I need help sing a 555, op-amp integrator, and comparator to generate a voltage controlled PWM signal
I don't quite understand how the op-amp integrates.


The left op-amp is used as an integrator, and is probably best left for you to research the theory behind how it works. But for simplicity sake, you can consider it as the active equivalent of a series RC circuit. For more information on how the integrator works, simply search "Op-Amp Integrator." 


  how do I calculate the R and C values for the op-amp?


You need to choose an RC value such that you obtain your desired triangle wave at 50 Hz.

As others have suggested, you will need a feedback resistor parallel with your capacitor, and that RC combination will determine your "cutoff" frequency, where any frequencies above this "cutoff" will be an integral output of the input. 

Below is your overall transfer function of the integrator alone:

$$
\frac{V_o}{V_{in}}=-\frac{R_F}{R_i}\times\frac{1}{1+j\omega(R_FC)}
$$

Take the magnitude of the above at 50Hz (or 314.16 rad/s), and that will be your gain at that frequency.

A resistor on the non-inverting input isn't entirely necessary, but is typically used to balance the bias currents and is set equal to the equivalent resistance seen by the inverting input. It doesn't not affect your gain or cutoff frequency.


  What should my input pulses look like


If you're talking about the clock, it should be a 50Hz with a 50% duty cycle, since the frequency of your pulses is 50Hz. If you're talking about the voltage control input, then they shouldn't be pulses. The voltage control determines the duty cycle of your output PWM. 

For example, suppose your triangle wave oscillates between 0V and 8.4V. For a 50% duty cycle (10ms pulse width in your case), set the voltage control to half of the max voltage of your triangle wave. (4.2V).

If you want 2.5ms, then 2.5/20ms = 12.5% duty cycle. Set your voltage control to 12.5% of your max voltage (1.05V).