Q:

Is my current understanding of the fundamental of calculus correct?

My current understanding:
part 1) means essentially the integral is the inverse of the derivatve
$$\frac {d}{dx} \int f'(x)dx = f'(x)$$
part 2) means essentially we can calculate the integral by evaluating the difference between $a$ and $b$ of the original function$$\int_{a}^{b} f'(x)dx = f(b) - f(a)$$
Is my current understanding correct? If not, a purely algebraic explanation is preferred (e.g. net change in functional value vs area under graph) - I really hate geometric interpretations.

Or am I missing some crucial point? What is really particularly bothering me is that (part 1) is usually written as follows in the textbook - which I don't really understand...
$$\frac {d}{dx} \int_{a}^{x}f(t)dt = f(x)$$
The change of variables between $t$ and $x$ seems redundant to me and I'm not sure if it's because of differences in personal taste or like it actually has an altogether different meaning than my interpretation - and I am actually wrong. (and this failure of understanding why unsettles me deeply)

A:

First of all, before we do anything with the Fundamental Theorem of Calculus, we have to be sure that whatever is going into the integral is continuous.
As you noticed, the first part of FTC is usually stated as 

If $f$ is continuous on $[a,b]$ and if $$g(x)=\int_a^x f(t)\,dt,$$ then for $a<x<b$, $g'(x)=f(x)$.

This seems more complicated then $$\frac{d}{dx}\int f(x)\,dx,$$ but there is a reason for it. The function $\int f(x)\,dx$ means the "the antiderivative of $f$," so saying $\frac{d}{dx}\int f(x)\,dx$ is obvious, and not super helpful. But the function $$\int_a^x f(t)\,dt$$ is a function which gives the area under the curve $f$ between $a$ and $x$. That this function's derivative should be equal to $f$ is not obvious. So that's why this theorem is helpful.
This is a concept that calculus students often misunderstand. The integral and the derivative are not opposites. We treat them like that sometimes because of FTC, and it often works out just fine. We get so used to that thinking that way that it can sometimes cloud our thinking on other types of problems. Just the other day, I asked my calculus students (who had just learned FTC) to solve this problem:
$$\frac{d}{dx}\int_0^{10}e^{e^x}\,dx=?$$
Only three of forty or so got it right on the first try.
On the advice of YoTangoUnLCD, here is the definition$^\ast$ of the integral for a continuous function $f$:
$$\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x_i,$$
where $a=x_0<x_1<x_2<...<x_n=b,$ and $\Delta x_i=x_{i}-x_{i-1}$. This limit (if it exists) is just a number, not a function.
$^\ast$ There's actually more technical definitions of the integral, but this one works for most of the functions you are likely to use.

A:

Without limits, $\int f(x) dx$ just means antiderivative, so there is nothing to prove!
The FTC says if we DEFINE the function $F(x) \colon = \int_a^x f(t) dt$, where $f$ is continuous, THEN $F$ is differentiable and furthermore, $F'(x)=f(x)$. 
The symbol $\int f(x) dx$ is defined as the antiderivative of $f$, so there is no theorem necessary. $\int f(x) dx$ is DEFINED as the (equivalence class of ) function such that $\frac{d}{dx}\int f(x) dx=f(x)$

A:

A good way to understand the FTC is this: sum of of the change in the middle is the total at the edges. Let's unravel that. Think about $f$ as the amount of some fluid. Then $f^\prime$ is the change in the amount of fluid. Ask the question: what is the total amount that the fluid changed? To calculate this, you would just add the amount of change at each point. Integration is really an infinite summation, and so the total amount of change in the interval $[a,b]$ is
$$ TotalChange= \int_{a}^{b} f^\prime(t) dt $$
But how are things changing? If there is a total change in the amount of fluid, it has to be leaving through the boundaries. Things change going over the $b$ boundary by going positive, and change by going over the $a$ boundary by going in the negative direction, and so the other way of calculating the total change is
$$ TotalChange = f(b) - f(a) $$
In total, this means
$$ \int_{a}^{b} f^\prime(t) dt = f(b) - f(a), $$
which is the Fundamental Theorem of Calculus. 
This generalizes. In Multivariable calculus you learn Green's Theorem and Stokes' Theorem. Then you learn Stokes' Theorem on manifolds. It's all the same idea: the sum of the change is equal to the sum along the boundary. We can write this succinctly as
$$ \int_{\Omega} d\omega = \int_{\partial \Omega} \omega $$
but it's really the idea that matters. Notice this is the same thing as before: the boundary of the interval $[a,b]$ is the points $a$ and $b$, so the sum of the change is the sum of the boundary. Mathematicians will say Stokes' theorem on manifolds is the true FTC, but it's all the same. If you understand this idea, then you understand the FTC.
Now for part two: why does that derivative formula work out? Let's just do it algebraically. Let $F$ be a function such that $F^\prime = f$. Then
$$ \int_{a}^{x} f(t)dt = f(x) - f(a) $$
by the FTC. But since $a$ is a constant, $f(a)$ is a constant, so taking the $x$ derivatives of both sides gives
$$ \frac{d}{dx} \int_{a}^{t} f(x)dx = f'(x). $$
What this is saying is that the antiderivative $\int$ is somewhat the same as integration from a constant $a$ to a variable $x$. The end is result may differ by some (arbitrary) constant, but when you take the derivative again, that constant drops out.
Note that I fluffed over all of the details about having $f^\prime$ continuous etc., once you know the idea you can work out the details.

