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I've been thinking recently and have confused myself a little bit as to the difference between sums and integrals.

I understand that an integral is a Riemann sum where you take the limit as the difference in whatever you are integrating with respect to becomes infinitesimally small.

However, I was thinking recently about how one would sum up all the values along a line/curve. For example, let's say I have a function f(x) that takes in a point and gives the electrical charge at that given point (Assume we are talking about a one dimensional system that just has x values). How would I sum up all the charges along a given interval [a,b].

I initially thought this would be an integral, but wouldn't that give the area under the curve because you are multiplying by $dx$, which isn't exactly what we are looking for? Unless it is, which is why I'm confused as I can't tell if that's what I would want for this situation.

Another thing that is obvious is to literally just add up all the values such as: $$f_1(x) + f_2(x) + f_3(x) + \ldots f_n(x).$$

This seems to be more of what I might be thinking about but what would be a way to do this "algorithmically" so to speak, where you don't have to literally add up an infinite amount of numbers.

Would you just us a normal sum, or would there be a different way?

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  • $\begingroup$ Perhaps you can elaborate on adding up charges. Because, at least to me, I lack the engineering background to be confident about a solution. A guess would be that one should add the charges by what proportion of the line they exert their charge. E.g. A line $[0,1]$ with charge $1$ on $[0,1/2]$ and charge $0$ on $(1/2, 1]$ would have a "summed up" charge of $1/2$ (that is the area or integral of the charge over $[0,1]$). This would also suggest the integral in general, as it informally, and sometimes exactly, gives the sum of values by their proportion on the line. $\endgroup$
    – Manifoldski
    Commented Aug 8, 2023 at 21:24
  • $\begingroup$ I feel quite confident saying that adding up the charge at all points is incorrect. If the charge is 1 at each point, you would by adding the charge at each point conclude that there is infinite summed up charge in the line…(would probably solve most energy crises in the world) $\endgroup$
    – Manifoldski
    Commented Aug 8, 2023 at 21:35

1 Answer 1

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If your function $f(x)$ represents the linear charge density (ie. the amount of charge per unit length), then the total amount of charge $Q$ along the line from $a$ to $b$ is given by the integral

$ Q=\int_a^bf(x)\,dx $

You can see this from dimensional analysis: since $f(x)$ has dimensions of $\mbox{charge/length}$, then the "area" under $f(x)$ (the integral above) correctly has dimensions of

$ (\mbox{charge\length})\times\mbox{length}=\mbox{charge} $

If you're familiar with the Dirac delta function, the "charge density" of $n$ discrete points located at $x_1,...,x_n$ between $a$ and $b$ and carrying charges $q_1,...,q_n$ is

$ f(x)=\sum_{i=1}^nq_i\delta(x-x_i) $

In this case, the integral gives

$ \int_a^bf(x)\,dx=\int_a^b\sum_{i=1}^nq_i\delta(x-x_i)\,dx=\sum_{i=1}^nq_i $

ie. it is just the total charge you would get by adding up all the individual charges.

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