vantuin ya kak,makasih

semoga membantu kak :)

a) titik potong kedua kurva :
x^2 - 1 = x + 1
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2 atau x = -1
Luas = (-1) int (2) [(x + 1) - (x^2 - 1)] dx
= (-1) int (2) [x - x^2 + 2] dx
= 1/2 x^2 - 1/3 x^3 + 2x | batas -1 sampai 2
= (1/2 . 2^2 - 1/3 . 2^3 + 2 . 2) - (1/2 (-1)^2 - 1/3 (-1)^3 + 2(-1))
= (2 - 8/3 + 4) - (1/2 + 1/3 - 2)
= 6 - 8/3 - 1/2 - 1/3 + 2
= 8 - 9/3 - 1/2
= 4 1/2
= 9/2

b) luas bagian kiri
= (-1) int (0) [(x + 1) - (x^2 - 1)] dx
= (-1) int (0) [x - x^2 + 2] dx
= 1/2 x^2 - 1/3 x^3 + 2x | batas (-1) sampai 0
= 0 - [1/2 (-1)^2 - 1/3 (-1)^3 + 2(-1)]
= 0 - 1/2 - 1/3 + 2
= (-3 - 2 + 12)/6
= 7/6
Luas bagian kanan
= (0) int (2) [(x + 1) - (x^2 - 1)] dx
= (0) int (2) [x - x^2 + 2] dx
= 1/2 x^2 - 1/3 x^3 + 2x | batas 0 sampai 2
= [1/2 . 2^2 - 1/3 2^3 + 2(2)] - 0
= 2 - 8/3 + 4
= 6 - 8/3
= 10/3 = 20/6
Perbandingan​ = 7/6 : 20/6 = 7 : 20

c)luas bagian bawah
= (-1) int (1) [-(x^2 - 1)] dx
= (-1) int (1) [-x^2 + 1] dx
= -1/3 x^3 + x | batas -1 sampai 1
= (-1/3 (1)^3 + 1) - (-1/3 . (-1)^3 + (-1))
= -1/3 + 1 - 1/3 + 1
= -2/3 + 2
= 4/3
= 8/6

Luas bagian atas
= (-1) int (1) [x + 1] dx + (1) int (2) [(x + 1) - (x^2 - 1)] dx
= (-1) int (1) [x + 1] dx + (1) int (2) [x - x^2 + 2] dx
= 1/2 x^2 + x | batas (-1) sampai 1 + (1/2 x^2 - 1/3 x^3 + 2x | bts 1 sampai 2
= (1/2 (1)^2 + 1) - (1/2. -1^2 + -1) + (1/2 . 2^2 - 1/3 . 2^3 + 2.2) - (1/2 . 1^2 - 1/3 . 1^3 + 2(1))
= (1/2 + 1 - 1/2 + 1) + (2 - 8/3 + 4) - (1/2 - 1/3 + 2)
= 2 + 4 - 7/3 - 1/2
= 6 - 7/3 - 1/2
= (36 - 14 - 3)/6
= 19/6

Perbandingan = 8/6 : 19/6 = 8 : 19