I recently learnt the derivation of radioactive decay formula and I am quite surprised about using integration to derive the formula.
But $N$ (the number of atoms) can only be discrete numbers (like 1,2 & not 1.5 or.9).
But won't using integration include all the continuous values of $N$?
So instead of using integration, shouldn’t some kind of summation be used to include the discrete values only?
This answer says something like this about the use of summation instead of integration. https://math.stackexchange.com/q/1509235/
$$
\begin{align}
\frac{dN}{dt} & = - \lambda N \\
\end{align}
$$
hence,
$$
\begin{align}
\int \frac{dN}{dt} & = - \lambda \int dt \\
\\ \ln{N} & = - \lambda t + C
\end{align}
$$
If the initial number of nuclei is $N_0$ and $N = N_0$ when $t = 0$
then (i becomes,
$$
\begin{align}
\ln N_0 = C
\end{align}
$$
substituting for C into (i
$$
\begin{align}
\ln{N} & = - \lambda t + \ln{N_0} \\
\ln{N} - \ln{N_0} & = - \lambda t \\
\ln({ \frac{N}{N_0} }) & = - \lambda t \\
\Rightarrow \frac{N}{N_0} & = e^{-\lambda t} \\
\Rightarrow N & = N_0 e^{- \lambda t}
\end{align}
$$