Sum of an infinite series of integrals between two bounds The Fundamental Theorem of Calculus states that:
$$\int_a^ b f(x) \, dx = F(b) - F(a)$$
We also know that any definite integral of any function bounded by the same two points is equal to zero.
$$\int_a^ a f(x) \, dx = 0$$
So, my question is, if we add up an infinite number of integrals of a function whose bounds are the same within every integral but incremented by a very, very small number between each subsequent integral, will that be equal to the integral bounded by the explicit two points or zero? Sorry if this question is confusing, hopefully the notation below will better describe it.
$$\int_a^ b f(x) \, dx = \int_a^{a} f(x) \, dx + \int_{a+n}^{a+n} f(x) \, dx + \int_{a+2n}^{a+2n} f(x) \, dx + ...+\int_{b}^{b} f(x) \, dx$$
Where n is a very, very small number.
This confuses me, because I would assume that each of these individual integrals are equal to zero.
Edit:
You can split an integral up like shown below:
$$\int_0^a f(x) \, dx = \int_{0}^{a/2} f(x) \, + \int_{a/2}^{a} f(x) \, $$
Why is it once the bounds are the same between each split up integral that suddenly it will be equal to zero?
 A: No.
You've written $[a, b]$ as a union of points, which is not right. You should write this as a union of open intervals.
A: Assuming the existence of a primitive you have written
$$
F(a)-F(a)+F(a+n)-F(a+n)+.....+F(b)-F(b)\ne F(b)-F(a)
$$
A: Actually, you'd be performing a limit.
See that
$$\int_0^af(x)dx=\int_0^{a_1}f(x)dx+\int_{a_1}^af(x)dx$$
$$=\int_0^{a_1}f(x)dx+\int_{a_1}^{a_2}f(x)dx+\int_{a_2}^af(x)dx$$
$$\vdots$$
$$=\int_0^{a_1}f(x)dx+\int_{a_1}^{a_2}f(x)dx+\dots \int_{a_n}^af(x)dx$$
For some sequence $0<a_1<a_2<a_3<\dots<a_n<a$.
To simplify, have $\epsilon>0$.
$$\int_0^af(x)dx=\underbrace{\int_0^{\epsilon}f(x)dx+\int_{\epsilon}^{2\epsilon}f(x)dx+\int_{2\epsilon}^{3\epsilon}f(x)dx+\dots \int_{a-\epsilon}^af(x)dx}_{a/\epsilon}$$
Particularly, we want to take the limit $\epsilon\to0^+$.  Doing this gives us
$$\lim_{\epsilon\to0^+}\underbrace{\int_0^{\epsilon}f(x)dx+\int_{\epsilon}^{2\epsilon}f(x)dx+\int_{2\epsilon}^{3\epsilon}f(x)dx+\dots \int_{a-\epsilon}^af(x)dx}_{a/\epsilon}=0\times\infty$$
which is a limit of the form $0\times\infty$.  This allows it to evaluate to some value.
Also note that evaluating the limit is equivalent to Riemann sums.

Your attempt in disguise:
$$\int_0^af(x)dx=\int_0^0f(x)dx+\int_a^af(x)dx$$
$$=\int_0^0f(x)dx+\int_{a/2}^{a/2}f(x)dx+\int_a^af(x)dx$$
$$\vdots$$
$$\int_0^a f(x) \, dx = \int_0^0 f(x) \, dx + \int_{n}^{n} f(x) \, dx + \int_{2n}^{2n} f(x) \, dx + ...+\int_{a}^a f(x) \, dx$$
See the difference in how we constructed the expansion?

Also note that your idea of the limit is incorrect, as you can see with the following limit:
$$\lim_{n\to\infty}\underbrace{\frac1n+\frac1n+\frac1n+\dots\frac1n}_n=1$$
despite the individual terms approaching $0$, the total limit is equal to $1$.
A: Your equation is not true.
In order to talk about integrals we must first define what one is.
Integration is often defined by Riemann sums, so I'll assume you're
familiar with the definition or can look it up.
In particular, an introduction to integral calculus is likely to
show examples of Riemann sums using a uniform grid, that is,
dividing the interval $[a,b]$ into some number of equal pieces.
I believe in a correct understanding of Riemann integration,
nobody really cares what the function is doing at each of the finite
number of points in the "grid" between $a$ and $b$.
What we really care about is what the function is doing at all the
points in between each pair of consecutive grid points.
The only reason we look at the grid points is because
(under certain reasonable assumptions) they give us
a clue about what the function is doing in between them.
What your sum of integrals represents is the integral of the function
at each of the input values $a, a+n, a+2n, \ldots, b$.
In other words, you've covered a set of points we don't really care about,
and left out the points in between, which we really do care about.
Simply presenting an infinite sequence of finer and finer grids doesn't
change this. Until you do something that covers some interval along the
real number line, you will have integrated essentially nothing.

To put it another way, there are simply too many real numbers $x$
between any two numbers $a$ and $b$ (with $a < b$) for you to
"add up" the the value of the integral in the way you're attempting.
Any sum of that form (with each integral starting and ending at the
same value of $x$) will look at $f(x)$ at only a finite number of
values of $x$ between $a$ and $b$
(because in standard analysis, if $n$ is not exactly zero
then $b < a + Qn$ for some integer $Q$),
which is precisely $0\%$ of all the values of $x$
that lie between $a$ and $b$.
You could extend this to an infinite number of values of $x$
by taking an infinite sequence of these sums with smaller and smaller
values of $n$, but in so doing you would still have covered
precisely $0\%$ of all the values of $x$
that lie between $a$ and $b$.
