So how do we extend this kind of average, eqn. 1.16 to continuous distributions? The normal procedure, replacing a sum by an integral.

$$\colorbox[rgb]{1,1,1}{$\u27e8$}\colorbox[rgb]{1,1,1}{$x$}\colorbox[rgb]{1,1,1}{$\u27e9$}\colorbox[rgb]{1,1,1}{$\equiv $}\colorbox[rgb]{1,1,1}{$\int $}\colorbox[rgb]{1,1,1}{$x$}\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$x$}\colorbox[rgb]{1,1,1}{$)$}\colorbox[rgb]{1,1,1}{$\mathit{d}$}\colorbox[rgb]{1,1,1}{$x$}$$ | (1.17) |

and replacing $\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$x$}\colorbox[rgb]{1,1,1}{$)$}$ above, with the probability density.

If you remember your physics, this should also remind you of taking the center of mass. In the discrete case, eqn. 1.16, it’s like you’re finding the center of mass, where the pass at a point $\colorbox[rgb]{1,1,1}{$x$}$ is $\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$x$}\colorbox[rgb]{1,1,1}{$)$}$ and the total mass is 1. In the continuous case, the definition looks the same but now $\colorbox[rgb]{1,1,1}{$P$}\colorbox[rgb]{1,1,1}{$($}\colorbox[rgb]{1,1,1}{$x$}\colorbox[rgb]{1,1,1}{$)$}$ represents a one dimensional density (mass per unit length).