Here is my question statement. I cannot understand the last equality.
Let $U=[-1,1]$. \begin{equation} \mathcal{U}[0, T] = \left\{ u:[0,T] \rightarrow U \mid u \text{ is } \{\mathcal{F}_t\}_{t\geq0}\text{-adapted} \right\} \end{equation} For $t\in [0,1]$ \begin{equation} \begin{cases} dX_t = u(t)dB_t \\ X_0 = 0 \end{cases} \end{equation} Then \begin{equation} X_t = \int_0^t u(s)dB_s. \end{equation} \begin{align} J(u(\cdot)) &= E \left\{ \int_0^1 X_t^2 - \frac{1}{2}u(t)^2 dt + X_1^2\right\}\\ &= E \left\{ \int_0^1 \left( \frac{3}{2}-t \right) u(t)^2 dt\right\}. \end{align}
I attached the reference where my question arise. On page 117 in this book. It just mention "simple calculation".
Yong, Jiongmin, and Xun Yu Zhou. Stochastic controls: Hamiltonian systems and HJB equations. Vol. 43. Springer Science & Business Media, 1999.
In summary
How is it possible? I want to know the procedure. \begin{align} E \left\{ \int_0^1 X_t^2 - \frac{1}{2}u(t)^2 dt\right\} &= E \left\{ \int_0^1 \left( \frac{3}{2}-t \right) u(t)^2 dt\right\}. \end{align}
Well, simple it is: $$E \left[ \int_0^1 X_t^2 dt\right] = \int_0^1 E\left[X_t^2\right] dt = \int_0^1 \int_0^t E\left[u(s)^2\right]ds\, dt \\= \int_0^1 \int_s^1 E\left[u(s)^2\right]dt\, ds = \int_0^1 E\left[(1-s)u(s)^2\right] ds = \int_0^1 E\left[(1-t)u(t)^2\right] dt.$$
Hence your equality follows, taking into account that $E\left[X_1^2\right] = \int_0^1 E\left[u(t)^2\right] dt$.