In some cases conversion from discrete sum to integral is straightforward and logical. Suppose ΣF.∆r is a sum of small but finite 'work done's. To convert this to an integral, we simply convert ∆r to dr and replace Σ by the integral sign, giving the total 'continuous' work done.

In other cases, such as when calculating the weighted average of a discrete random variable, we write Σx.P(x). After converting this to the continuous analogue of the discrete summation, we get ∫x.f(x)dx.

In the first case, the discrete sum as well as the integral have the units of work done (Newton-metre).

Let's say, in the latter case, x has unit of length, f(x) and P(x) have the units of mass. On converting the discrete sum into an integral, wouldn't we get different units because of the extra dx term? So instead of getting kg-m in the discrete case, we get kg-m² in the integral form.

Please help me wrap my head around this.

  • $\begingroup$ Does this answer your question? $PDF$ taking values $>1$: what are the "units"? $\endgroup$
    – K B Dave
    Nov 20 at 19:05
  • $\begingroup$ Be very careful here: $\sum xP(x)$ has units of length times mass, where as $\int xf(x)~\mathrm{d}x$ has units of length. If $P$ is a mass function, then $f(x)$ is the related density function. The units of mass and density differ by the units of volume. $\endgroup$
    – whpowell96
    Nov 20 at 19:08


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