Given is the ODE initial value problem
$$ \frac{d}{dt} x(t) = f(t, x(t)), \quad x(t_0) = x_0. $$
By integrating we obtain
$$ x(t) = x_0 + F(t, x(t)) - F(t_0, x(t_0)) = x_0 + \int_{t_0}^{t} f(s, x(s))\,\mathrm{d}s. $$
In order to solve the ODE numerically on $[t_0, t_f]$, we use a grid $G = \{t_0 < t_1 < \ldots < t_f \}$, so we have
$$ x(t_{i+1}) = x(t_0) + \int_{t_0}^{t_{i+1}} f(s, x(s)) \,\mathrm{d}s \tag{1} $$
and by applying a arbitrary quadrature rule for the integral, we obtain a numerical scheme. However, I noticed that in most books it reads
$$ x(t_{i+1}) = x(t_i) + \int_{t_i}^{t_{i+1}} f(s, x(s)) \,\mathrm{d}s \tag{2} $$
instead. Did I miss something? Or can I derive (2) from (1)?
This is just because integration is additive for subdivisions of the integration interval. \begin{align} x(t_{i+1}) &= x(t_0) + \int_{t_0}^{t_{i+1}} f(s, x(s)) \,\mathrm{d}s \tag{1} \\ &=x(t_0) + \int_{t_0}^{t_i } f(s, x(s)) \,\mathrm{d}s + \int_{t_i}^{t_{i+1}} f(s, x(s)) \,\mathrm{d}s \\ &=x(t_i) + \int_{t_i}^{t_{i+1}} f(s, x(s)) \end{align}
This motivation is not really helpful for more than the first and second order Runge-Kutta methods. With linear multi-step methods one can go a little farther with the integral form, but usually the methods are constructed using Taylor series and B-series enumerated by Butcher trees.