Solution: At first assume that $\lambda =0$. This implies we have the equation \begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}x^2} = 0 \implies u(x) = ax + b. \end{align*} Applying the given conditions, we obtained \begin{align*} u(0) = 0 & \implies b = 0 \\ \int _0^{\pi /2} u(t) \mathrm{d} t = 0 & \implies \frac{a}{2}\left( \frac{\pi ^2}{4} \right) = 0 \implies a = 0. \end{align*} Thus, $u(x) = 0$. So $\lambda =0$ does not give any nontrivial solution. So let us assume $\lambda \neq 0$.

If $\lambda\neq 0$, then the given differential equation has a solution given by \[ u(x) = a \cos \lambda x + b \sin \lambda x. \] Applying the conditions, we have \begin{align*} u(0) = 0 & \implies a = 0 \\ \int _0^{\pi /2} u(t) \mathrm{d} t = 0 & \implies \int _0^{\pi /2} b\sin \lambda x \mathrm{d} x = 0 \\ & \implies \frac{b}{\lambda } \big[-\cos \lambda x\big]_0 ^{\pi /2} = 0\\ & \implies b \left( 1 - \cos \frac{\lambda \pi }{2} \right) = 0 \\ & \implies b =0 \text{ or } \cos \frac{\lambda \pi }{2} = 0. \end{align*} If $b = 0$, then again we will obtain a trivial solution. So, a nontrivial solution to \eqref{eq:24May2024-1} exists if \begin{align*} \cos \frac{\lambda \pi }{2} = 1 & \implies \frac{\lambda \pi }{2} = 2n\pi ,\quad n \in \mathbb{Z}\setminus \{ 0 \} \\ & \implies \lambda = 4n,\quad n\in \mathbb{Z} \setminus \{ 0 \} . \end{align*} Thus, the values of $\lambda ^2$ for which the given boundary value problem has a nontrivial solutions are \[ \lambda _n = 16n^2, \quad n \in \mathbb{N} . \]