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Why we can apply calculus in cases where discrete quantities take place?

Suppose we have a box that has two partitions, namely A and B (look at the figure below). Suppose we know the rate that particles pass from A to B. That is we know the rate $r=\frac{dN}{dt}$. Then we calculate the number of particles that passed from A to B from $t_i$ to $t_f$ as following:

$$N = \int_{t_i}^{t_f}rdt$$

Suppose that the rate is constant and equal to $2$ particles per second (or just $2$ per second) and the interval is equal to $2.6$ s. Then the amount of particles that went from A to B is $5.2$. But how is it possible the amount of particles to be a real number and not an integer? I just used that example for particles but it can generalized to charge, photons etc.

I was trying to understand what the instantaneous rate of transfer of a discrete quantity means but I just can't. I thought one could measure that rate like we measure the power of some engine ($\frac{dE}{dt})$. But the main difference is that $dE$ is an infinitesimal of a continuous quantity.

What is the justification that allows us to use calculus in case of discrete quantities?

                                                                 

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  • $\begingroup$ Calculus is used for continuous functions and your function is not continuous in time. $\endgroup$ Jul 10 at 23:15
  • $\begingroup$ The rate of particles can be non integer, multiply it by time again gives back an integer $\endgroup$
    – User688539
    Jul 11 at 2:06
  • $\begingroup$ When you wanted to avoid calculus you need to fiddle around with a lot of expressions involving discrete sums and differences instead of integrals and derivatives. Calculus is a lot more convenient. For systems with discrete numbers $N$ it is typically applied when $N$ is very large, say, $10836.6$. This is clearly $\approx 10836$ so we can safely apply calculus for problems with large $N$. $\endgroup$
    – Kurt G.
    Jul 11 at 4:27
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We can use calculus in problems involving populations of discrete objects if we are averaging over a large number of objects or instances and if we realise that by modelling the number of objects by a continuous variable we have made an approximation that will affect the precision and interpretation of our results.

In your example, a uniform rate of $2$ particles per second means that in a random interval of $2.6$ seconds there could be either $5$ or $6$ particles passing from A to B. The result of $5.2$ particles that you get from a continuous approximation means that over a large number of $2.6$ second intervals the average number of particles passing from A to B in each interval will be $5.2$ - but the result in any individual interval will be either $5$ or $6$.

It is like saying that the average family has $2.1$ children - but no individual family actually has $2.1$ children.

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