(t)\,\mathrm dt$ for a continuous function $f$, then you will note that $F(x+h)-F(x)=\int_x^{x+h}f(t)\,\mathrm dt$ can be estimated by a simple Riemann sum with a single rectangle. The lower sum is a rectangle of width $h$ and heght $\min\{ f(t)\mid x\le t\le x+h\}$ and the upper Riemann sum is the same with $\max$. Therefore the difference quotient
$\frac{F(x+h)-F(x)}h$ is a number between the min and max of $f$ on $[x,x+h]$. As $h$ becomes smaller, both min and max converge to $f(x)$ if - as we required - $f$ is continuous. In other words, the derivative of $F$ is $f$.
A: I'm not a mathematician nor a calculus practitioner; nor is this a proof. But this is how I've come to terms with anti-derivative being the area under the derivative curve. I don't even know if this is the right way to think about it, but here it is:
Let's say I'm given a function, f(x), which I'm told is the derivative of some unknown function F(x). My task, should I choose to accept it, is to construct the function F(x). I'm additionally given the value of F(k), at some k, as a starting point. These are the only two things I'm given—I don't know how F(x) looks, for example.
Conveniently, let's assume that f(x) is a constant positive function. What this is telling me is that, whatever F(x) represents, it's gaining magnitude at a constant rate as it moves to the right. Also, for convenience, let's picture a discrete version of the derivative:
The discrete derivative function, f(x):
 |
 |
 |-.-.-.-.-.-.
 | | | | | | |

The derivative, f(x), encodes changes to its anti-derivative, F(x). Given a starting value for F(x) at k, I can (re-)construct the anti-derivative by taking discrete steps to the right, and then copying the previous value across and additionally stacking up the change incurred on top of it. Obviously, the previous value carries with it all the differences before it.
The construction of anti-derivative, F(x), starting at some k:
   |
   |
   |          o
   |        o |     Incrementally stacking up changes ("differences")
   |      o | |     up until any given point, to construct the anti-derivative.
   |    o | | |
   | .o.-.-.-.-     ---
   | ??????????      ^
   | ??????????     F(k)
   | ----------      v

 ->| k |<-


Writing these out:
F(k + 0) = given
F(k + 1) = F(k) + f(k + 0)
F(k + 2) = F(k) + f(k + 0) + f(k + 1)
F(k + 3) = F(k) + f(k + 0) + f(k + 1) + f(k + 2)

Thus:
F(k + ni) = F(k) + sum of every f(k + i) for all 0 <= i < n
          = F(k) + Area under a segment of the derivative function.

What this is telling me is that the anti-derivative function only gives the appearance of a real function. Under the hood, at any given point, it's only a "dummy function" accumulating all the differences (obtained by consulting the derivative) until that given point.
Come to think of it: the derivative function itself might be an anti-derivative of some other function, meaning f(x) is also a dummy function taking orders from somewhere else. So it's turtles all the way down.
