We have known already the proposal of Archimedes to cover the parabolic segment, but it gives an approximation of the area. On this page, we will see how obtain the exact area of the surface.

Area of the parabolic segment= Area of BCA +Area of DAB + Area of BGC +Area of ODA +Area of MBD + Area of BIG +Area of GKC +... [br][br]Moreover, you can confirm that these equalities:

So, if we introduce these equivalences in the first equation, we will get the following result:

In conclusion, the area of the parabolic segment is 4/3 of the area of the first triangle inside it