To show that $u^* = 3t$ is a successful control (in the sense that it satisfies the boundary conditions), we can plug it into the state equation, giving $\dot{x} = 3t^2$. Integrating $\dot{x}$ to get $x^*(t)$ gives $x^*(t) = \int_{0}^{t} 3\tau^2 d\tau = t^3$. Then we see that $x^* = t^3$ (as expected from the problem statement). Plugging $t = 0$ and $t = 1$ into $x^*$, we have $x^*(0) = 0$ and $x^*(1) = 1$ which together mean that $u^*$ is a successful control.

To find the value of $J^*$, we examine $J^* = \int_{0}^{1} u^2dt = \int_{0}^{1} 9t^2dt = 3t^3\bigg|_{0}^{1} = 3$.

Now take $u = u^* + v$. If this is a successful control, then it must also generate a state trajectory which satisfies the boundary conditions. Then
\begin{equation}
x(t) = \int_{0}^{t} t(u^* + v)dt = \int_{0}^{t} tu^* dt + \int_{0}^{t} tvdt = t^3 + \int_{0}^{t}tvdt
\end{equation}
where we have used the fact that $\int_{0}^{t} tu^*dt = t^3$. This clearly satisfies $x(0) = 0$. To check $x(1) = 1$, we have
\begin{equation}
x(1) = 1 + \int_{0}^{1} tvdt = 1.
\end{equation}
Then it must be the case that $\int_{0}^{1}tvdt = 0$.

Looking at $J = \int_{0}^{1} (u^* + v)^2dt$, to minimize $J$, we'd like to minimize the term $(u^* + v)^2$ at each point in time while still satisfying the boundary conditions and the condition that $\int_{0}^{1}tvdt = 0$. We can achieve this by setting $v = 0$ for all time. Then $J = 3$ in this case as well. Then no other successful control has a lower cost than $u^*$ and $u^*$ is indeed the optimal control.