It's natural that the Fundamental Theorem of Calculus has two parts, since morally it expresses the fact that differentiation and integration are mutually inverse processes, and this amounts to two statements: (i) integrating and then differentiating and (ii) differentiating and then integrating get us (essentially) back where we started.
On the other hand, many people have noticed that the two parts are not completely independent: e.g. if $f$ is continuous, then (ii) follows easily from (i). However, for discontinuous -- but Riemann integrable -- $f$, the theorem still holds, and this is what requires a nontrivial additional argument. See page 8 of
http://math.uga.edu/~pete/243integrals1.pdf (Wayback Machine)
for some discussion of this point.
I can't tell from your question how squarely this answer addresses it. If yes, and you have further concerns, please let me know.
The area under the curve for a continuous function $y = f(x)$ between $x$ and $x + h$ could be computed by finding the area between $0$ and $x + h$, then subtracting the area between $0$ and $x$. Thus the area of the strip would be $A(x+h) - A(x)$.
Now $f(x) \cdot h$ is the linear approximation to the area of the strip. This approximation improves as $h$ becomes smaller, i.e. $A(x+h) - A(x) \approx f(x) \cdot h$. The approximation becomes an equality as $h$ approaches zero in the limit.
Now divide both sides by $h$, then $f(x) \approx (A(x+h) - A(x))/h$. As $h$ approaches zero in the limit, then the RHS becomes the derivative of the area function $A(x)$.
Thus, what the fundamental theorem of calculus says is this: "The derivative of the area under the curve $f(x)$ is the curve $f(x)$". Hence differentiation and integration are inverse operations. This is an informal (non-rigorous) proof. But I hope it gives you an intuitive understanding of the theorem.
The inverse relationship between integration and differentiation can be further elaborated (intuitively) as follows:
What are we essentially doing, when we are differentiating? We are taking the change in $f(x)$, over an interval of length $h$ units, and looking at how this change is distributed over the interval (i.e. we are calculating the change over an interval of unit length). If $f(x)$ was a linear function, then we would just divide this change by the length of the interval. $(\Delta f(x)/h = m)$. (where $m$ is the slope of $f(x)$).
If $f(x)$ is a non-linear function, then, $f'(x)$ = $\lim_{h\to0} \Delta f(x)/h$.
When we integrate a function $f(x)$ over an interval of length $h$, we are accumulating the function $f(x)$ over the length of the interval. If $f(x)$ was a linear function then we would just multiply the change in unit length by the length of the interval and add it to the value of $f(x)$ at the starting point of the interval $f(x)$ = $f(x_0) + \Delta f(x)$, where
$\Delta f(x) = m \cdot h$ (where $m$ is the slope of $f(x)$).
If $f(x)$ is a non-linear function, then, $f(h)$ = $f(x_0) + \int_{x_0}^h f'(x)dx $. (Note that here we are integrating (accumulating) the change in unit length).
Now, division (repeated subtraction) and multiplication (repeated addition) are inverse operations, and provided, that the error between the linear approximation and the actual area or rate of change is an infinitesimal (i.e. it can be made as small as we please), then this inverse relationship holds even in the non-linear case.
The primary reason for the confusion, regarding the inverse relation between differentiation and integration, is the interpretation of integration as measuring an area. One has to understand that this area is only a measure. If $f(x)$ represented an area, and $x$ length, then the 'area' under $f(x)$ would actually represent a volume. The area is thus a measure of the volume.
It would be better, to think of the integral as accumulation of a quantity (in our discussion above, this quantity is the change over an interval of unit length). This definition is consistent, with the interpretation of integral as an area. Let's consider a rectangle of length $4$ units and breadth $3$ units. If $x$ represents length, then, $f(x)$ = $3$. What are we doing, when we are calculating its area? We are accumulating this breadth over an interval of length $4$ units. $A$ = $3 + 3 + 3 + 3 = 3 \cdot 4 $ = $\int_0^4 3 dx$ = $[3x]_0^4$ = $12$ square units.
Now lets differentiate this function. To differentiate, we have to take its accumulated change over any interval and calculate its distribution. But this function doesn't change - $f(x)$ = a constant = $3$. So there's no change, and hence the derivative is equal to zero. $f'(x) =0$.
It is intuitively obvious then, that if we first accumulate a quantity over an interval and then distribute this accumulation over the same interval, then we should end up with our original quantity. Let's turn to the rectangle example again. To calculate the area we accumulate $f(x)$, over an interval of length $4$ (where $f(x)$ = $3$), then we get the area as $12$ square units. Now if we distribute this area over the interval of $4$ units, then what do we get? $12/4$ = $3$ units - ($f(x) = 3$) - which is the distribution of this accumulation(area) over over the interval of $4$ units. (Derivative of $3x$ = $3$). This is the inverse relationship aspect of the FTC.
There may not be any visible connection between calculating an area and calculating a rate of change. But there is an inverse relationship between accumulation and distribution.
Best Answer
Yes, $\int_a^bf(t)\,\mathrm dt$ is a number. But if you change $a$ or $b$ (or both), you usually get a different number. So, $(a,b)\mapsto\int_a^bf(t)\,\mathrm dt$ is a function of $a$ and $b$ (and $f$). And, in particular, for $a$ (and $f$) fixed, $x\mapsto\int_a^xf(t)\,\mathrm dt$ is a function. And the Fundamental Theorem of Calculus states that, if $f$ is continuous, then $F$ is differentiable and $F'=f$.