The unperturbed Hamiltonian is that of a particle in a one-dimensional infinite potential well of width (L), with eigenfunctions and eigenvalues: [ \psi_n^{(0)}(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right), \quad E_n^{(0)} = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}, \quad n = 1,2,3,\dots ] The perturbation is [ H_1(x) = V_0 \sin\left(\frac{\pi x}{L}\right). ] --- First-order correction: The first-order energy correction is [ E_1^{(1)} = \langle \psi_1^{(0)} | H_1 | \psi_1^{(0)} \rangle = \int_0^L \psi_1^{(0)}(x) , H_1(x) , \psi_1^{(0)}(x) , dx ] [ E_1^{(1)} = \int_0^L \left(\sqrt{\frac{2}{L}} \sin\frac{\pi x}{L}\right) V_0 \sin\frac{\pi x}{L} \left(\sqrt{\frac{2}{L}} \sin\frac{\pi x}{L}\right) dx ] [ E_1^{(1)} = \frac{2 V_0}{L} \int_0^L \sin^3\frac{\pi x}{L} , dx ] Use the identity (\sin^3 \theta = \frac{3 \sin \theta - \sin 3\theta}{4}): [ E_1^{(1)} = \frac{2 V_0}{L} \int_0^L \frac{3 \sin(\pi x/L) - \sin(3 \pi x/L)}{4} , dx = \frac{V_0}{2L} \int_0^L [3 \sin(\pi x/L) - \sin(3 \pi x/L)] , dx ] Integrate term by term: [ \int_0^L \sin(\pi x/L) dx = -\frac{L}{\pi} \cos(\pi x/L) \Big|_0^L = -\frac{L}{\pi} [\cos \pi - \cos 0] = -\frac{L}{\pi}[-1 - 1] = \frac{2L}{\pi} ] [ \int_0^L \sin(3\pi x/L) dx = -\frac{L}{3\pi} [\cos 3\pi - \cos 0] = -\frac{L}{3\pi}[-1 - 1] = \frac{2L}{3\pi} ] [ E_1^{(1)} = \frac{V_0}{2L} \left[ 3 \cdot \frac{2L}{\pi} - \frac{2L}{3\pi} \right] = \frac{V_0}{2L} \cdot \frac{(6 - 2/3) L}{\pi} = \frac{V_0}{2L} \cdot \frac{16 L}{3 \pi} = \frac{8 V_0}{3 \pi} ] So the first-order correction is [ \boxed{E_1^{(1)} = \frac{8 V_0}{3 \pi}} ]