I apply formulas for stray capacitances etc....

Typically simple things like transmission line equivalent L and C, comparing impedances, that sort of thing. I doubt I'll ever find a place to use an integral in layout as such, though I have done a few in nearby fields (thermal conduction).
E&M field problems normally have way too many boundary conditions, so that if you cannot fit it into an already solved case (e.g., microstrip), it's somewhere between ridiculous to impossible to solve analytically (i.e., on paper), and you need a simulator.
The two examples that I have integrated, are: the temperature distribution along an evenly heated bar cooled from one end, and a finite panel of some radius, with an isothermal source of some radius in the middle, and uniform cooling capacity over the surface of the panel.
The first case is easy: it's a separable differential equation, and the solution is a parabola, with the vertex at the far (uncooled) end of the bar, and the steepest side at the cooled end (which makes sense: the temperature drop across any given segment of the bar is the total power hanging off that end, which is linear with position, so the temperature distribution is quadratic with position).
The second case is hard, but in an easy sort of way. The double dependency of heat flow with temperature, in a circularly symmetrical system, leads to a differential equation with solutions of the Bessel functions. IIRC, it was the first half-cycle of the normal Bessel function, with the origin in the center, and the first zero being the periphery of the disk. The shape of this function is oscillatory, but in an unpredictable way (the zeroes are irregular), not quite like a sine wave. This makes sense, as around the origin, there's not much surface area to remove heat, so the temperature remains high; the fact that the heat is spreading out over ever-greater circumferences means the thermal conductivity is good (except for in the very center, where you can imagine a pinpoint source might get very hot indeed, so it's important to have a wide heat-spreading area). As you go out in radius, eventually the area becomes considerable, and temperature starts falling off, and eventually it reaches ambient at infinity, or some nonzero value at the edge of a finite disk. This gives a length scale where most of the heat is dissipated, and diminishing returns are had for larger panels. The diameter of the hot spot is given by the lateral heat spreading (thermal conductivity) of the panel and the heat dissipation rate (the air flow, as it were).
Plugging in typical values for, say, a copper clad PCB in still air, we get a radius of 2-4cm and, for typical semiconductors and ambient temperatures, a power dissipation capability of about 5W.

Tim