What is a functional? And how is it defined for the length?

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Im reading about Calculus of varations and there is a lot of references to "the functional" i.e we want to find the minimum of the functional etc. From what i have read, "the functional" is simply the function we are integrating, is this correct - but then if so, what is the difference?

Secondly, a number of times, the lentgh of a curve is given as $$ \int_a^b \sqrt{(1+y'^2)} dx$$ with functional therefore $$\sqrt{(1+y'^2)} $$ and am unsure on why this is. Thank you very much.

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A functional is a mapping from a vector space to the underlying scalar field. So, for example one could define a functional from $ C^{1}([a,b])$ to $\mathbb{R}$ by

$J[u]=\int_{a}^{b}\sqrt{1+(u')^{2}} dx$

Of course, there are other choices for the domain.

For a derivation of the arc length formula you could consult the Wikipedia page on arc length (http://en.wikipedia.org/wiki/Arc_length). This would also be in a calculus textbook (e.g. Stewart's Calculus). If you have any specific questions about the derivation you could come back and ask. You essentially partition the interval, approximate the curve with a polygon, express the length of the polygon as a Riemann sum and then take a limit.

Edit: Based on the comment I'm going to try to elaborate on what a functional is.

A functional is a specific kind of function. It takes a vector (element of a vector space) as an input and returns a scalar (element of the underlying scalar field). In the calculus of variations you will deal with functionals that map from some function space to the real numbers. For example, $C^{1}([a,b])$ is the vector space consisting of all continuously differentiable (real-valued) functions on $[a,b]$. The underlying scalar field is $\mathbb{R}$. $J[u]$, as defined above, is a functional that takes elements of $C^{1}([a,b])$ and maps them to real numbers.

Another example would be to consider

$I[f]=\int_{0}^{1} f~ dx$

where $f \in C([0,1])$. This would be a functional mapping $C([0,1])$ to $\mathbb{R}$.