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(x-3)(x-1)
jadi batas atas dan batas bawahnya adalah 3 dan 1
∫((x²-4x+3)-(x-1))dx
∫(x²-5x+4)dx
=(1/3)x³-(5/2)x²+4x
=[(1/3)3³-(5/2)3²+4(3)]-[(1/3)1³-(5/2)1²+4(1)]
=(9-(45/2)+12)-((1/3)-(5/2)+4)
= (-3/2 - 11/6)
= (-9/6 - 11/6)
= (-20/6)
= -10/3
karena Luas maka hasilnya selalu positif L= 10/3 satuan luas