If the area of the region \(x,y):0 y \2x,6x-x2 \ \ is A, then 12A is equal to .
Answer
304
Solution
2x=6x-x2 x=0,4; y 0 needs 0 x 6
[figure]
=2x on [0,4], =6x-x2 on [4,6] (shaded region)
A1= 12 4 8=16
A2= 46 (6x-x2 )dx= [3x2- x33 ]46=36- 803= 283
A=16+ 283= 763 12A=304
Answer: 304