differentiable on open interval zero elsewhere When thinking about another question I came across this and wonder if it's true:
If $F$ is differentiable on $(a, b)$ and continuous at $[a, b]$, then given $g (x) = F^{\prime} (x)$ on $(a, b)$ and $g (a) = g (b)
= 0$, is it true that $\int_a^b g (x) d x = F (b) - F (a)$?
[edit] Am I really just using the definition of indefinite integrals here?

Suppose $F:[a, b] \to \mathbf{R}$ is continuous, differentiable in $(a, b)$, and that $g(x) = F'(x)$ if $a < x < b$, and $g(a) = g(b) = 0$. Fix a point $x_{0}$ in $(a, b)$ arbitrarily. By the fundamental theorem of calculus,
$$
\int_{x_{0}}^{x} g(z)\, dz
  = \int_{x_{0}}^{x} F'(z)\, dz
  = F(x) - F(x_{0})\quad\text{for all $x$ in $(a, b)$.}
$$
Since $F$ is continuous at $b$,
$$
\int_{x_{0}}^{b} g(z)\, dz
  = \lim_{x \to b^{-}} \int_{x_{0}}^{x} g(z)\, dz
  = \lim_{x \to b^{-}} F(x) - F(x_{0})
  = F(b) - F(x_{0}).
$$
Similarly,
$$
\int_{a}^{x_{0}} g(z)\, dz = F(x_{0}) - F(a).
$$
That is, in the sense of improper integrals,
$$
\int_{a}^{b} g(z)\, dz
  = \lim_{x \to a^{+}} \lim_{y \to b^{-}} \int_{x}^{y} g(z)\, dz
  = F(b) - F(a).
$$
