Let f(x) be some PDF, and F(x) be its CDF. Shouldn't F(x)=.5 give us the expected value of f(x)? I was playing around in R and have gotten myself very confused about the relation between probability distributions, their expected values, and their cumulative distribution functions.
Say we're interested in the expected value of the gamma distribution. Here's a hacky way to calculate it in R:
library("pracma")
m=2
s=1
grid = seq(0,50,.01)
y = dgamma(grid, m, s)
(expected = trapz(grid,grid*y))
[1] 2

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. 
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. (EDIT: If I had to pinpoint where my misconception occurred, I would say it was here. Center of mass isn't just about proportion of weight, as the answers below have reminded me.)
I decided to follow this hunch:
(halfway  = qgamma(.5, m, s))
[1] 1.678347

Nope, my hunch was wrong. The numerical difference seems to be too systematic to be the result of using numerical integration (seems to happen on any asymmetric distribution, to varying degrees, try for yourself). 
How am I thinking about this incorrectly?
 A: In the cases where it (the mean) exists, a nice way to look at this is that the mean is the value $M_2$ which minimizes
$$
\int_{\text{support of $x$}} |x-M_2|^2 f(x) dx.
$$
That is, the the distance from the mean is considered unevenly over the support.. all else equal the farther a point $x$ is from the mean the higher its 'cost' (if the distance is > 1).
The value for which $F(x) = 0.5$ is in fact the median, which is the value $M_1$ which minimizes 
$$
\int_{\text{support of $x$}} |x-M_1|^1 f(x) dx.
$$
Distributions which are skewed* in one direction will have medians which are not in the same place as their means.. because the means will gravitate towards the longer tail to compensate for the squared distance, which penalizes further values more than the absolute difference does (when the distance is > 1).

*As rightly pointed out by @Glen_b skewness is used here as defined by Pearson's 2 skewness coeficient. That is, skew is $3\frac{\text{mean} - \text{mode}}{\text{std. deviation}}$.
A: 
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. 

The first part of 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 identical 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 $X$".
A: $F(x)=0.5$ gives $x$ as the median of the distribution, which you could say is the centre of probability mass. 
I think your understanding of the expected value being the 'centre of mass' of a distribution is a little misguided. Expected value is more of a weighted average than a centre of mass, in the discrete (and finite) case what you're doing is weighting the possible values your distribution can take on by the probability of that value occurring. In continuous distributions the analogy isn't as clear, but you're still weighting possible values your distribution can take on by their probability.
