The notation of the second fundamental theorem of Calculus I am self studying calculus, and just finished the lesson on the second fundamental theorem of calculus.
the way the theorem is described is:
$\Large\frac{d}{dx}(\int_a^x f(t)\,dt) =  f(x)$
and it was told that the meaning is that the derivative of an integral of a function is the function itself.
I don't get how you can get that from this. the expression that I would think suggests this is:
$\Large\frac{d}{dx}(\int f(x)\,dt) =  f(x)$
so the derivative of an indefinite integral (as oppose to integrating over a range) of a function is the function itself.
another interpretation of the FToC2 I read here, is that it means that the derivative of the functions that gives the area under the curve of a different function is the different function.
this is also something I don't understand how the FToC2 suggests of?
to me, it seems like what this means:
$\Large\frac{d}{dx}(\int_a^x f(t)\,dt) =  f(x)$
is how a very small change in $x$ affects that area under $f(t)$ between $a$ (a constant) and $x$. how do I get from that to the right interpretation?
 A: It should be
$$\frac{\text{d}}{\text{d}x} \int_a^x f(t)\:\text{d}t=f(x).$$
The variable $t$ has no meaning outside the context of the integral.
Concerning your question about indefinite integrals, they may be a convenient abuse of notation if you're trying to solve an ODE, but they are confusing at best and misleading at worst and aren't/shoudn't be a thing. The integral is designed to produce a number given a function (and maybe a set to integrate over, but that one can be done away with), and conditioning students to mechanically add a $+C$ to integrals does way more harm than good in my experience.
A: The two are the same statements just expressed in different ways:
Use your first theorem as $\newcommand{\d}{\text{d}}$
$$\int_a^b f(x)\,\text{d}x=F(b)-F(a)$$
Set $b=x$ and take derivatives:
$$\frac{\text{d}}{\text{d}x}\int_a^xf(x)\,\d x=f(x)-0=f(x)$$
which is the mathematical expression of the quoted sentence of the theorem.
The point is, the antiderivative $F(x)=\displaystyle\int f(x)\,\d x$ attains a constant value for the second limit and is the same function at the first limit; the derivative of constant is zero, which produces the result.

You may wonder why it is expressed so indirectly.
The first and second theorem have many different expressions. For example, an other version of the second theorem states

If $f$ is a function and $c$ is any constant, then there exists an antiderivative $A$ such that $A(c)=0$ and is given by
$$A(x)=\int_c^xf(x)\,\d x$$

So, don't worry about the wording. Just understand the meaning!
Hope this helps. Ask anything if not clear :)
EDIT
You may wish to visit this as well.
A: Suppose $ f $ is continuous on $ I \subset\mathbb{R}$, let $ a,x_{0}\in I $.
Let $ \varepsilon >0 $, there exists some $ \eta >0 $, such that $ \left(\forall y\in\left(x_{0}-\eta,x_{0}+\eta\right)\cap I\right),\ \left\vert f\left(y\right)-f\left(x_{0}\right)\right\vert<\varepsilon$
We have for any $ x\in \left(x_{0}-\eta,x_{0}+\eta\right)\cap I $ : \begin{aligned}\left|\int_{a}^{x}{f\left(y\right)\mathrm{d}y}-\int_{a}^{x_{0}}{f\left(y\right)\mathrm{d}y}-\left(x-x_{0}\right)f\left(x_{0}\right)\right|&=\left|\int_{x_{0}}^{x}{\left(f\left(y\right)-f\left(x_{0}\right)\right)\mathrm{d}y}\right|\\&\leq\int_{\min\left(x,x_{0}\right)}^{\max\left(x,x_{0}\right)}{\left|f\left(y\right)-f\left(x_{0}\right)\right|\mathrm{d}y}\\ &\leq\left|x-x_{0}\right|\varepsilon\end{aligned}
Since $ x_{0} $ is arbitrary, for any $ x_{0}\in I $, we have : $$ \int_{a}^{x}{f\left(y\right)\mathrm{d}y}=\int_{a}^{x_{0}}{f\left(y\right)\mathrm{d}y}+\left(x-x_{0}\right)f\left(x_{0}\right)+\underset{\overset{x\to x_{0}}{}}{\scriptsize{\mathcal{O}}}\left(x-x_{0}\right) $$
Which means $ x\mapsto f\left(x\right) $ is the derivative of $ x\mapsto\int_{a}^{x}{f\left(y\right)\mathrm{d}y} $.
