Averages of continuous variables So how do we extend this kind of average, eqn. 1.16 to continuous distributions? The normal procedure, replacing a sum by an integral. $$\langle x \rangle \equiv \int x P(x) dx$$ (1.17) and replacing $P(x)$ 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 $x$ is $P(x)$ and the total mass is 1. In the continuous case, the definition looks the same but now $P(x)$ represents a one dimensional density (mass per unit length). As an examples, consider the average speed of a car over some period $0$ to $T$ seconds. Denoting the speed as a function of time $S(t)$, then $$\langle S \rangle = \int_{-\infty}^\infty S P(S) dS$$ (1.18) Here $P(S)$ is the distribution of speeds over this time interval. But we can also view this another way in terms of a the limit of the sum of $N$ different speeds measured at equal time intervals $S_1, S_2, \dots, S_N$. Looking back at eqn 1.15 this can be written as $$\langle S \rangle = \lim_{N\rightarrow\infty} {1\over N} \sum_{i=1}^N S_i$$ (1.19) But if we take the limit as $N \rightarrow \infty$ then this becomes an integral, in the usual calculus way so that $$\langle S \rangle = {1\over T} \int_0^T S(t) dt$$ (1.20) These two formulas 1.18, and 1.20 are equivalent. If you have don't have the distribution of speeds then the last formula might be handier, but if you do, as often happens in statistical mechanics, then the first formulation is a better way to go.