Q: Givens : $f(x)=x^2$ , $F_1 (x)=\int_0^x f(t)dt$, $F_2(x)=F_1(x)+4$. Question : express $F_2$ with the form $ F_2=\int_b^x f(t)dt$, i.e. find $b$. Givens : $f(x)=x^2$ , $F_1 (x)=\int_0^x f(t)dt$, $F_2(x)=F_1(x)+4$. Question : express $F_2$ with the form $ F_2=\int_b^x f(t)dt$, i.e. find $b$.
For a better undersanding of this example : https://www.desmos.com/calculator/wmr887ecoh

The title of my question may not be as clear as desirable.
Here is what I mean.
Givens of the problem :
$(1)$ Suppose that $f$ is a continuous function, let's say on $\mathbb R$.
$(2)$ Suppose that both $F_1$ and $F_2$ are primitives of $f$, and that $F_2(x) = F_1(x)+C $ with $C\neq 0$ and $C$ known to us.
$(3)$ Suppose also that both $F_1$ and $F_2$ are integrals of $f$.
Note : I add this hypothesis explicitly since not all primitives of a function are also integrals of this function, if, as it seems to me, the converse of the FTC is not true.
$(4)$ Suppose that we know the lower limit of integration of $F_1$ , I mean, we know the value of $a$ in the true statement
$$ F_1 = \int _a ^x f(t)dt.$$
( I claim the statement is true for some $a$, since  , by hypothesis, $F_1$ is an integral of $f$).
(5) Since $F_2$( by hypothesis) is also an integral of $f$, it must be true for some $b \space ( \neq a)$ that :
$$ F_2 = \int_b^x f(t)dt.$$
My question ( in case the situation I've described actually makes sense) is :

assuming we know the value of $C$  (with $C = F_2(x)-F_1(x)) $) , and the value of $a$ as said above  , is it possible
to determine the value of $b$?

Or, in short :

how to relate the way $F_1$ and $F_2$ differ as primitives , and the
way they differ as integrals ( the first difference being a matter of
" constant $C$" and the second difference being a matter of lower
limit of integration).


A: If you write the difference:$$F_2(x)-F_1(x)=\int_b^xf(x)dx-\int_0^xf(x)dx$$
you can use $$\int_\alpha^\beta f(x)dx=-\int_\beta^\alpha f(x) dx$$
Then $$F_2(x)-F_1(x)=\int_b^0f(x)dx$$
You can then just plug in $f(x)$ to calculate $b$.
