Well,
Let $M$ a compact manifold and $dM = \star 1$ the volume element, $f : M \to \mathbb{R}$. Define the inner product of $p-forms$ like: $(\omega,\theta) = \int_M \langle \omega, \theta \rangle dM = \int_M \omega\wedge \star \theta$ Is true that :
$$\int_M \Delta f \star 1 = 0$$
because $(\Delta f,1) = (f,\Delta 1) = 0$.
But, if we put $f : [0,1] \to \mathbb{R}$ been $f(x) = x^3$ . So, $\Delta(f) = 6x$. Then, $\int_{[0,1]} f(x)dx = 3(1)^2 = 3$. So, what is wrong ? Is this theorem false ?
The Laplacian is self adjoint only if you apply it to functions with suitable boundary conditions. Try to show that the Laplacian is self adjoint on a Euclidean domain or an interval and you will see boundary terms appear. This is what happens in your example.
If you work with a closed manifold, there is no boundary and hence no issue.