Why do we equate an indefinite integral to a specific value?

Question

Suppose we want to obtain a displacement vector defined as $\mathbf s(t) = x(t)\mathbf i + y(t)\mathbf j + z(t)\mathbf k$ from the components of a velocity vector $\mathbf v(t) = \dot x(t)\mathbf i + \dot y(t)\mathbf j + \dot z(t)\mathbf k=\mathbf 0$. According to my notes, this can be done by equating each scalar component of the displacement vector to the indefinite integral of the corresponding scalars of the velocity vector, i.e. $$ \mathbf s(t)=\begin{pmatrix} x(t)=\int \dot x\ dt \\ y(t)=\int \dot y\ dt \\ z(t)=\int \dot z\ dt \end{pmatrix} $$ But, as $\int f(x)\ dx=\{F(x): \frac {dF}{dx}=f(x)\}$, this should be syntactically wrong, because we're implying that a number is equal to an infinite set of numbers, or am I missing something?

Moreover, this also leads to a weird equation when solving the integral; for example, by taking into consideration the $x$-component of $\mathbf s$, we would have that $$ x(t)=\int \dot x\ dt=c_1 $$

Which is correct, but it would also mean that the $x$-component of the velocity could be equal to any value belonging to $\mathbb R$. Because of that, we substitute $c_1$ with the initial condition and we equate it to zero, giving it a specific value: $c_1=0$. But, to me, this sounds like a break of the definition of indefinite integrals, as $\int \dot x\ dt=c_1=0$ would basically mean that an indefinite integral is one, specific function.

I know this may be a very stupid question, and maybe it has to do with the same shortcuts that make us not specify "$\forall c \in \mathbb R$" when adding the constant $c$ in the solutions of an indefinite integral, but this doubt is really challenging me and I still don't understand whether I'm missing some point or it should actually be written $x(t)=c_1=0 \in \int \dot x\ dt$. Thanks a lot in advance!

Answer 1

In physics we frequently leave off the limits of the integral when the limits can be figured out from the context. So, in the first case, the actual relation is: $$x(t) = x(0) + \int_0^t \dot{x} \operatorname{d}t'.$$

Most often, though, when the limits are left off the implied limits are over all possible values of the dummy variable. For example: $$ Q = \int \rho(\mathbf{x}) \operatorname{d}x^3$$ is understood to be the integral over all of space.

Answer 2

You are being confused by the shortcut that people took who wrote that expression. They mean for the integral to be taken between definite limits.

It would be more proper to say

$$x(t) = x(0) + \int_0^t \dot{x} dt$$

But that gets longwinded. Most people, when seeing the expression as you gave it, will understand it to mean what I wrote. But technically, it's not the same.