In my Optimization class, we are messing around with some Calculus of Variations in an effort to find functions which minimize functionals. In these cases, the spaces we're working with are spaces where the vectors are themselves functions. This is particularly interesting because (when we are working with a space which is additionally an inner product space) we can take the inner product of 2 functions and get some notion of an angle between functions. A notion of an angle between 2 functions? That's a neat idea.
The reason I hesitate to say yes-- that ALWAYS a definite integral is a functional-- is because I'm wondering if there's a situation where the space I'm working in might be one where the vectors are themselves functions. In that case, maybe the definite integral would return a function, I think? But, then, maybe in that case the function would be considered a scalar relative to the space I'm working in?
Other than that situation, which I'm not knowledgeable enough to give a definite (pun) answer, I'd say yes-- all definite integrals are functionals.
It depends on your definition/understanding of "definite integral".
If you think of the Lebesgue definite integral, the answer is yes, because each function get mapped to a scalar value and this action is linear. However, this action may not be bounded.
There are, however, more general ideas of definite integrals, e.g., the Bochner integral. Think of a measure space over $\Omega$ and an arbitrary Banach space $X$. Then, you can integrate (some) functions mapping $\Omega \to X$, and the definite integral is a vector in $X$. Hence, this definite Bochner integral is not a functional, but a linear operator.