Brain Teaser - Area enclosed by a parabola and a line

[Gerson W. Barbosa]

(09-13-2015 04:23 PM)Thomas Klemm Wrote:  Thus:
\(x-u+2u(y-u^2)=0\)

From the equation above we can find the formula for \(g(x)\):

\(y=\frac{u-x}{2u}+u^2\)

Things start to get complicated for the biquadratic parabola, if I've done it right:

\(y=\frac{4u^{7}+u-x}{4u^{3}}\)

u: 0.760000000000 Area: 1.23320954839
u: 0.750000000000 Area: 1.23009174283
u: 0.740000000000 Area: 1.22972179352
u: 0.745000000000 Area: 1.22956112013
u: 0.744000000000 Area: 1.22953780108
u: 0.743750000000 Area: 1.22953629609
u: 0.743687500000 Area: 1.22953619031
u: 0.743671875000 Area: 1.22953618081
u: 0.743664062500 Area: 1.22953617857
u: 0.743660156250 Area: 1.22953617806
u: 0.743659179690 Area: 1.22953617801
u: 0.743658691410 Area: 1.22953617799
u: 0.743658203125 Area: 1.22953617797
u: 0.743656250000 Area: 1.22953617799
u: 0.743652798750 Area: 1.22953617828
u: 0.743652343750 Area: 1.22953617836
u: 0.743648437500 Area: 1.22953617913
u: 0.743640625000 Area: 1.22953618202
u: 0.743625000000 Area: 1.22953619273
u: 0.743500000000 Area: 1.22953652222
u: 0.743000000000 Area: 1.22954217016

Dull figures... Hopefully I'm mistaken :-)

Cheers,

Gerson.

Edit to include correct values. The table above is wrong, except for u=0.74, due to a silly error in my first Pascal program. I chose a numeric method because it's easy to extend the problem to x^6, x^8, x^10... The result is accurate to 8 digits, but more are possible (Three pairs of values are used to obtain the equation of the parabola that encompasses them, then its minimum is used as a new u).


u: 0.730000000000 Area: 1.232203807010

u: 0.740000000000 Area: 1.229721793524

u: 0.741869418802 Area: 1.229582549556

u: 0.743490000000 Area: 1.229543561785

u: 0.743514882370 Area: 1.229543552881

u: 0.743515349953 Area: 1.229543552884

u: 0.743543197140 Area: 1.229543564411

u: 0.744136760955 Area: 1.229549111764

u: 0.75000000000o Area: 1.230145031369