Confusion about calculus of variations

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I am currently struggling with variational calculus and hope for some clarification.

The goal of variational calculus is to find a specific path/function $y(x)$, for which the integral

$I[q] = \int_{x_1}^{x_2} F(y(x),y'(x),x) dx$

either has a maximum or a minimum. In order to find this specific path, we introduce

$\tilde{y}(x) = y(x) + \epsilon \eta(x)$,

with the additional contraint $\eta(x_1) = \eta(x_2) = 0$. Using this information, we can introduce

$I_{\eta}[q] = \int_{x_1}^{x_2} F(y(x)+\epsilon \eta(x),y'(x)+\epsilon \eta'(x),x) dx$.

In order that the integral $I$ has an extremum for the path $y(x)$,

$\frac{dI}{d\epsilon}\vert_{\epsilon = 0} = 0$

must be true. The expression above naturally leads to the Euler-Lagrange-equations:

$\frac{dI}{d\epsilon}\vert_{\epsilon = 0} = \int_{x_1}^{x_2} \Big( \frac{\partial F}{\partial y} - \frac{d}{dx}\frac{\partial F}{\partial y'} \Big)\eta(x) dx$.

My question is wether the displacements $\eta(x)$ can only be equal to zero for $x = x_1$ and $x = x_2$ or also for any $x_n \in (x_1,x_2)$, since I am not sure, wether the latter case also produces the Euler-Lagrange equations.