Equation for the work done to bring in an small amount of free charge in a dielectric

Griffiths says:

Suppose the dielectric material is fixed in position, and we bring in the free charge, a bit at a time. As $rho_{f}$ increased by an amount $triangle rho_{f}$, the polarization will change and with it the bound charge distribution; but we’re interested only in the work done on the incremental free charge:
$$
Delta W=intleft(Delta rho_{f}right) V d tau
$$

How is the equation given correct? Why is he integrating?
To me the small work should be, the charge brought ,times the voltage there is i.e

$Delta W=Q V$ ,hence

$Delta W=Delta rho_{f} cdot V cdot d tau$