I would like to understand what has to be differential and integral form of the same function, for example the famous equations of James Clerk Maxwell:
How to know where to apply each way? Excuse the ignorance, but always confused me the head.
Edit
Remembering that, I know that the concept of integration and derivation, is something like:
The equations are entirely equivalent, as can be proven using Gauss' and Stokes' theorems.
The integral forms are most useful when dealing with macroscopic problems with high degrees of symmetry (e.g. spherical or axial symmetry; or, following on from comments below, a line/surface integrals where the field is either parallel or perpendicular to the line/surface element).
The differential forms are strictly local - they deal with charge and current densities and fields at a point in space and time. The differential forms are far easier to manipulate when dealing with electromagnetic waves; they make it far easier to show that Maxwell's equations can be written in a covariant form, compatible with special relativity; and far easier to put into a computer to do numerical electromagnetism calculations.
I would think that these three points generalise to any system of differential vs integral forms in physics.
Your confusion lies in failing to recognize that they are exactly the same equations. Take for example Gauss's law:
$$ \vec \nabla \cdot \vec E = \dfrac{\rho}{\epsilon_0}$$ You can see that there $\rho$ is the charge distribution, and in general, can be a function of the position.
Now consider a volume $V$, you can just integrate the density to obtain the total charge in the volume, but you can also integrate the electric field gradient to get a measure of the electric field $$ \vec \nabla \cdot \vec E = \dfrac{\rho}{\epsilon_0} \hskip 30pt / \int_V dV$$
$$ \int_V \vec \nabla \cdot \vec E dV = \int_V \dfrac{\rho}{\epsilon_0} dV $$
Now you can use the Divergence's Theorem on the left side. $$ \vec \nabla \cdot \vec E = \iiint_V \vec \nabla \cdot \vec E = \iint_S E\cdot d\vec S $$
Here $S$ is the surface of the volume and $d\vec S$ is the area element of the surface pointing perpendicularly to the surface out from the volume so the equation final form is
$$ \iint_S E\cdot d\vec S = \iiint_V \dfrac{\rho}{\epsilon_0} dV $$
Which convention's aside is the same equation as shown in Wikipedia. You can do exactly the same to each of the Maxwell's equations using the Divergence's Theorem and Stokes' Theorem
Now, which one is which. If you look at the equations you will see that every equation in the differential form has a $\overrightarrow{\nabla}$ operator (which is a differential operator), while the integral form does not have any spatial differential operator, but it's integrating the terms of the equations.
Finally, as to which one to use, it doesn't matter, because they are the exact same equations and to solve many problems you end up integrating the equations anyway. That being said I like to start from the differential form and if needed integrate the equations.
All this tells you is that the fields satisfies both the inategral and the differential equations. The two are related by the mathematical identities called the divergence theorem and Stokes' theorem.
So which do you apply? Well, which ever one you want! If you run into an integral, you use the integral form, and if you're ever asked for the divergence or curl of the electromagnetic fields, you know exactly what it is, without even trying!
