When do two functions differ by a constant throughout an interval (Fundamental Theorem of Calculus) I'm reading the proof of the Fundamental Theorem of Calculus here and I don't understand the following parts (at the bottom of page 2):

I don't know how to conclude that $G(x)-F(x)=C$ for a $x \in [a,b]$.
How do I prove the above statement and does it rely on another theorem not mentioned in this proof?
I tried to figure this out by looking at the definitions of $G(x)$ and $F(x)% but only the definition of $G(x)$ is provided.
 A: You should be able to reach this conclusion by the following:

Prove: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and $f'(x)=0$ for all $x\in (a,b)$, then $f$ is constant.

And if this is a typical sort of analysis book, then you should already have learned the following:

Differentiation distributes across addition: if $f$ and $g$ are differentiable on $(a,b)$, then $f+g$ is differentiable on $(a,b)$, and $(f+g)' = f'+g'$.

With this, the conclusion in the proof is simple: $G'(x)=F'(x)$ on $(a,b)$ implies that $(G-F)'(x) = G'(x)-F'(x)=0$ for all $x\in (a,b)$. Thus $G(x)-F(x)$ must be constant on $[a,b]$. If we call this constant $C$, then $G(x)-F(x)=C$, i.e. $G(x)=F(x)+C$ on $[a,b]$.
A: This is a consequence of the following general fact:

If $f'(x) = 0$ for all $x$ in an interval $[a, b]$, then $f$ is constant on $[a, b]$.

One way to prove this is by the Mean Value Theorem: If there were to exist $x_1$ and $x_2$ in the interval for which $f(x_1) \ne f(x_2)$, there would exist a $c$ between $x_1$ and $x_2$ for which
$$0 \ne \frac{f(x_1) - f(x_2)}{x_1 - x_2} = f'(c)$$
contradicting the fact that $f' \equiv 0$.
A: Theorem 1 (first page) says that
$$\int_a^b F'(t) dt = F(b) - F(a)$$
so we have
$$\begin{array}{c}G(x) = \int_a^x F'(t) dt = F(x) - F(a) \\
G(x) - F(x) = -F(a)\end{array}$$
But $a$ is constant here, so $F(a)$ is constant. Let $C=-F(a)$ to get the result.
