For an open set $\Omega \subset\mathbb{R}^n$, consider the following problem:
$$\left\{\begin{matrix} - \Delta u + u=f & \mbox{ in }\ \Omega \quad \quad \quad \quad \quad \\ \ \ \ \quad \quad \ u=0 & \ \mbox{ in }\ \partial \Omega \quad \quad \quad \quad \quad \end{matrix}\right.$$
and its weak formulation $$\int_{\Omega} \nabla u \nabla v +\int_{\Omega}uv= \int_{\Omega} fv.$$
The variational approach to solve this PDE involves solving the weak formulation by means of Lax-Milgram theorem. That is, we want some $u \in H_0^1(\Omega)$ which verifies the last equation for every $v \in H_0^1(\Omega)$. Then, Lax-Milgram theorem applied to the bilinear form $$a(u,v)= \int_{\Omega} \nabla u \nabla v+ \int_{\Omega} uv$$ and the linear functional $\varphi(v)= \int_{\Omega} fv$ yields a unique solution to the weak equation.
Then it is needed to show the equivalence of weak and classical formulation to show existence and uniqueness of solution.
Now, the question is:
Why is that precise weak formulation chosen? Why not use, for example, $$\int_{\Omega} \Delta u v +\int_{\Omega}uv= \int_{\Omega} fv \ ?$$
As far as I know, it is for the sole purpose of lowering the requeriments of $u$. For the first one we need $u$ to have (at least) one derivative. For the second one we need (at least) two derivatives. Is there any other reason I am missing?