Intuitively, the expected value of the distribution is something like its 'center of mass': a point where half the weight of the distribution is on one side, and half is on the other side.
This is correct, it is a center of mass. But beware about how you interpret that; you can't just say "half the mass is each side".
If this reasoning is right (I thought to myself), then this point should be identical to the point where the cumulative distribution function is equal to .50. Why? Well, because at the X value where the CDF is .5, a random variable has 50% change of being less than X and a 50% chance of being greater than X-- that is, X is the point where half the weight of the distribution lies on either side.
No, that's the median. They'll be the same when the distribution is symmetric (assuming the mean exists at all, naturally), but when it's non-symmetric, the two rarely* coincide.
* some people mistakenly assert 'never'. It's unusual, but possible
The moment has two components - the sign (which side you're on) and the magnitude - how far away you are. It's not sufficient to say there are the same fraction on either side**, so we must be at the mean - it depends on how far away they are.
**(or the same number of values on either side for a sample)
Consider a sample, like 1,2,18 (or a distribution with 1/3 probability at each point, it doesn't matter). Let's take "2" as being our candidate for the mean - after all, by your logic, there's the same proportion either side, so they should balance, right?
But the thing is, by putting a value 16 units above the proposed mean, it exerts 16 times the moment (in the physical sense) as the value one unit below it - just as a person sitting at the far end of a see-saw doesn't balance someone of the same weight sitting partway along the other arm:
So it's not having half the distribution either side that determine the mean.
However, if you weight those proportions by their distance you will get the balance you're thinking of (quite literally - take a very light, stiff rod and hang weights at 1, 2 and 18 distance units along the rod and find where they balance ... which will be at 7 units. If you work out the signed distances (-6,-5,+11) you can see they balance out.
If you check various distributions on Wikipedia, you'll find that typically for asymmetric distributions the mean and the median differ.
Indeed, the difference between them is used as the basis of a measure of skewness (with appropriate rescaling) - the second Pearson skewness measure.
Another note about your question: when you say "the expected value of $f(X)$", you misspeak. You intend to say "the expected value of $X$, where $X$ has density $f$". The expected value of a continuous $X$ is $\int x\,f(x) dx$. The expected value of $f(X)$, on the other hand is the expected value of the transformed random variable $Y=f(X)$, which by the law of the unconscious statistician is $\int f(x).f(x) dx$. That's a different object altogether. Just stick with "the expected value of $X$".