The Calculus of Variations starts with a definition of functional
Such an expression, the argument of which is a function, is called a functional.
Particularly, they say that $J = \int_a^b{y(x)dx}$ is a functional. Yet, when $y$ is expanded with another variable, $\alpha$, they say that $J(\alpha) = \int_a^b{y(x,\alpha)dx} (5.5)$ is not a functional "because it is a function of parameter $\alpha$"! So, $J$ which has no arguments is considered as functional but $\int_a^b{y(x,\alpha)dx}$, which has function $y$ as its argument is not considered a functional. Why? Is it because name of parameter $\alpha$? In my opinion, both J do the same thing - compute the length of the path. I do not see how adding a free variable $\alpha$ to choose the path undoes this fact.