I am reading Modern Many-Particle Physics by Lipparini. In chapter 1.5 he talks about the matrix elements of the one-body operator: $$F_1 = \sum_{i=1}^{N}f(x_i)$$ He mentions that the matrix elements are given by: $$ \langle \phi | F_1|\psi \rangle=\int dx_1...dx_N \phi^*(x_1,...,x_N)\sum_{i=1}^{N}f(x_i) \psi(x_1,...,x_N)=N\int dx_1...dx_N \phi^*(x_1,...,x_N)f(x_1) \psi(x_1,...,x_N)$$ and that this is true for both fermions and bosons. My doubt is in the equality between the two integrals. I understand how they are equal in the case for bosons, but why the equality holds for fermions?
For example, if we have $N=2$ then $\psi(x_1,x_2)=\phi_1(x_1)\phi_2(x_2)-\phi_1(x_2)\phi_2(x_1)$. Then, we would have: $$ \sum_{i=1}^{N}f(x_i) =[f(x_1)+f(x_2)] \psi(x_1,x_2) =\big( [f(x_1)\phi_1(x_1)]\phi_2(x_2)-\phi_1(x_2)[f(x_1)\phi_2(x_1)] \big) +\big( \phi_1(x_1)[f(x_2)\phi_2(x_2)]-[f(x_2)\phi_1(x_2)]f(x_1)\phi_2(x_1) \big) $$ inside the first integral. But this is not equal to twice the first term (only applying $f(x_1)$.
Of course, am assuming I am doing something wrong, but what is it?