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Is there a way to simplify an integral such as:

$\lgroup{\int^T_0 r(t)u(t)dt}\rgroup ^2$

The expression appears in an optimal control problem where we seek to minimize the following cost function subject to a control $u(t)$ such that $0 < u(t) < 1 $:

$\lgroup{\int^T_0 r(t)u(t)dt}\rgroup ^2 + \lgroup{\int^T_0 s(t)u(t)dt}\rgroup $

Here $r(t)$ and $s(t)$ are predetermined functions in $t$ and you must choose an admissible trajectory $u(t)$ that will minimize the cost function above.

In theory, one can find such a $u(t)$ using continuous time optimal control, except that the cost function first has to expressed in the form:

$h(x(T)) + \int^T_0 g(x(t),u(t))\space dt$

This will inevitably involve simplifying the squared integral expression in the cost function.

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.

Is there a way to simplify an integral such as:

$\lgroup{\int^T_0 r(t)u(t)dt}\rgroup ^2$

The expression appears in an optimal control problem where we seek to minimize the following cost function:

$\lgroup{\int^T_0 r(t)u(t)dt}\rgroup ^2 + \lgroup{\int^T_0 s(t)u(t)dt}\rgroup $

The issue is that I have to express this function as in the form:

$h(x(T)) + \int^T_0 g(x(t),u(t))\space dt$

Is there a way to simplify an integral such as:

$\lgroup{\int^T_0 r(t)u(t)dt}\rgroup ^2$

The expression appears in an optimal control problem where we seek to minimize the following cost function subject to a control $u(t)$ such that $0 < u(t) < 1 $:

$\lgroup{\int^T_0 r(t)u(t)dt}\rgroup ^2 + \lgroup{\int^T_0 s(t)u(t)dt}\rgroup $

Here $r(t)$ and $s(t)$ are predetermined functions in $t$ and you must choose an admissible trajectory $u(t)$ that will minimize the cost function above.

In theory, one can find such a $u(t)$ using continuous time optimal control, except that the cost function first has to expressed in the form:

$h(x(T)) + \int^T_0 g(x(t),u(t))\space dt$

This will inevitably involve simplifying the squared integral expression in the cost function.

Is there a way to simplify an expression such as:

$\lgroup{\int^A_0 r(x)u(x)dx}\rgroup ^2$

More specifically, is there a way to express such term in the form:

${\int^A_0 s(x)u(x)dx}$

or

$f(s( \space A \space )) + {\int^A_0 s(x)u(x)dx} $

Is there a way to simplify an integral such as:

$\lgroup{\int^T_0 r(t)u(t)dt}\rgroup ^2$

The expression appears in an optimal control problem where we seek to minimize the following cost function:

$\lgroup{\int^T_0 r(t)u(t)dt}\rgroup ^2 + \lgroup{\int^T_0 s(t)u(t)dt}\rgroup $

The issue is that I have to express this function as in the form:

$h(x(T)) + \int^T_0 g(x(t),u(t))\space dt$