I am only answering your first question which poses the problem of the language of the physicist when he uses the integral calculus.
Take the simplest possible example: calculate the distance traveled between instant $0$ and instant $t_0$ by an object whose speed $v(t)$ varies. In physics, we will use the following language:
Between $t$ and $t+dt$ , during the small time interval $dt$, the speed practically does not vary and the distance traveled is calculated as if the speed were constant: $dx=v(t)dt$. Then we say that the total distance traveled is the sum of the elementary distances, sum which is an integral $L=\int_{0}^{t_0}dx=\int_{0}^{t_0}v(t)dt$ ( Do not forget that the sum sign in the integral is indeed a distorted S!)
It's all very simple but quite subtle. To calculate the elementary distance, we could just as well take the speed at the end of the interval $dx=v(t+dt)dt$ or perhaps better, the speed in the middle of the interval $dx=v( t+dt/2)dt$.
In reality, even if the interval $dt$ is small, none of these formulas is "exact". Obviously the error is all the smaller as $dt$ is small but one could worry: an infinite sum of very small errors could give a finite error. If I have a tape measure that is 10% too long, I might decide to do a lot of small measurements, each with a small error but in the end the total error will be the same!
The key to the problem is that the errors are of the second order in $dt$: with our language, we should write something like $dx = v(t)dt + O(dt^2)$. The number of intervals of width $dt$ between $0$ and $t_0$ is $N=t_0/dt$ so that the total error is in $N O(dt^2)=(t_0/dt)O(dt ^2)=O(dt)$ and this total error tends to $0$ when $dt$ tends to $0$.
We have the same kind of problem when we want to calculate the area under a curve $f(x)$. In physics, we would say: the area under the small interval $dx$ is $dS=f(x)dx$, calculated as if it were a rectangle. Obviously, it's not a rectangle. But the error is second order and an infinite sum of second order errors gives a first order error which tends to $0$ when $dx$ tends to $0$. The area under the curve is therefore indeed $S=\int_ {0}^{x_0} f(x)dx$ without any approximation.
Mathematicians don't have this problem because they don't use slicing as explicitly. In mathematics class, we studied Riemann sums and proved that for an integrable function, the area is well defined in this way. But experience shows that the physicist's language is sometimes difficult to replace and it is good to know it in some detail.
Edit, to complete the answer :
Showing that the error is of the second order is precisely the job of the mathematician. This can easily be seen by assuming to illustrate that the function $f(x)$ is regular and increasing between $x$ and $x+dx$. Between the rectangle with area $dS = f(x)dx$ and the curve, we have an error. This error is less than the area of a small rectangle with area $(f(x+dx)-f(x))dx$.
But the mathematics course (Mean value theorem) tells us that there exists $c$ in the interval $[x,x+dx]$ such that $f(x+dx)-f(x)=f'(c)dx$ So the error is less than $f'(c)dx^2$ which is second order.
The questions you are asking yourself are close to the philosophical questions that arose with the invention of integral calculus ("evanescent" quantities that are difficult to define). In the language of the physicist, we act as if $dx$ were finite but in the end, the limit $dx \rightarrow 0$ is still taken. So the error is zero. This is the miracle of integral calculus.
If you used numerical methods to calculate an integral, the step $dx$ would not be zero and there would remain an error in the numerical result.
Hope it can help and sorry for my poor english.
