# Tag Info

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 \...