Payoff from an option contract
When integrals look different than what pops into your head, often the reason is integration by parts. For your example note that
$$\int_R^1 (\theta -R) g(\theta) d \theta + \int_R^1 G(\theta) d \theta = (1-R) - 0,$$
where the right-hand side is equivalent to $\int^1_R 1 d\theta$.
Hence, the two expressions you consider are equivalent.
It's of the form
$$\int u(x) v'(x) dx + \int u'(x) v(x) dx = \int u(x) v(x) dx.$$