Brain Teaser - Area enclosed by a parabola and a line

[Gerson W. Barbosa]

(09-15-2015 02:59 PM)fhub Wrote:  Well, I get a different result:
u=0.743514881896 with A=1.22954355289

I think I can blame it on my Simpson 3/8 integrator in Pascal (too many iterations required!). Your result is indeed a minimum, as checked in W|A:

solve 0.608258988304(1.24590156402-x) = x^4 to 12 digits

integrate 0.608258988304(1.24590156402-x) - x^4, x=-1.09201722595,0.74350488189 to 12 digits


solve 0.608234446071(1.24595886588-x) = x^4 to 12 digits

integrate 0.608234446071(1.24595886588-x) - x^4, x=-1.09201232966,0.74351488189 to 12 digits

solve 0.608209905131(1.24601617162-x) = x^4 to 12 digits

integrate 0.608209905131(1.24601617162-x) - x^4, x=-1.09200743451,0.74352488189 to 12 digits


u = 0.743504881896 Area = 1.22954355433

u = 0.743514881896 Area = 1.22954355289

u = 0.743524881896 Area = 1.22954355432


Gerson.