TITLE: Why can't the second fundamental theorem of calculus be proved in just two lines?
QUESTION [71 upvotes]: The second fundamental theorem of calculus states that if $f$ is continuous on $[a,b]$ and $F$ is an antiderivative of $f$ on the same interval, then:
$$\int_a^b f(x) dx= F(b)-F(a).$$
The proof of this theorem in both my textbook and Wikipedia is pretty complex and long. It uses the mean value theorem of integration and the limit of an infinite Riemann summation. But I tried coming up with a proof and it was barely two lines. Here it goes:

Since $F$ is an antiderivative of $f$, we have $\frac{dF}{dx} = f(x)$. Multiplying both sides by $dx$,  we obtain $dF = f(x)dx$. Now, $dF$ is just the small change in $F$ and $f(x)dx$ represents the infinitesimal area bounded by the curve and the $x$ axis. So integrating both sides, we arrive at the required result.

First, what is wrong with my proof? And if it is so simple, what is so fundamental about it?
Multiplying the equation by $dx$ should be an obvious step to find infinitesimal area, right? Why is the Wikipedia (and textbook) proof so long?
I have also read that the connection between differential and integral calculus is not obvious, making the fundamental theorem a surprising result. But to me, it seems trivial. So, what were the wrong assumptions I made in the proof and what am I taking for granted?
It should be noted that I have already learnt differential and integral calculus and I am being taught the "fundamental theorem" in the end and not as the first link between the two realms of calculus.
In response to the answers below: If expressing infinitesimals on their own is not "rigorous" enough to be used in a proof, then what more sense do they make when written along with an integral sign, or even in the notation for the derivative? The integral is just the continuous sum of infinitesimals, correct? And the derivative is just the quotient of two. How else should these be defined or intuitively explained? It seems to me that one needs to learn an entirely new part of mathematics before diving into differential or integral calculus. Plus we do this sort of thing in physics all the time.

REPLY [3 votes]: When you see infinitesimals ($dx, dy$) in an expression, it is helpful to think of them as small positive numbers ($\Delta x, \Delta y$), together with the understanding that you are not finished until you take the limit (i.e. where $\Delta x$ goes to zero).  
This is basically what we do in calculus proofs--we work with deltas and then take the limit of the resulting expression.  Before taking the limit, we are just working with numeric quantities.  So, in some cases, there may be common delta factors in the numerator and denominator that are both going to zero at the same rate and can be cancelled out.  If you can get the expression reduced to one where setting the delta values to zero will not lead to a singularity or an indeterminate expression, then you can safely replace them with zero to take the limit. 
Example: 
$$\frac{d}{dx}x^2 = \frac{d(x^2)}{dx}$$
$$= \lim_{\Delta x\to 0}\frac{(x+\Delta x)^2 - x^2}{\Delta x}$$
$$= \lim_{\Delta x\to 0}\frac{(x^2 + 2x\Delta x + \Delta x^2) - x^2}{\Delta x}$$
$$= \lim_{\Delta x\to 0}\frac{x^2 + (2x\Delta x + \Delta x^2) - x^2}{\Delta x}$$
$$= \lim_{\Delta x\to 0}\frac{2x\Delta x + \Delta x^2}{\Delta x}$$
$$= \lim_{\Delta x\to 0}  ( 2x + \Delta x )$$
$$= 2x$$
As long as $\Delta x$ is not zero, you can divide by $\Delta x$, which allows you to factor the common $\Delta x$ from numerator and denominator.
In the remaining expression, $\Delta x$ is just one term of the sum, and now, if it goes to zero, it can simply be dropped.
This may help explain why "multiplying by $dx$" seems to work, since, before you actually take the limit, it is valid to multiply by $\Delta x$.  But at some point, you need to take the limit, and the pivotal question is whether you can do that without having to perform an invalid operation such as dividing by zero.  
Note that you can always turn a false equation, such as $3=5$, into a true one by multiplying both sides by zero, but it doesn't prove anything about the original expression to do that.  So "multiplying both sides by $dx$" does not necessarily accomplish anything meaningful.