TITLE: Can the fundamental theorem of calculus be proved without an appeal to mean value or Rolle's theorem or its immediate consequences?
QUESTION [10 upvotes]: I think the answer is in the negative. Here are two of the ways I know. Both of them use
the Mean Value Theorem. The first one use in an indirect way, and the second uses
it more forthrightly. The first proof goes something like this.
Prove that $F(x) = \int_a^x f(t) dt, a \le x \le b$ is a particular anti derivative and if $G$ is an anti derivative that is $G^\prime =(x) = f(x)$. Then apply the Mean Value Theorem to $F-G$ on any interval $(c,d) \subset (a,b)$ to conclude that $F(c) - G(d) = 0$. Now keep $c$ fixed and move $d$ in the interval
$[a, b]$. Use $F(b) - G(b) = F(a)-G(a)$ to conclude $G(b) - G(a) = F(b) - F(a) = F(b) = \int_a^b f(x)dx$.
The second proof, which I prefer, goes something like this:
$G(b) - G(a) = \int dG = (G(b) - G(x_{n-1})) + (G(x_{n-1}-G(x_{n-2}) )+ \ldots (G(x_1) - G(a))$ for any partition $\{a, x_1, x_2,\ldots, x_{n-1}, b \}$ of $[a, b] $. Now
you use the Mean Value Theorem to write replace each term $G(x_k) - G(x_{k-1})$
by $G^\prime(c_k)(x_k - x_{k-1})$ to get a Riemann sum which converges to $\int_a^b f(x)dx $.
I think one has to use some sort of theorem like mvt which gives you global information $f(b) - f(a)$ of $f$ using the derivative which can only provide local information on $f$.

REPLY [7 votes]: As others have said, in the version of FTC in which we agree to integrate / antidifferentiate only continuous functions, the only place where a consequence of MVT occurs is when we want to assert the uniqueness of the antiderivative up to a constant, which -- as N.S. points out -- is equivalent to the Zero Velocity Theorem (ZVT):
Let $f: [a,b] \rightarrow \mathbb{R}$ be differentiable with $f'$ identically equal to zero.  Then $f(x) = f(a)$ for all $x \in [a,b]$.  
It is traditional to prove ZVT by noting that its contrapositive form is that a nonconstant differentiable function has a point at which the derivative is not equal to zero, and that this follows immediately from MVT.  This is the way I proved ZVT in my recently completed Spivak Calculus class.  Just for fun though, let me give an alternate proof using Real Induction (see e.g. page 14 of 
these course notes).
Note that it is no loss of generality to assume that $f(a) = 0$ and show that $f$ is identically zero on $[a,b]$.  For $\epsilon > 0$, let $S(\epsilon)$ be the set of $x$ in $[a,b]$ such that for all $y \in [a,x]$, $|f(y)| \leq (x-a)\epsilon$.  If we can show that $b \in S(\epsilon)$ for all $\epsilon > 0$, then we have that $|f(y)| \leq (b-a)\epsilon$ for all $y \in [a,b]$ for all $\epsilon > 0$, so $f \equiv 0$.  Now:
(RI1) Since $f(a) = 0$, $a \in S(\epsilon)$.
(RI2) Suppose that for some $a \leq x < b$, $x \in S(\epsilon)$.  Since $f'(x) = 0$, there exists $\delta > 0$ such that $|y-x| \leq \delta$ implies $|f(y)-f(x)| \leq  
\epsilon |y-x| \leq \epsilon \delta$.  Thus for all $0 < \delta' \leq \delta$ and $y \in [x,x+\delta']$,
$|f(y)| \leq |f(x)| + |f(y) - f(x)| \leq |f(x)| + \epsilon \delta' \leq (x-a)\epsilon + \delta' \epsilon = (x+\delta'-a)\epsilon$, 
so $[x,x+\delta] \subset S(\epsilon)$.  
(RI3): Similarly, suppose $a < x \leq b$ and $[a,x) \in S(\epsilon)$.  Choose $\delta > 0$ such that $|y-x| \leq \delta$ implies $|f(y) - f(x)| \leq \epsilon |y-x| \leq \ \delta \epsilon$.  For $y \in [x-\delta,x]$, we have 
$|f(y)| \leq |f(y) - f(x-\delta)| + |f(x-\delta)| \leq \delta \epsilon + 
(x-\delta-a)\epsilon = (x-a) \epsilon$.  
I can't think of a situation in which I would rather give this proof than prove the Extreme Value Theorem and use that to to prove the Mean Value Theorem, but I find arguments by real induction to have a certain charming directness to them...