Question

Luas daerah yang dibatasi oleh kurva y = x^2 dan y = 4x-x^2

Answer 1

Y = x^2
y = 4x - x^2
y1= y2

x^2 = 4x - x^2
2x^2 - 4x = 0
a = 2
b = -4
c = 0

D = b^2 - 4ac
D = (-4)^2 - 4(2)(0)
D  = 16

Luas = D√D/(6 a²)
L = 16√16/ (6.2²)
L = 16(4)/ (6.4)
L = 16/6 = 8/3