Intuition Behind E(X) of a distribution In most undergrad classes, E(X) is taught to be the "expected gain" by playing a game of bets. 
For instance: If we consider a Bernoulli distribution with $p=0.3$ (Winning) and if you win, you get \$10 and you lose, you pay \$3, the E(X) = 10 x 0.3 - 3 x 0.7 = 3 - 2.1= \$0.9. 
Is there a way this intuition carries forth to a continuous distribution?
I understand that E(X) can be considered are centre of mass for a distribution but I don't seem to understand what it means intuitively to say that a mean of a continuous distribution is 5. 
 A: For simplicity, assume that the density function $f(x)$ is non-zero over the interval $[a,b]$. Divide the interval from $a$ to $b$ using the equally spaced points $x_0=a$, $x_1$, $x_2$, and so on up to $x_n=b$.
Think of $n$ as large.  The probability that our random variable $X$ takes on values in the interval $[x_i,x_{i+1}]$ is approximately the length of the interval, namely $\dfrac{b-a}{n}$ times the density at $x_i$.
Now imagine all this probability as concentrated at the point $x_i$. So now we have a discrete distribution. Its mean, by the usual "finite" rule you quoted, is approximately
$$\sum_{i=0}^{n-1}x_i \frac{b-a}{n}f(x_i).$$
Finally, let $n\to\infty$. Then the above Riemann sum  approaches
$$\int_a^b xf(x)\,dx.$$
A: Maybe the easiest approach is to start with the uniform distribution over $(4,6)$.  It seems reasonable to say that the expected value of this is $5$.  The official definition is $E[X]=\int X\cdot P(X) dX$ and you can see this is satisfied.  Then if you think about distributions that have stairsteps, that are constant over a couple of lengths and see that the center of mass (think of cutting the region out of a piece of graph paper) is a reasonable expected value.
