Question about an integration technique in Tveito-Winther's Introduction to Partial Differential Equations I do not understand the last line in the following computation, taken from pgs 40-41 of Tveito and Winther's book Introduction to Partial Differential Equations: A Computational Approach [TW]:

...we define $$F(y) = \int_0^y f(z) \; dz,$$ and observe that
\begin{align*}
\int_0^x \left( \int_0^y f(z) \; dz \right) &= \int_0^x F(y) \; dy \\
&= [yF(y)]_0^x - \int_0^x y F'(y) \; dy \\
&= xF(x) - \int_0^x yf(y) \; dy \\
&= \int_0^x (x - y)f(y) \; dy.
\end{align*}

It seems as though the "missing steps" connecting the last two lines are:
\begin{align}
\tag{1} xF(x) - \int_0^x yf(y) \; dy &= x\int_0^x f(y) \; dy - \int_0^x f(y) \; dy \\
\tag{2} &= \int_0^x x f(y) \; dy - \int_0^x yf(y) \; dy \\
&= \int_0^x (x - y) f(y) \; dy,
\end{align}
where in (1), I am using the definition $F(y) = \int_0^y f(z) \; dz$, but rewriting it as $F(x) = \int_0^x f(y) \; dy$...if that is allowed, I do not understand why. And then in (2), I am moving the variable $x$ inside the integral sign and, again, if that is allowed I do not understand why.
Any help will be greatly appreciated!
 A: Some of the topics mentioned in your comments are quite delicate, and may (or may not) be studied in honors courses in calculus/analysis, but what you have in your main question is much simpler.
It should be familiar from basic calculus that
$$\int_a^b f(x) \, dx = \int_a^b f(u) \, du = \int_a^b f(t) \, dt = \cdots$$
(since all those integrals represent the same “area below the graph”), and this is what is meant by saying that $x$ (or $u$ or $t$) is a “dummy variable” – it can be replaced by any other currently unused variable without changing the meaning of the expression. And the fact that a formula like $f(y) = y^2$ defines the same function $f$ as the formula $f(x) = x^2$ should be pre-calculus knowledge.
So the formula $$F(y) = \int_0^y f(z) \, dz$$ defines the same function $F$ as the formula $$F(x) = \int_0^x f(z) \, dz$$ (since you have just replaced $y$ with $x$ everywhere), and this is the same thing as $$F(x) = \int_0^x f(y) \, dy$$ (since you have just replaced the dummy variable $z$ with $y$, which is fine now that $y$ isn't used in the formula any longer; note that writing $F(y) = \int_0^y f(y) \, dy$ would generally be frowned upon).
Also, when you're integrating with respect to the variable $y$, the variable $x$ is just like any other constant, and can therefore be freely moved in or out of the integral, just like the factor $2$ in $\int_a^b 2 \cos y \, dy = 2 \int_a^b \cos y \, dy$, for instance.
