luas daerah yang dibatasi kurva f(x)=-x^2+4x+12 dan g(x)=x^2-2x-8 adalah... A.162 2/3 satuan luas B.114 1/3 C.96 2/3 D.81 1/3 E 58 2/3

f(x) = -x^2 + 4x + 12
g(x) = x^2 - 2x - 8

Mencari titik potong kurva
f(x) = g(x)
-x^2 + 4x + 12 = x^2 - 2x - 8
2x^2 - 6x - 20 = 0
(2x + 4)(x - 5) = 0
2x + 4 = 0 atau x - 5 = 0
2x = -4 atau x = 5
x = -4/2
x = -2
Luas daerah dar x = -2 hingga x = 5
L = (b)integral(a) (f(x) - g(x))dx
L = (5)integral(-2) (-x^2+4x+12 - (x^2-2x-8))dx
L = (5)integral(-2) (-2x^2 + 6x + 20)dx
L = [-2/3 x^3 + 3x^2 + 20x](5)<->(-2)
L = [-2/3(5)^3 + 3(5)^2 + 20(5)] - [-2/3(-2)^3 + 3(-2)^2 + 20(-2)]
L = [-250/3 + 75 + 100] - [16/3 + 12 - 40]
L = [-250/3 + 225/3 + 300/3] - [16/3 + 36/3 - 120/3]
L = 275/3 - (-68/3)
L = (275 + 68)/3
L = 343/3
L = 114 1/3
Jadi luas daerah tersebut adalah
114 1/3 satuan luas (B)