Flawed Proof of Fundamental Theorem of Calculus? I am currently learning calculus, have not learned much about antiderivatives, and nothing about indefinite integrals. I came by the 2nd Fundamental Theorem of Calculus:
$$\int_a^b f(t)dt = F(b) - F(a)$$ where $F$ is the antiderivative of $f$. I was trying to understand why this theorem works and I came up with a possible proof, albeit informal, of the above theorem. Assuming (1) the antiderivative, F(x), is equivalent to $\int_c^x f(t)dt$ where $c$ is any constant and (2) $f(t)$ is continuous and therefore integratable everywhere:
$$F(b) - F(a) = \int_c^b f(t)dt - \int_c^a f(t)dt = \int_c^b f(t)dt + \int_a^c f(t)dt = \int_a^b f(t)dt$$ I think that this proof might be too simple, possibly neglecting important facts. My questions are: (1) Is the proof above correct or incorrect and why? (2) Is it true that any antiderivative $F(x) = \int_c^x f(t)dt$, given $f$ is continuous?
 A: What you have written is true, but it is not the fundamental theorem of calculus. You've just rearranged the definition
$$
F(x) = \int_c^x f(t) dt
$$
using integral properties.
The First Fundamental Theorem of Calculus
If we define $F(x) = \int_c^x f(t) dt$, then one result worthy of the name "fundamental theorem of calculus" says
$$
F'(c) = f(c)
$$
whenever $f$ is continuous at $c$. Note that we need $f$ to be continuous at $c$ or else the result can fail to be true.
Why is this true? You've made the key insight.
$$
F'(c) = \lim_{h \to 0} \frac{F(c + h) - F(c)}{h} = \lim_{h \to 0} \frac{\int_c^{c + h} f(t) dt}{h}
$$
To finish the proof, you just need to explain why that last limit is equal to $f(c)$. This is where you will have to use continuity at $c$.
The Second Fundamental Theorem of Calculus
The other result that goes by the name "fundamental theorem of calculus" says that if we have a function $F$ such that $F' = f$ and $f$ is integrable on $[a,b]$, then
$$
\int_a^b f(t) dt = F(b) - F(a)
$$
The classic proof uses a completely different trick. We use the mean value theorem with $F$ and $f$ and the integrability of $f$ to set up a telescoping sum.
People will often give a simpler proof which is more properly a corollary of the first fundamental theorem. If we assume $f$ is continuous on $[a,b]$, then we can argue as in RafGaming's answer to prove the same result. But note that the mean value theorem proof, though more complicated, proves the result without making any assumptions about the continuity of $f$.
Comparing the Two Theorems
Suppose $f$ is integrable on $[a,b]$ and $c \in (a,b)$
Schematically, in the first theorem, we have
$$
\left[ F(x) = \int_c^x f(t) dt  \quad \& \quad \lim_{x \to c} f(x) = f(c) \right] \implies F'(c) = f(c)
$$
Note, again, that we need continuity of $f$ at $c$. The statement $F(x) = \int_c^x f(t) dt \implies F'(c) = f(c)$ is not true.
In the second theorem, we have
$$
F' = f \quad \implies \int_a^b f(t) dt = F(b) - F(a)
$$
Though similar looking, these results are not the same. In fact, they are not even exactly converses. They're two distinct results that both formalize the intuition that "differentiation undoes integration and vice versa."
A: $\int_{a}^{b}f(t)dt$ expresses the area below the curve $y=f(t)$ between the lines perpendicular to the $x$-axis at points $P(a, 0), Q(b, 0)$. You can define a function $G(t)$ to denote the area between points $P(a, 0)$ and $Q(x, 0)$: \begin{equation*}G(t)=\int_{a}^{x}f(t)dt\end{equation*}
Also let $F(t)$ be an antiderivative of $f(x)$ ($F'(x)=f(x)$). Now, if you remember the 1st fundamental theorem of calculus, you'll know that two andiderivatives of the same function only differ by a constant. Thus, $G(x)=F(x)+C$. If $x=a$, $G(a)=0$ so $F(x)=-C$. If $x=b$, $G(b)=F(b)+C=F(b)-F(a)$. Hence \begin{equation*}\int_{a}^{b}f(t)dt=F(b)-F(a)\end{equation*}
