So how do we extend this kind of average, eqn. 1.16
to continuous distributions? The normal procedure, replacing
a sum by an integral.
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 is and the total mass is 1. In the
continuous case, the definition looks the same but now
represents a one dimensional density (mass per unit length).