a) y = x² - 1 y = x + 1 kurangkan 0 = x² -x - 2 atau x² -x - 2 = 0 a = 1 , b = -1 , c= -2 D = b²-4ac D = (-1)² - 4(1)(-2) D = 9 Luas = D√D/(6a²) L = 9√9 / (6)(-1)² L= 27/6 = 9/2 = 4,5satuan
b. Luasan terbagi dengan sumbu y (gambar lampiran 1) y1 = y2 x + 1 = x² - 1 -x² + x + 2 = 0 -(x -2)(x +1)= 0 x = -1 , x = 2 L1 dengan batas x =-1 dan x = 0 L1 = ₋₁⁰∫ -x² + x + 2 dx L1 = [ - 1/3 x³ + 1/2 x² + 2x]⁰₋₁ L1 = -1/3 (0+1) + 1/2 (0 -1) + 2(0+1) L1 = - 1/3 - 1/2 + 2 = 7/6 satuan
Verified answer
Luas dengan integrala)
y = x² - 1
y = x + 1
kurangkan
0 = x² -x - 2 atau x² -x - 2 = 0
a = 1 , b = -1 , c= -2
D = b²-4ac
D = (-1)² - 4(1)(-2)
D = 9
Luas = D√D/(6a²)
L = 9√9 / (6)(-1)²
L= 27/6 = 9/2 = 4,5satuan
b. Luasan terbagi dengan sumbu y (gambar lampiran 1)
y1 = y2
x + 1 = x² - 1
-x² + x + 2 = 0
-(x -2)(x +1)= 0
x = -1 , x = 2
L1 dengan batas x =-1 dan x = 0
L1 = ₋₁⁰∫ -x² + x + 2 dx
L1 = [ - 1/3 x³ + 1/2 x² + 2x]⁰₋₁
L1 = -1/3 (0+1) + 1/2 (0 -1) + 2(0+1)
L1 = - 1/3 - 1/2 + 2 = 7/6 satuan
L2 dengan batas x = 0 dan x =2
L2 = ₀²∫ -x² + x + 2 dx
L2 = [ - 1/3 x³ + 1/2 x² + 2x]²₀
L2 = -1/3 (8) + 1/2 (4) + 2(2)
L2 = - 8/3 + 2 + 4
L2 = 10/3 = 20/6
L1 : L2 = 7/6 : 20/6 = 7 : 20
terbukti
c) luasan terbagi dua bagian oleh sumbu x (gambar lampiran 2)
L1 luasan diatas sumbu x
L1 = ₋₁¹∫ x+1 dx + ₁²∫ -x² + x + 2 dx
L1 = [1/2 x² + x]¹₋₁ + [ -1/3 x³ + 1/2 x² + 2x]²₁
L1 = 1/2(1-1) + (1+1) - 1/3 (8-1) + 1/2 (4-1) + 2(2-1)
L1 = 0 + 2 - 7/3 + 3/2 + 2
L1 = 19/6
L2 luas dibawh sumbu x
L2 = - [₋₁¹∫ x² -1 dx]
L2 = - [ 1/3 x³ - x ]¹₋₁
L2 = - [ 1/3(1+1) - (1+1)]
L2 = - [ 2/3 - 2]
L2 = 4/3 = 8/6
L1 : L2 = 19/6 : 8/6 = 19 : 8