3
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
$$\...
Only top voted, non community-wiki answers of a minimum length are eligible
