Optimal CFM for kerosene to burn?
This is something of a standard reaction stochiometry question, although the situation is a bit more involved than the ones you normally see in textbooks.

The first step is to work out what the chemical reaction we're talking about actually is. One complication is that "kerosene" doesn't name a single molecular species, but is instead a mix of hydrocarbons. But the hydrocarbons are similar enough that we can probably approximate the mixture as a "typical" hydrocarbon of roughly the right type and mass. Wikipedia approximates kerosene as dodecane, and even gives a combustion reaction (saving us the effort of balancing one):

$$\ce{2 C12H26(l) + 37 O2(g) → 24 CO2(g) + 26 H2O(g)}$$

Now that we have stoichiometry, we just need to convert the quantities of reactants involved to numbers of molecules (moles). We're dealing with rates of consumption for both reactant, so let's just run the numbers for a fixed period of time, say 1 minute. 

We have an amount of kerosene we want to burn ($\ce{X\; mL*s^{-1} * 60s}$) and a conversion from volume to mass (the density), and all we need to get the "moles of kerosene" would be the molecular weight of kerosene. Since we're approximating kerosene as dodecane, use the number for dodecane ($170.34\;  \ce{g*mol^{−1}}$).

Using the knowledge from the balanced chemical reaction equation that 2 moles of dodecane require 37 moles of oxygen to combust completely, we can determine how many moles of oxygen we need to burn the given quantity of kerosene.

Now we need to convert moles of oxygen to a volume of air. If we assume oxygen and air is an ideal gas (a safe assumption under normal conditions), we can use the ideal gas law to interconvert moles ($n$) into volume ($V$): $PV = nRT$, where $P$ is the pressure, $T$ is the temperature, and $R$ is the ideal gas constant (about $8.314\;\ce{L*kPa*K^{−1}*mol^{−1}}$). Or, if we're dealing with standard temperature and pressure (about 1 atmosphere at sea level and about room temperature), then 1 mol of ideal gas takes up about $22.4\;\ce{L}$. So we can find out what volume of oxygen we need.

But we're not dealing with pure oxygen, so we have to divide by 0.21 to get the volume of air which would contain that volume of oxygen. (This works because the 21% is a volumetric percentage.) The final little bit is to convert the number of liters of air used per minute into cubic feet used per minute.

So there's really no calculus involved - it's just lining up the conversion factors and multiplying through. The tricky bit is probably figuring out the chemical reaction formula you want to use (e.g. the assumption of dodecane). It also helps to know that things are easiest if you do most of the calculations in SI (metric) units, and only convert to non-SI at the end.

One final caveat is to keep in mind that we assumed that kerosene was well represented as dodecane. If your kerosene is substantially different from dodecane, the reaction will have to be adjusted. We also likely assumed that the input air was at standard temperature and pressure, and was dry (the 21% value is for dry air). Warmer/cooler air, air at altitude, and humid air will all change the calculation. Also keep in mind that the calculation is for combustion under ideal conditions - incomplete mixing and non-ideal combustion issues might mean you need more air in order to ensure complete combustion.