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(0, -2) --> (x1, y1)
(2/3, 0) --> (x2, y2)
(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)
(y - (-2))/(0 - (-2)) = (x - 0)/(2/3 - 0)
(y + 2)/2 = x/(2/3)
(2/3)(y + 2) = 2x (kali 3 kedua ruas)
2(y + 2) = 6x
2y + 4 = 6x
2y = 6x - 4
y = 3x - 2
Batas x yg dibatasi adalah 1 sampai 3
Maka luas daerahnya adalah
= ∫ y dx (dari 1 sampai 3)
= ∫ (3x - 2) dx (dari 1 sampai 3)
Jawaban: B