Proving the fundamental theorem of calculus without MVT In this question Proof of the Mean Value Theorem for Integrals, someone proves the MVT for integrals using the fundamental theorem of calculus. I have seen another proof like this as well:
If $f$ is continuous over $[a,b]$ and differentiable over $(a,b)$, there exist a point $c$ in $(a,b)$, such that:
$$f'(c)=\frac{f(b)-f(a)}{b-a}$$
by the fundamental theorem of calculus:
$$\int^b_a f'(x) \, dx = f(b)-f(a)$$
so:
$$f'(c)=\frac{1}{b-a} \int^b_a f'(x) \, dx$$
let $g(x) = f'(x)$:
$$g(c)=\frac{1}{b-a} \int^b_a g(x) \, dx$$
The proof for the first fundamental theorem of calculus, or at least the only one I know, requires the mean value theorem for integrals this can be seen here, so the above would be circular.
Is there another way to prove the fundamental theorem of calculus without this theorem?
 A: The MVT for integrals is not necessary for the first FTC. Suppose $f:[a,b]\to\Bbb{R}$ is Riemann-integrable and is continuous at $c\in [a,b]$. Define $F:[a,b]\to\Bbb{R}$ as $F(x):=\int_a^xf$. We wish to show $F$ is differentiable at $c$ and $F'(c)=f(c)$.
For concreteness, suppose $c\in (a,b)$. Now, given $\epsilon>0$, continuity of $f$ at $c$ shows there is a $\delta>0$ such that for any $t\in [a,b]$
\begin{align}
|t-c|\leq \delta \implies|f(t)-f(c)|\leq \epsilon.\tag{$*$}
\end{align}
So, $0<|h|\leq \delta$ implies
\begin{align}
\left|\frac{F(c+h)-F(c)}{h}-f(c)\right|&=\left|\frac{1}{h}\int_c^{c+h}[f(t)-f(c)]\,dt\right|\\
&\leq \frac{1}{|h|}\int_{\text{min}(c,c+h)}^{\text{max}(c,c+h)}|f(t)-f(c)|\,dt\\
&\leq \frac{1}{|h|}\cdot \epsilon|h|\tag{by $*$}\\
&= \epsilon.
\end{align}
Since $\epsilon>0$ is arbitrary, this shows $F$ is differentiable at $c$ with $F'(c)=f(c)$ (very slight rephrasing proves the cases $c=a$ and $c=b$).
A: The fundamental theorem of calculus is not needed to prove the Mean Value Theorem for Integrals.
Since $f$ is continuous on $[a,b]$, by the extreme value theorem, there exist points $u, v \in [a,b]$ such that for all $x \in [a,b], f(u) \le f(x) \le f(v)$. Intuitively, it is obvious that $$f(u)(b-a) \le \int_a^bf(x)\,dx \le f(v)(b-a)$$
For a rigourous argument from the definition of the Riemann integral, if $P = (\{x_i\}, \{t_i\})$ is any tagged partition of $[a,b]$, then
$$\sum_i f(t_i)(x_i - x_{i-1}) \ge \sum_i f(u)(x_i - x_{i-1}) = f(u)(b-a)$$
Taking the limit as $|P| \to 0$, the integral also satisfies $$\int_a^b f(x)\,dx \ge f(u)(b-a)$$ The other inequality follows similarly.
Thus $$f(u) \le \dfrac{\int_a^bf(x)\,dx}{b-a} \le f(v)$$
As $f$ is continuous on the interval $[u,v]$ (or $[v,u]$, depending on which of $u, v$ is larger), there is some $c$ in that interval with $$f(c) = \dfrac{\int_a^bf(x)\,dx}{b-a} $$ by the intermediate value theorem.
