In relation to this answer, why is the integrand
$$f(x,z-x)=4x(1+x-z)$$
when trying to get the pdf of $Z=X+Y$ as
$$f_Z(z)=\int 4x(1+x-z)dx$$
if $X$ and $Y$ are independent random variables where $X$ has a pdf given by $f_X(x)=2xI_{(0,1)}(x)$ and $Y$ has a pdf given by $f_Y(y)=2(1-y)I_{(0,1)}(y))$?
And of course, it is immediate by just hitting the right prompt on the YT archives. In this case, this MIT lecture.
The deal is to integrate over lines of constant $z=x+y,$ and this is done with this recipe:
$$f_Z(z)=\int_{-\infty}^{\infty}f_X(x)f_Y(z-x)\rm dx$$
which gets rid of the $y$ within the integral, and will result on a function of $z$ after integrating out $x$. This is the convolution
$$(f_X*f_Y)_Z(z)$$
The recipe is not obvious, and results from deriving the joint pdf of $X$ and $Z$ and then integrating $X$ out to get the marginal pdf of $Z$ (the desired result). The joint pdf $f_{(X, Z)}$ will call for a conditional pdf, i.e. $f_{Z\mid X}(z\mid x)$ times $f_X(x).$ The crux is to see that
$$f_{Z\mid X}(z\mid x)=f_Y(z-x)$$
which is arrived noting the iid premise.
The rest is just mindless
$$f_X(x)f_Y(y)= 2x\left(2(1-y)\right)=f_X(x)f_Y(z-x)=2x\left(2(1-(z-x) \right)$$