I have the following functions
$g(x) = 16x^2(1-x)^2$ on $[0,1]$
$f(x) = 1 - g(x)$ on $[0,0.5]$ and $f(x) = 0$ on $[0.5,1]$
$h(x) = 0$ on $[0,0.5]$ and $h(x) = 1 - g(x)$ on $[0.5,1]$
As one can infer from the definitions, $f(x)$ and $h(x)$ are piecewise functions that are continuous and differentiable (to emphasize, even at $x=0.5$).
Are the following analytic statements true about the functions and their derivatives ($f^\prime$, $g^\prime$, $h^\prime$)? (Note: $f(x)$ is denoted by $f$, and likewise for other functions)
$f + g + h = 1$ on $[0,1]$
$f + g = 1$ on $[0,0.5]$
$g + h = 1$ on $[0.5,1]$
$f^\prime + g^\prime + h^\prime = 0$ on $[0,1]$
$f^\prime + g^\prime = 0$ on $[0,0.5]$
$g^\prime + h^\prime = 0$ on $[0.5,1]$
If the above are true, can I apply them to write the following definite integrals?
$\int_{0}^{1} g(x) dx = \int_{0}^{1} [1-f(x)-h(x)] dx = \int_{0}^{0.5} [1-f(x)] dx + \int_{0.5}^{1} [1-h(x)] dx$
$\int_{0}^{1} g^\prime(x) dx = \int_{0}^{1} [-f^\prime(x)-h^\prime(x)] dx = \int_{0}^{0.5} [-f^\prime(x)] dx + \int_{0.5}^{1} [-h^\prime(x)] dx$
Saying that $f+g=1$ on an interval (or any set for that matter) just means that $f(x)+g(x)=1$ for every $x$ from that set. You just need to make sure that both functions are defined there, which you certainly did.
As far as I can tell (having verified them myself), all of your statements are true.