tentukan luas bidang yang dibatasi oleh kurva y^2=4x dan garis 4x-3y=4

4x - 3y = 4 ----> x = (3y + 4)/4

y^2 = 4x
y^2 = 4(3y + 4)/4
y^2 = 3y + 4
y^2 - 3y - 4 = 0
D = (-3)^2 - 4(1)(-4) = 9 + 16 = 25
Luas = D√D/6a² = 25√25 / 6(1)² = 125/6