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The area of the region enclosed by the parabolas y=4x-x2 and 3y=(x-4)2 is equal to :
(4) 6
[figure]
Intersection: 3(4x-x2)=(x-4)2 x2-5x+4=0 x=1, 4
Area = 14 [(4x-x2)- (x-4)23 ]dx
= [2x2- x33- (x-4)39 ]14
= (32- 643-0 )- (2- 13+3 )= 323- 143=6
Answer: (4)
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If the area of the region \(x,y):0 y \2x,6x-x2 \ \ is A, then 12A is equal to .