Recent content by igorrn

Thank you very much, but isn't it? \int_0^{X} \ 2cos x - X \,dx where X = \cos X. To discount the overlapping area? And could you explain me what is "numerically'? (The int should be the integral symbol. Really don't know how to use it)

I tried this: X = cos(y) → y = arccos(x) for x E(-1,1) and y E (0,2) Then: There's a point I(Xi,Yi) in which: Cos(Xi) =Arccos(Xi) Then I said area1 (file: A1) A1 = ∫cosx dx definite in 0, Xi And A2 (file:A2): A2 = ∫cosy dy definite in 0, Yi And the overlapping area as A3 (file: A3): A3 = ∫Yi dx...