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y₂ = x + 1
* Mencari batas-batas
y₁ = y₂
3x² + 4x + 1 = x + 1
3x² + 4x + 1 - x - 1 = 0
3x² + 3x = 0
x² + x = 0
x(x + 1) = 0
x = 0
x = -1
* Mencari luas
Luas = x² + x dx
= ⅓x³ + ½x²
= [⅓(0)³ + ½(0)² ] - [⅓(-1)³ + ½(-1)² ]
= 0 - (-⅓ + ½)
= 0 - (-²/₆ + ³/₂)
= 0 - ¹/₆
= ¹/₆ satuan luas
Jadi, luas daerahnya adalah ¹/₆ satuan luas
Jadi: