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Verified answer
ApliKasi InTegraLMengHitung Luas DaeRah
Batas IntegraL
y = y
-x² - 4x + 5 = x + 5
-x² - 5x = 0
x(x + 5) = 0
x = 0 ; x = -5
Luas daerah
= ∫((-x² - 4x + 5) - (x + 5)) dx [0...-5]
= ∫(-x² - 5x) dx
= -1/3 x³ - 5/2 x²
= 0 - (1/3 . 5³ - 5/2 . 5²)
= 125/2 - 125/3
= 125/6 satuan luas
Atau
x² + 5x = 0
D = b² - 4ac = 5² - 0 = 25
Luas = D√D /6a²
L = 25√25 / 6 . 1²
L = 125/6 satuan luas