How to prove these basic concepts about integration? While we were introduced to integration, we were told about some basic concepts that, as we were told, could not be proved based on our level of sophistication. They are as follows:


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*$$\int_a^b \! f(x) \, \mathrm{d}x=\phi(b)-\phi(a),$$ where $\phi$ is a primitive of $f$ in $[a,b]$

*$$\int_b^a \! f(x) \, \mathrm{d}x=-\int_a^b \! f(x) \, \mathrm{d}x$$


When I learned Riemann Integral, I thought I would be able to prove them, as they didn't seem to be too out-of-the-earth type. So, my question is what is the concept of a primitive according to Riemann Integration? And how can it be used to prove $(1)$? If there isn't any, then where should I look?
Also, how can I even conceptualise $(2)$? I mean, Riemann Integration is defined using Sums. How can I Sum in the opposite way? Does it even matter? And how does it become negative?
I'm just a curious high school student, familiar with the basic concepts of Riemann Integration. So, spare me if my questions are too dumb. And it'd be great if you can suggest some study material where I should look for these type of concepts. Thank you in advance.
 A: The first question is the fundamental theorem of calculus: it says that if there is a differentiable function $F$ with $\frac{dF}{dx} = f$, and $f$ is Riemann integrable on $[a,b]$, then $\int_{a}^b f(x) dx = F(b) - F(a)$. This can be proved from the definitions relatively easily, but not obviously. It will be covered in any book on elementary real analysis, a good book on calculus, or Wikipedia (NB: When I was a curious high school student, Wikipedia was a godsend for explaining things that didn't make sense or weren't explained in calc class!)
The second statement should be considered a definition. For $a < b$, we define the integral $\int_{a}^{b} f(x) dx$ to be the integral of $f$ with respect to the interval $[a,b]$ (note that the definition of the integral only really depends on the interval!). Then, for any $a,b,c$ with $a < b < c$, we have $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx$ (which is easy to see from the definitions). Now, we want to assign meaning to the symbol $\int_b^a f(x) dx$ with $a \leq b$. We can do this in a unique way such that the addition formula above holds for any $a,b,c$. First, plugging in $b = c$, the formula reads $\int_a^b f(x) dx + \int_b^b f(x) dx = \int_a^b f(x) dx$, so $\int_b^b f(x) dx = 0$. Then, if $a < b$, we can plug in $c = a$ to get $\int_a^b f(x) dx + \int_b^a f(x) dx = \int_a^a f(x) dx = 0$, so $\int_b^a f(x) dx = - \int_a^b f(x) dx$. 
A: You cannot associate any physical interpretation to $\int_b^a f(x) \ dx$
You're right, the sum would be the same irrespective of whether you sum it forward or backward. But $\int_b^a f(x) dx$ doesn't connote taking the sum in the opposite manner. If it were then $\int_b^a f(x) dx$ would have been equal to $\int_a^bf(a+b-x)dx$ which doesn't happen.
However, this is a useful concept since it allows us to write $\int_a^bf(x) dx = \int_a^cf(x) dx + \int_c^bf(x) dx$ even when $a < b < c$
A: $\int_a^b \! f(x) \, \mathrm{d}x=g(b)-g(a)$ is a result obtained directly from the fundamental theorem of calculus.
Say there's a function $f(x)$
Now the area under the curve from $0$ up til $x$ can be found using Reimann sums. Let A(x) = $\int_0^x f(t) dt$ where A(x) is the area function that gives you the area under the curve of $f(x)$ from $0$ up to an $x$ value
It can be easily shown that $A '(x) = f(x)$ (https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_meaning)
Which means, $A(x)$ is an antiderivative of $f(x)$
Using this result, we can compute areas under the curve of a function easily.
Depending on your choice of antiderivative, you can calculate the area from any point $a$ up to $x$ under the curve. For e.g if you want to find the area from $0$ up to $x$, you will choose an antiderivative g(x) that satisfies the relation g(0) = 0.
Now say you want to find the area from $a$ up to $b$.
To make things easier for us, we can choose any antiderivative g(x) we like, evaluate the value of $g(b)$ ie. find the area under the curve up till $b$ and from it subtract $g(a)$ ie the area under the curve up to $a$ which will leave us with the area from $a$ to $b$
The proof for the above can be found here -https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Proof_of_the_second_part
