Rough modeling of rows of Pascal's triangle as a distribution

I found a rough correspondence between the formula under the Consider: and Pascal's triangle...

I tried a few different rows and it was always close, I'm not sure whether it becomes a better approximation for a row of Pascal's triangle for larger rows but I think it does get closer...

Also, one can solve for the middle number approximately:

for r = 11

the right answer is 252 so pretty close. 

One practical reason for this approach is for example finding the middle number of row r of pascal's triangle, ordinarily you would use the combinatoric formula r!/((r-1!)*(r!)) but eventually the factorial involves numbers too large for Maple to calculate, but using my integral formula for the middle number only involves calculating 2^r which it can do more easily.