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If we suppose that we have the problem of finding the derivative for some function, say $f(x)$ at a point $p$, can we use an integral to calculate it?

My incomplete idea is that we can take an integral: $$\int_a^p{f(x) dx}$$

which gives us the area under the function $f(x)$. We then divide the area by the $x$ distance, which gives us the average slope of $f(x)$:

$$\frac{\int_a^p{f(x) dx}}{p-a}$$

We then take the limit as $a$ approaches $p$ to find the slope at that point:

$$\lim_{a \to p}{\frac{\int_a^p{f(x) dx}}{p-a}}$$

Is this correct? If not, how could we similarly simulate a derviate using an integral?

I am also interested in finding the function of the derivative using an integral (at all points, instead of just one).

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If we take that limit, that's going to be exactly $f(p)$, in case $f$ is continuous. – Berci Feb 24 '13 at 2:09
It doesn't work for constant functions. – Marra Feb 24 '13 at 2:11
up vote 3 down vote accepted

You are not so far from the truth...

The fundamental Newton-Leibniz theorem states basically that the integral function $F:=p\mapsto \int_a^pf(x)dx$ of a continuous function $f$ is differentiable, and $F'=f$. (And not $f'$ -- except for $f(x)=a\cdot e^x$).

However, in complex analysis there is a formula of a derivative by line integral around a circle, known as Cauchy's integral formula: $$f^{(n)}(p)=\frac{n!}{2\pi i}\int_\gamma\frac{f(z)}{(z-a)^{n+1}}dz $$ where $\gamma$ is an appropriate circle around $p$ on the complex plane. The proof (say, for $a=0$) basically relies on the fact that the integral over the circle vanishes for all $z^n$ functions $(n\in\Bbb Z)$ except for $n=-1$, where $\int \frac1zdz=\ln z$ and this can have multiple values (because $e^{2\pi i}=1=e^0$), and in fact the integral of $1/z$ over the circle gives $2\pi i$. Then, use this fact and the Taylor series of $f$..

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A note for those at home: this is very similar to the vector calculus version I posted above. (Pay attention to the $n=1$ case, and realize that, when you choose a circular arc, $2\pi (z-a)^2$ is just the area of that circle. Choosing an analytic function and using this construction just makes the limit process unnecessary. – Muphrid Feb 24 '13 at 2:56

There is a general result that says, for instance, the gradient of a scalar field is given by

$$\nabla \psi = \lim_{V \to 0} \frac{1}{V} \oint_{dV} \psi \hat n \, dS$$

And similarly for divergence and curl. Applying this in one dimension, however, yields

$$f'(x) = \lim_{a \to 0} \frac{1}{a} [f(x+a/2) - f(x-a/2)]$$

Which is just a disguised (centered differenced) form of the usual limit definition of the derivative. "Integration" over the boundary is, in this case, just evaluation on the two endpoints of an interval.

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which gives us the area under the function f(x). We then divide the area by the x distance, which gives us the average slope of f(x):

Wrong. Divide the area underneath the graph and you get the function value. Divide again to get the slope.

What you are doing there is basically "I don't want to differentiate, therefore I integrate once and then differentiate twice. But I don't call it differentiation"

The integral in the numerator might evaluate to the difference of the antiderivative. And the limit of two differences does look close to the definition of the derivative, no? From Wikpedia: from wikipedia

I said might because: We can differentiate nearly every function, but Integration is much more complex. Differentiation is work, integration is art ;-)

For example take this Function: $f(x) := \sqrt{1+ln(1+x^2)}$

Derivative is no problem, integral more so.

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