# Approaches to solve definite integral equation

What approaches can be used to solve equations like this $\int_0^L{\sqrt{1 + (f'(x)) ^ 2}}dx = s$ where s and L are known constants?

The question is whether it is possible to express $f(x)$ or its derivative i.e. just get rid of integral and have the function expression that takes into an account s and L?

UPDATE

I see that analytical solution that expresses f(x) just over s and L is not possible and it was my fault that I didn't mentioned this in my question. The question was whether it is possible to get rid of integral and come to differential equation or so that operates with s, L, f and derivatives of f. By doing so I hope to take into an account known limitations that is put on f (known length of curve) and substitute the resultant expression to another differential equation I'm already aware of.

Besides that it is known, that $s > L$ - thanks to Daryl for pointing on that.

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There are infinitely many $f$ which will have the same integral equal to $s$ (this is simply the arc length of the curve $\{(t,f(t)) \;;\; 0\leq t\leq L \}$). And infinitely many parameterized curves have the same arclength. So there is no hope to express $f$, or $f'$, in terms of $s$ and $L$. –  1015 Feb 12 '13 at 19:56

Generally, it seems no possible to find $f(x)$ since integral $\int\limits_0^L{\sqrt{1 + (f'(x)) ^ 2}}dx$ is a length of graph of $f(x)$ between vertical lines $x=0$ and $x=L$ with fixed length $s$.
In general, it is impossible to express $f(x)$ in terms of $L$ and $s$. However, in some specific instances, you can say something about $f(x)$.
Noting that $\sqrt{1+x^2}\geq1>0$, you also have $$\int_0^L\sqrt{1+(f'(x))^2}\,dx\geq\int_0^L1\,dx=L.$$ Thus, if $s<L$, then no such function $f(x)$ exists. Furthermore, if $s=L$, you must have $f'(x)=0$, which implies $f(x)=c$.
Now, consider $s>L$. As an example for why there are infinitely many functions which satisfy the integral equation for a given $s$, consider a given periodic function $f(x)$ with period $L$. There exist functions of the form, for any k, $$g(x)=f(x+k)$$ which all have the same value of the integrand.
Thank you for your answer. I've added an important update to my question and yes, we can assume $s > L$ –  Alex Feb 13 '13 at 13:49