I am reading the book Topics in Advanced Quantum Mechanics by Holstein. In Chapter 3, section 3 he discusses the Aharonov-Bohm effect, but before doing so he discusses the ordinary double slit experiment. This book is based on the path integral approach and what he does is as follows: First we assume that the two slits are the same distance $d_0$ from the particle source (located at $x_i$), and that the first slit, at a point $x_1$, is a distance $d_1$ from the detection point and the second slit at $x_2$ is a distance $d_2$ from the detection point, call it $x_f$. Now the action for the classical particle is simply the free particle action $S = \int_0^{t_f}dt\frac 12 m \dot x^2$. Holstein then states, that assuming that the classical path is dominant, the contribution to the path integral from the path through slit 1 is $$\exp(i\int_0^{t_f}dt\frac 12 m \dot x^2)\approx \exp(i\frac{2\pi d_0}{\lambda}+i\frac {2\pi d_1}{\lambda}),\tag{3.3}$$ and similarly for the path through slit 2, just replacing $d_1$ with $d_2$, and where $\lambda$ is the de Broglie wavelength.
My question is, how is this derived? Holstein simply states it, offering virtually no justification. I understand how, using this formula, one can derive the diffraction pattern for the double slit experiment, but I don't understand how this formula is derived.
Edit: $\hbar =1$ throughout.