I have solved the problem. In the process I have found that the book Understanding Physics by <PERSON> is much clearer than <PERSON> and <PERSON>.
Basically, there seem to be two points of confusion: 1. What does $\int$ really mean? 2. What are the initial conditions?
First of all, what $\int$ really means is $\int_c^x$ where $c$ is an arbitrary constant. This is why we get a constant of integration, because $\int x^2 dx$ is really $\int_c^x x^2 dx = \frac{x^3}{3}-\frac{c^3}{3}$ where $-c^3/3$ is an arbitrary constant, which is why we just let it be $+C$ when we integrate indefinitely. I.e. $\int x^2 dx = \frac{x^3}{3}+C$.
How does this apply to mechanics? In mechanics, the arbitrary constant $c$ cannot be anything. This is because our variable $t$ cannot be anything, since it depends on when we started measuring. For example, I want to find the velocity of a body. 474
If I only started measuring at $t=t_0$, it follows that I have no information about $t<t_0$. This means I can only have a lowest limit of $t_0$. Thus, in physics, it is not appropriate to just say $\int$, we must rather specify the limits: $\int_{t_0}^t$.
So using $a=\frac{dv}{dt}$ we get $\int_{t_0}^t a dt = \int_{t_0}^t \frac{dv}{dt} dt$, ie $v(t)-v(t_0)=a(t-t_0)$ or $v(t)=v(t_0)+a(t-t_0)$, where $t_0$ is not necessarily $0$, as explained above.
At this point, someone clever might say: "You can now let $t=0$ to get $v(0)=v(t_0)+a(-t_0)$, and plugging this back into the equation gives $v(t)=v(0)+at$, which is simpler and neater."
The reason this is wrong, is because he started by saying let $t=0$. As I explained above, since we only started measuring the body (by assumption) from $t=t_0$, it follows that we have no information about $t=0$. Therefore, $v(0)$ is meaningless, so $v(t)=v(0)+at$ is actually un-usable for us, since we don't know $v(0)$: we only started at $t=t_0$! However, if someone were to come and tell us: "I started observing the body before you, and I find that $-t_0$ second before you started measuring, ie at your $t=0$, the body was going at $v(0)=k$", and he tells us an actual numerical value $k$. If this happened, then indeed, $v(t)=v(0)+at=k+at$ is simpler than our equation, and is now usable for us, whereas before it wasn't.
Ok, that clears up the velocity equation. Now we simply integrate it again to get $\int_{t_0}^t v(t)-v(t_0) dt = \int_{t_0}^t a(t-t_0) dt$, ie $x(t)-x(t_0)-v(t_0)(t-t_0)=\frac{1}{2}a(t-t_0)^2$, hence $\Delta x=v(t_0)\Delta t + \frac{1}{2} a(\Delta t)^2$.
This is all general, we still haven't set $x_0$, $v_0$ or $t_0$. So everything is clear: we just go with the convention at this stage and set $t_0=0$, ie we started measuring the body at the same time as we started measuring time. Also, we let $x_0=0$. Therefore, the equations become:
$v(t)=v(0)+at$ and $x(t)=v(0) t+\frac{1}{2} at^2$. 563