Question

Luas daerah terbatas di parabola y = -x + 4 dan di atas garis lurus y = 3x adalah

Answer 1

Verified answer

Aplikasi InTegraL
(luas)

y1 = -x² + 4
y2 = 3x

Batas IntegraL
y1 = y2
-x² + 4 = 3x
x² + 3x - 4 = 0
(x + 4)(x - 1) = 0
x = -4 ; x = 1

Luas
= ∫(y1 - y2) dx [1...-4]
= ∫(-x² - 3x + 4) dx
= -x³/3 - 3x²/2 + 4x
= (-1/3 - 3/2 + 4) - (64/3 - 24 - 16)
= -11/6 + 4 + 56/3
= 101/6 + 4
= 125/6 satuan luas


Cara lain
y1 = y2
x² + 3x - 4 = 0
D = b² - 4ac = 3² - 4.1(-4) = 25

Luas = D√D /6a²
L = 25√25 /6
L = 125/6 satuan luas