Proving that there exists a unique f(x) given Area and arc-length of f(x) on a given interval I've been suggested to this site by some nice people at mathoverflow.net
Before I get started, let me tell you a little about myself. I’m a fourth year Mechanical Engineering student at the University of Michigan, Dearborn Campus. This said, I am not a mathematician, however I do know a bit about math, and I also enjoy doing math.
Please excuse any redundancies I may make and my “lack of rigor”.
The problem I’m working on now is an interesting one, and it goes as follows:
Suppose we have some unknown function of some real variable x, namely f(x), which is “well-behaved” on some known, finite interval, {a,b}. What is meant by “well-behaved” is that it is at least once differentiable, once integrate-able, smooth, and continuous on {a,b}.
Suppose further that we are given some finite area-under-the-curve A, and some finite arc-length S. Suppose finally that we are given f(x=a) and f(x=b).
The problem is this:
Given the above restraints on f(x), prove that there always exists some unique function f(x) that “fits the bill”.
For example:
Say f(x) = Sin(x), a = pi/2, and b = pi.
We know that:
f(a) = 1, f(b) = 0, A = 1, S = Sqrt[2]*EllipticE[1/2] ≈ 1.9101.
The problem in this case would be to prove that f(x) = Sin(x) is the only function that has the known area A, and known arc-length S.
Side Note: f(x) in this case could also be Cos(x - pi/2), however, this would not constitute two unique functions, f(x).
A corollary question could be posed if a proof to the first problem was given, and is as follows:
Given the same criterion for f(x) as before, devise a method of determining what f(x) would have to be to “fit the bill”. That is, like in our previous example, given: A = 1, S = 1.9101; devise a method of determining that f(x) must equal Sin(x).
Another Side Note: 
I have realized (with help from mathoverflow.net) that f(x)=(x-1)^2(x+1) and f(-x) have the same Area and arc-length on [-1,1]. Making a potentially erroneous proposition, let us say that for all polynomials of nth order P(x), P(-x) will have the same Area and arc-length on any [-c,c]. 
This would prove difficult to prove analytically due to the "ugly" expression for arc-length. 
While in general f(x) and f(-x) are unique functions, there must be some way of thinking about them as similar enough to "lump" them into one unique function... Just a thought 
Thank you all. I look forward to any insight you can shed on the problem.
 A: Consider the infinite family of functions $\sin(nx)$ on $[0,2\pi]$.  Each will have an integral of $0$, and will have the same endpoints, namely $0$ at each end.  Can we choose a constant so the arclength's are the same?  Yes we can.  
Take the slightly different family of functions $\frac{1}{n}\sin(nx)  $.  Again, the endpoints are $0$ and the integrals are the same.  But this time the arclength is the same for any $n$.  We have $$\operatorname{arclength}(\frac{1}{n}\sin(nx))=\int_0^{2\pi} \sqrt{1+(\cos nx)^2}dx$$ making the substitution $x=\frac{u}{n}$ this becomes  $$\frac{1}{n}\int_0^{2n\pi}\sqrt{1+\cos^2 x}dx=\int_0^{2\pi}\sqrt{1+\cos^2 x}dx$$ which is independent of $n$. (and where the last equality follows from the periodicity of $\cos x$)
Thus the answer to your question is no.  The example above is an infinite family of analytic functions with the same arc-length, integral and endpoints.
A: As explained by Eric and phv, such a function is not unique in general.  However, there are some cases in which it is unique.  For example, if the arclength is $\sqrt{(b-a)^2 + (f(b)-f(a))^2}$, then the only continuous function from (a,f(a)) to (b,f(b)) is the straight line.  Then there is only one possibility for the are under the curve, (b-a)(f(a)+f(b))/2
For any bigger choice of arclength from (a,f(a)) to (b,f(b)), there is a maximum possible area under the curve and a minimum possible area under the curve.  (These are symmetric problems-- if you solve one, you'll solve the other.)  For any area between these maximum and minimum values, there will be infinitely many solutions.
It is an interesting question to identify these curves with fixed arclength that give maximum and minimum area under the curve.  Can you think of any solutions, or perhaps the general solution?  The straight line is the most trivial example.
A: Perhaps I don't understand the problem, but is this proposition true at all.
Suppose let a=0, b=1, and f(x) equal 1 everywhere except on some short interval (x0, x1) where there is a smooth bump. The bump could be anywhere on the interval. 
