For the EVT, the statement is clearly potentially false if you delete endpoints. E.g., the function $f(x) = \frac{1}{x}$ on $(0,1)$ has neither a max nor a min. If you try to get cute and give it domain $(0,1]$ then it has a min but still no max. If you want to be guaranteed both, you need a closed and bounded interval.
For RT, consider the upper half of a circle. Let's take $f(x) =\sqrt{1-x^2}$ for definiteness. For the "right" RT, note that this is continuous on $[-1,1]$ and differentiable on $(-1,1)$. Connecting the dots at the ends, RT says there should be a horizontal tangent somewhere. Well, sure, it's at the north pole $(0,1)$. However, you cannot demand a version of RT that requires differentiability on $[-1,1]$ and retain this example: the function is not differentiable on the closed interval $[-1,1]$ as the derivative blows up at the endpoints (vertical tangents). This is what user nextpuzzle was getting at when they said the original version is stronger.
As you've noted, RT is roughly equivalent to the MVT, so the same reason suffices.
Consider the non-trivial case where $\int_a^bg(x) \, dx \neq 0$ and let $q = \int_a^bf(x) g(x) \, dx \, / \, \int_a^b g(x)\, dx$. Let $m = \inf_{x \in [a,b]}f(x)$ and $M = \sup_{x \in [a,b]}f(x)$. There are three cases to consider. In all cases we find $c \in (a,b)$ such that $f(c) = q$ and $\int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx.$
Case 1: Suppose $m < q < M$. By properties of the infimum and supremum, there exists $a < \alpha < \beta < b$ such that
$$m < f(\alpha) < q < f(\beta) < M \quad \text{or}\quad m < f(\beta) < q < f(\alpha) < M$$
Since $f$ is continuous there exists by the intermediate value property $c \in [\alpha,\beta] \subset (a,b)$ such that
$$f(c) = q = \frac{\int_a^b f(x) g(x) \, dx}{\int_a^b g(x) \, dx}$$
Case 2: Suppose $q = m$ and assume without loss of generality that $g(x) \geqslant 0$. Since $f(x) \geqslant m$, we have
$$\int_a^b |f(x) - m| \, g(x) \, dx = \int_a^b (f(x) - m) \, g(x) \, dx = (q -m) \int_a^b g(x) \, dx = 0,$$
and it follows that $(f(x) - m) \, g(x) = 0$ almost everywhere. In this case where $\int_a^b g(x) \, dx > 0$ we have $g(x) > 0$ almost everywhere and there must be a point $c \in (a,b)$ such that $f(c) = m$.
In Case 3 we have $q = M$ and the proof is similar to Case 2.
Best Answer
Let $f:[a,b] \to \mathbb R$ a continuous function, which is differentiable on $(a,b)$. If $f(a)=f(b)$, then Rolle's theorem says: there is $s \in (a,b)$ such that $f'(s)=0$.
Look at a proof of this theorem, then you will see:
and