The instantaneous frequency of a signal \$A\cos(\phi(t))\$ where \$\phi(t)\$ is an arbitrary function of time is defined as the derivative of \$\phi(t)\$ if you want to measure frequency in radians per second and as \$\frac{1}{2\pi}\$ times the derivaive_ of \$\phi(t)\$ if you want to measure frequency in Hertz. Of course, in the common case of a
fixed frequency this corresponds to the familiar
\$\phi(t) = \omega_c t+\phi_0 = 2\pi f_c t + \phi_0\$.
The standard definition of a frequency-modulated signal is one in which the
deviation of the instantaneous frequency (at time \$t_0\$, say),
from the carrier frequency is proportional to the value \$x_m(t_0)\$ of the modulating signal \$x(t)\$ at time \$t_0\$. The constant of proportionality is denoted by
\$f_{\Delta}\$ in your notation: a \$1\$ volt signal creates a deviation of \$f_{\Delta}\$
Hz. Thus, if \$A\cos(\phi(t))\$ is the FM signal, then we have that
$$ \left.\frac{\mathrm d}{\mathrm dt}\phi(t)\right|_{t=t_0}
= 2\pi f_c + 2\pi f_{\Delta} x_m(t_0)
$$
so that the deviation of the instantaneous frequency \$f_c + f_{\Delta} x_m(t_0)\$
from the carrier frequency \$f_c\$ is \$f_{\Delta} x_m(t_0)\$, just as we want it to
be. It then follows from the fundamental theorem of calculus that
$$\phi(t_0) = \int_{0}^{t_0}2\pi f_c + 2\pi f_{\Delta} x_m(t_0)\, \mathrm dt
= 2\pi f_c t_0 + \int_{0}^{t_0} 2\pi f_{\Delta} x_m(t_0)\, \mathrm dt$$
or, with a slight change in notation, the FM signal can be expressed as
$$A\cos\left(2\pi f_c t + \int_{0}^{t} 2\pi f_{\Delta} x_m(\tau)\, \mathrm d\tau\right)$$
the way you have it. Note that \$A\$ is the amplitude of the FM signal and is
fixed; it is the frequency that is varying. Surely we need to distinguish
between the FM signal when it is created using a voltage-controlled
oscillator with an amplitude of \$1\$ volt and when it comes out of the
power amplifier and goes to the antenna with a power of 10 kW?
This answer is focussing on radio receivers such as AM and FM.
If you are only interested in receiving a signal from one station you may not need to have or use an intermediate frequency. You can build your receiver to tune in to just that frequency - the tuning needs to be sharp - you need to reject all possible other sources that may pollute the signal you want.
This is done by a bunch of band pass filters that together, have a passband that is wide enough to cope with the signal you wish to receive but not so wide that it lets others in.
Now say you wanted to tune in to 2 stations - you'd have to re-align all this filtering to coincide with a new station. Historically radios were simple and moving a bunch of tuned band pass filters to a new centre frequency would be hard.
It was a lot easier to have a bunch of fixed band-pass filters that did the majority of all the unwanted channel rection rather than trying to align them as you tuned the dial.
Thus super-heterodyne receivers were conceived. The incoming broad range of many radio stations were "mixed" with an oscillator that can be simply tuned with a dial - this produced sum and difference frequencies and usually the difference frequency became the new "wanted" frequency. So for FM (88MHz to 108MHz), the I.F. frequency became 10.7MHz and the oscillator would be (typically) at 98.7MHz for tuning 88MHz signals and at 118.7MHz for tuning 108MHz signals.
Don't hang me on this - it could equally be at 77.3MHz rising to 97.3MHz to produce the same set of difference frequencies. Maybe someone can modify my answer or advise me on this.
It's a small matter though because the point is that once you were able to manipulate the incoming signal's carrier frequency you can feed the result through a tightly tuned fixed set of band-pass filters before you demodulate.
A bit more info about the VHF FM band
It goes from 88MHz to 108MHz and has an IF that is just slightly bigger (10.7MHz) than half the frequency range it covers. There is a sensible reason - if the oscillator were exactly tuned to pick up 88MHz (i.e. osc = 98.7MHz) the difference frequency it would produce from the top of the band at 108MHz would be 9.3MHz and this would be just out of band of the tuning centred at 10.7 MHz and therefore "rejected".
Of course if someone started transmitting just outside the FM band you may pick this up but I believe that legislation prevents this.
Following recent activity in this question I remembered that there is another valid reason for using an intermediate frequency. Consider that the signal from an antenna might be in the order of 1 uV RMS and then consider that you'll probably want the radio circuit to amplify this to something like 1V RMS (forgive the hand waving) at the demodulator. Well, that's a gain of 1 million or 120 dB and, no matter how hard you might try, having a circuit board with a gain of 120 dB is a recipe for feedback disaster i.e. it will oscillate and turn into a "theramin".
What an IF gets you is a break in the signal chain which prevents oscillation. So, you might have 60 dB of RF gain then convert to your IF and have 60 dB of IF gain - the signal at the end of the chain is no longer frequency compatible with what happens at the antenna and therefore, there is no theramin effect!
Some radios might have two intermediate frequencies - for just this reason alone you can reduce the RF gain to 40 dB and each IF stage can have a gain of 40 dB and NO theramin.
Best Answer
I don't think an "image rejection" answer is complete enough. There are other reasons.
For example, AM band case that ranges from \$530\:\textrm{kHz} \le f_{RF}\le 1700\:\textrm{kHz}\$, using \$f_{IF}=455\:\textrm{kHz}\$. Assuming upper LO injection, there will be in-band images and this fact puts pressure on having a very selectable and tunable 1st band-pass filter. (Lower LO injection will also have in-band images.) The reason here for using upper LO injection isn't as much about image rejection (the argument is moot) but instead about the required dynamic range of the LO. Upper LO injection: \$985\:\textrm{kHz} \le f_{LO}\le 2155\:\textrm{kHz}\$ for a dynamic range of only about a factor of 2 or so. Relatively easy to do. Lower LO injection: \$75\:\textrm{kHz} \le f_{LO}\le 1245\:\textrm{kHz}\$ for a dynamic range of a factor of 16 or so. Much, much harder to do well.
In this case, I think the argument for the upper LO injection is more about tuning the LO than about image rejection.
Now assume \$f_{IF}=10.7\:\textrm{MHz}\$. Lower LO injection: \$10.17\:\textrm{MHz} \ge f_{LO}\ge 9\:\textrm{MHz}\$. Upper LO injection: \$11.23\:\textrm{MHz} \le f_{LO}\le 12.4\:\textrm{MHz}\$. The worst case image frequency for lower LO injection is \$19.7\:\textrm{MHz}\$ and for upper LO injection is \$21.93\:\textrm{MHZ}\$. A relatively simple 2-pole low pass filter is kind of easy for 40 db per decade and given that it has to pass the entire AM band, should be about \$-40\:\textrm{dB}\$ at \$17\:\textrm{MHz}\$. This suggests about \$-42.5\:\textrm{dB}\$ for lower LO injection and about \$-44.4\:\textrm{dB}\$ for upper LO injection.
In this case, tuning the LO isn't the issue at all. Now it's more about image rejection.