Select the nonnegative branch of an integrated equation

$\int y\,dy=\int(2x-1)dx$ gives $y^2/2=x^2-x+A$. Thus $y^2=2x^2-2x+C$, with $C=2A$. The nonnegative branch is chosen on intervals where it is real and differentiable; an implicit algebraic identity alone does not justify a nondifferentiable join at a zero.

At $(0,1)$, $C=1$. Since $y\ge0$, $y=\sqrt{2x^2-2x+1}=\sqrt{2(x-1/2)^2+1/2}$. It is positive for every real $x$, symmetric about $x=1/2$, passes through $(0,1)$ and $(1,1)$, and has minimum $(1/2,1/\sqrt2)$. The negative square-root branch is excluded. The radical is visible on the source page even though extraction drops it.