An IV problem using d'Alembert's solution: how does this integration work? I'm using Roger Knobel's An Introduction to the Mathematical Theory of Waves to study for a Mathematics BSc third-year module in waves. In Section 8.2, Knobel

construct[s] the solution of the following initial value problem for the displacement $u(x,t)$ of an infinitely long vibrating string:
\begin{align*}
\text{PDE:}\hspace{1em}& u_{tt}=c^2u_{xx}, \hspace{1em} -\infty<x<\infty,\;t>0, \\
\text{IC:}\hspace{1em}& u(x,0)=f(x),\\&u_t(x,0)=g(x).
\end{align*}

He continues,

To solve this problem one can start with the general form for solutions of the wave equation $$u(x,t)=F(x-ct)+G(x+ct).$$ Substituting into the initial position condition $u(x,0)=f(x)$ gives \begin{align}F(x)+G(x)=f(x),\end{align} while substituting into the initial velocity condition $u_t(x,0)=g(x)$ results in $$-cF'(x)+cG'(x)=g(x).$$ Dividing this last equation by $c$ and integrating from $0$ to $x$ provides us with a second equation for $F(x)$ and $G(x)$, \begin{align}-F(x)+G(x)=-F(0)+G(0)+\frac{1}{c}\int_0^xg(s)\;\text{d}s.\end{align}

If I understand right, $F'$ and $G'$ are $t$-derivatives. So why does integrating them from $0$ to $x$ w.r.t the dummy variable $s$ retrieve $F$ and $G$, as if we'd integrated w.r.t. $t$? I guess this is a basic calculus question.
 A: Generally, the name of a variable of integration doesn't actually matter, it's just an identifier. It only really matters when there's more than one variable involved that you could be integrating with respect to, and in this problem there isn't. You can think of $s$ as "the single argument of $g$", which you could give any name you want, provided it isn't already bound in the surrounding scope, so not $x,F,G,g$ or $c$. (So you could technically call it $f$, even though there is an $f$ elsewhere in the problem. Please don't, though.)
Note that the name of a variable limit of integration does matter, because the definite integral depends on that limit. Thus for example here the use of $x$ as the upper limit of integration is not a completely stylistic choice; whatever letter you use has to be the same letter that is used for the arguments of $F$ and $G$ on the LHS.
The big thing going on in the background here surrounds the difference between an expression and a function. An expression can have named variables in it. A function, strictly speaking, does not have named variables in it, it just has arguments and their positions. This means that strictly speaking a function can't be differentiated with respect to a named variable, only with respect to an argument position. But no one writes like that, unless their audience is a computer.
This results in rather convoluted things happening under the hood. For example, in this context, the symbol $\frac{\partial u}{\partial t}$ is a shorthand for $(x_0,t_0) \mapsto \frac{\partial}{\partial t} \left. \left [ u(x,t) \right ] \right |_{x=x_0,t=t_0}$. Here we take $u$, a function, turn it into $u(x,t)$, an expression, turn that into $\frac{\partial}{\partial t} [u(x,t)]$, also an expression, and then finally convert that back into a function. Then we even change the name of the arguments of that function back to $(x,t)$. Similar things are going on with integration.
