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?
Best Answer
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 $c=b$).