Luas daerah yang dibatasi oleh y=x²-5x+6,sumbu x, dan 0

y = x² - 5x + 6
x² - 5x + 6 = 0
(x - 3)(x - 2) = 0
x = 3 dan x = 2
- (x² - 5x + 6) dx
-(⅓x³ - (5/2)x² + 6x)
- { {⅓(3³) - (5/2)(3²) + 6(3)} - {⅓(2³) - (5/2)(2²) + 6(2)} }
= - ({9 - (45/2) + 18} - {(8/3) - 10 + 12})
= - ({27 - (45/2)} - {(8/3) + 2})
= -(25 - (45/2) - (8/3))
= -( 25 - (135/6) - (16/6))
= -(25 - (151/6))
= -(150/6 - 151/6)
= -(-1/6)
= 1/6