Integral definition of derivative After seeing the integral definitions of div, grad, and curl, I'm left to wonder if we can define the regular derivative of a function $f: \Bbb R \to \Bbb R$ as the limit of an integral.  For reference, here's the integral definition of grad
$$\operatorname{grad} f := \lim_{V\to 0} \frac{\oint_S f(x)d\sigma}{V}$$ where $S$ is the boundary of the volume $V$.
In this same vein, is the following definition also correct?
$$\frac{df}{dx}:=\lim_{b\to a} \frac{\int_a^bf(x)dx}{b-a}$$
where $f$ is continuous in $[a,b]\subseteq \Bbb R$.  If so, how could it be proven?

No, it is not. By Lebesgue's differentiation theorem, for any $f\in L^1(\mathbb{R})$ and for almost every $x\in\mathbb{R}$ the following limit exists and it equals $f(x)$:
$$ \lim_{r\to 0}\frac{1}{2r}\int_{|t-x|<r}f(t)\,dt.$$
If $f$ is a continuous function it is absolutely continuous over any compact interval, hence:
$$ \lim_{b\to a}\frac{1}{b-a}\int_{a}^{b}f(t)\,dt = f(a), $$
i.e. any point is a Lebesgue point for a continuous function.

On the other hand, it is also true that for any entire function $f(z)$ we have:
$$ f'(z) = \frac{1}{2\pi i}\oint_{\|w-r\|=r}\frac{f(w)}{(z-w)^2}\,dw, $$
that is Cauchy's integral formula. Moreover, it is very common to define (pseudo)differential operators through particular integrals. See, for instance, the Wikipedia page about fractional calculus.
