Let's say I want to calculate the height of a thrown ball: Let $x''(t)=-g$ and $x(0)=x(T)=0$ and $x'(0)=v_0$.
One could then integrate 2 times and it is done. My professor told me to write it this way, and I just don't get it: Could somebody help me understand it?:
$$x(t)=\int_0^t\left(\int_0^s x''(r) \mathrm{d}r \right)\mathrm{d}s=-\int_0^t(gs+C_1)\mathrm{d}s=-\frac{1}{2}gs^2-C_1t-C_2$$
Why from $0$ to $s$? And why are there integration constants?
Put it in this way for easy to grab each step:
$$ \tag{1} x(t) = x(0) + \int_0^t v(\tau) d\tau. $$ and $$ \tag{2} v(\tau) = v(0) + \int_0^\tau a(\xi) d\xi. $$
These are basic kinematic relations. Substitute Eq.(2) into Eq.(1):
$$ \tag{3} x(t) = x(0) + \int_0^t \left( v(0) + \int_0^\tau a(\xi) d\xi \right) d\tau. $$
This is the form for double integrrals. Your post expression is not quite right. The integral constant comes out very strangely (because you wrote a definite integal, there should have no integral constants.).
or, you may write an indefinite integrals, then leave the integral constants to be determined by initial conditions.
$$ \tag{3} x(t) =\int^t d\tau \left( \int^\tau a(\xi) d\xi \right). $$
The choice of integration variables ($r$ and $s$) is just to distinguish which variables are currently being integrated. The choice of letter is arbitrary since $$\int_0^1 f(x)dx = \int_0^1 f(y)dy.$$ Choosing different letters for different integrals keeps them separated into different operations. The variable $r$ denotes the time when the ball has a certain acceleration. The variable $s$ denotes the time when the ball has a certain velocity. And $t$ denotes the time when the ball has a certain position. You could have also chosen $t_a$ for the acceleration time instead of $r$ and $t_v$ for the velocity instead of $s$ and everything would have worked out the same.
Your result has integration constants because you didn't use your initial conditions after each integral.
