hitunglah luas daerah yang dibatasi oleh kurva y=x2 -6x +8 dan garis y= -2x +5

y = x² - 6x + 8
y = -2x + 5
x²- 6x + 8 = - 2x + 5
x² - 4x + 3= 0
a=1 , b = - 4, c = 3
D=b²- 4ac = 4
Luas = D√D/(6a²)= 8/6 = 4/3

InTegraL
>> Luas

y1 = -2x + 5
y2 = x² - 6x + 8

Batas IntegraL
y1 = y2
-2x + 5 = x² - 6x + 8
x² - 6x + 8 + 2x - 5 = 0
x² - 4x + 3 = 0
(x - 1)(x - 3) = 0
x = 1 ; x = 3

Luas
= ∫(y1 - y2) dx [3...1]
= ∫(-x² + 4x - 3) dx
= -x³/3 + 2x² - 3x
= (-9 + 18 - 9) - (-1/3 + 2 - 3)
= 0 - (-1 1/3)
= 1 1/3 satuan luas