# Expressions for the Square of an Integral

Is there a way to simplify the following expression:

$\lgroup{\int^A_0 x(s)ds}\rgroup ^2$

I'm looking for an expression that can possibly get rid of the squared term, so that I can have just an integral of the first order.

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If $I$ is the value of the integral, why can't you just take $s(x) = I^2/(Au(x))$? This is probably not the answer you were looking for, so can you be more specific? –  Justin Shih Oct 19 '10 at 19:27
Yes, unless the OP is OK with an s depending on r and u. I agree: the question should be more precise, and maybe redirected to math.stackexchange.com –  Pietro Majer Oct 19 '10 at 19:52
It's still confusing. What are $r(t)$, $s(t)$? And you minimize the cost function under what? –  Hung Tran Oct 20 '10 at 1:22

\begin{align} &\min_{u(t) \in \Omega(t)} \, J = z(T)^2 + \int_{0}^{T} s(t)u(t)dt \\ s.t. &\frac{dz(t)}{dt} = r(t)u(t),\quad z(0) = 0 \end{align}
For instance, the derivatives wrto $A$ of the two expressions coincide choosing $s(x):=2r(x)\int_0^xr(\xi)u(\xi)d\xi$. So the two expressions coincide for all $A$ since they both vanish at $A=0$.