Find the area of the region RR bounded by
y=sin(x), y=cos(x), x=−π/3, x=13/6.
The curves intersect at -3π/4, π/4, 5π/4
cosx >= sinx on [-π/3,π/4]
sinx >= cosx on [π/4,13/6] is that a typo?
Anyway, this means that the geometric area (as opposed to the signed algebraic area) between the two curves is the sum of
∫[-π,π/4] cosx-sinx dx
+ ∫[π/4,13/6] sinx-cosx dx