Why Does a Functional Depend on the Curve and Its Derivative?

["Don't panic!"]

First of all, apologies if this isn't quite in the right section.

I've been studying functionals, in particular pertaining to variational calculus. My query relates to defining a functional as an integral over some interval x\in [a,b] in the following manner I[y]= \int_{a}^{b} F\left(x, y(x), y'(x)\right)dx
Clearly from this we see that I is not dependent on x, but instead it depends only on the function y(x). I is a functional and as such it defines a mapping from the set of all functions y(x) satisfying y(a)=0=y(b) to \mathbb{R}.

My question really, is why the integrand a function of the set of curves y(x) (as defined above) and their derivatives y'(x) (I've kept it to first-order for simplicity, but I know that in general it can be dependent on higher orders)?

Is this because, as I is depends on every single value that y(x) takes in the interval x\in [a, b], and not just its value at a single point, we must consider how y(x) changes (i.e. we must consider it's derivatives) over this interval as we integrate over it. Thus, this implies that the integrand should be a function of the curve and it's rate of change?

Please could someone let me know if my thinking is correct, and if not, provide an explanation.

Thanks for your time.

 

["Don't panic!"]

the set of all functions y(x) satisfying y(a)=0=y(b) to \mathbb{R}.

Sorry, I meant the set of all functions y(x) satisfying \delta y(a)=0=\delta y(b) to \mathbb{R}.