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 $\mathbb{R}$.