Indefinite integral of a density function Suppose that $\rho(x)=\frac{dm}{dx}$ is the linear density of a rod. Can we find the mass at each point of the rod by integrating $\rho(x)$, so that:$$m(x)=\int\rho(x)dx.$$ Can we do the same with probability density in quantum mechanics, so that:$$P(x)=\int|\Psi|^{2}dx$$ (assuming one dimensional wavefunction). In the case of probability density I think we can't because the probability in every point would be 0 because the position is a continuous variable. Any ideas?
 A: It's neither possible to find the mass of a point nor the (quantum) probability in such a point.
It is possible to find the mass of a small interval $\delta x$, located at $x$ as:
$$m(x,x+\delta x)=\int_x^{x+\delta x}\rho(x)\text{d}x$$
Similarly:
$$P(x,x+\delta x)=\int_x^{x+\delta x}|\Psi|^{2}\text{d}x$$
Note that in both cases, when $\delta x=0$, the integral returns $0$.
A: Well, you can calculate indefinite integral of
$$ m(x)=\int\rho(x)dx $$
For simplicity lets say density is uniform along all rod, so integration result will be :
$$ m(x) = \rho\, x + m_o $$
where $m_o$ would be integration constant. The meaning of equation would be rod mass until point $x$ + integration constant. So semantically this integral is equivalent (up to integration constant) to a definite integral
$$\int_0^x\rho(x)dx$$
So in principle I would recommend too to work on definite integrals more, because in such case you


*

*Would not have to think about integration constant. In this simple example integration constant is just some value, but in general case it can be an arbitrary expression. 

*You always will be forced to think about physical meaning first
, because you must think about boundary conditions, i.e. integration limits. In most general case you can use definite integral $\int_{-\infty}^{+{\infty}}$ (For example probability of finding a particle in a whole universe, etc.)

