 Solution

Take compactly supported metric variations, or impose boundary conditions that remove the integration by parts terms. Let $F=f^{\prime }(R)$ and $\square ={\nabla }^a{\nabla }_a$. Since $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}{h}_{ab}$, the gravitational action varies asApplying integration by parts twice to the derivative terms givesFor the usual stress-energy tensor definition $T_{ab}=-2(-g{)}^{-1/2}\delta S_m/\delta g^{ab}$, varying the covariant metric tensor gives $\delta S_m=\frac{1}{2}\int \sqrt{-g}{T}^{ab}{h}_{ab}{d}^{4}x$. Thus the action's stated normalization impliesThere is no implicit $1/(16\pi )$ in this gravitational action. This is the normalization of the metric f(R) field equation, rather than the commonly normalized version with $8\pi T_{ab}$ on the right.

Finally, the chain rule gives ${\nabla }_a{\nabla }_{b}F=f^{\prime \prime }{\nabla }_a{\nabla }_{b}R+f^{(3)}{\nabla }_{a}R{\nabla }_{b}R$ and $\square F=f^{\prime \prime }\square R+f^{(3)}(\nabla R{)}^2$. Substitution yieldsThis is a metric f(R) gravity variation: the connection is always the Levi-Civita connection of the varied metric tensor, not an independent variable.