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y1 = y2
x² - 2x = 2x - 3
x² - 4x + 3 = 0
(x - 3)(x - 1) = 0
x = 3 dan x = 1
∫dengan batas atas = 3 dan batas bawah = 1 dari (x² - 4x + 3 ) dx
substitusi batas atas = 3 dan batas bawah = 1 ke ( (x³)/3 - 2x² + 3x )
=((3³)/3 - 2(3)² + 3(3) ) - ( (1³)/3 - 2(1)² + 3(1))
= ( 9 - 18 + 9 ) - ( 1/3 - 2 + 3)
= (0) - (4/3()
= -4/3
karena luas daerah bernilai positif
maka luas daerah = 4/3 satuan