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...