soal yang no. 2 b , c , d ya..... saya masih kurang ngerti

Luas dengan integral

2b)
y = x - 1
y = 3- x
kurangkan
0 = 2x - 4
2x = 4
x = 2  

y = x - 1
y = 0 --> x = 1

y= 3 - x
y = 0 --> x = 3

luas daerah = L
L = ₁²∫ x -1 dx + ₂³∫ 3 - x dx
L = [1/2 x² - x]²₁ + [3x- 1/2 x²]³₂
L = 1/2 (4-1) - (2-1)  + 3(3-2) - 1/2(9-4)
L = 3/2 - 1 + 3 - 5/2
L = 1 satuan
(gambar  lampiran 1)

C)
y = -x² + 2
y = -x
kurangkan
-x² + x + 2 = 0
-(x-2)(x+1) = 0
x = 2 atau x = - 1
L = ₋₁²∫ -x² + x + 2 dx
L = [- 1/3 x³ + 1/2 x² + 2x]²₋₁
L = - 1/3(8+1)  + 1/2 (4-1) + 2(2+1)
L = -3 + 3/2 + 6
L = 9/2 = 4,5 satuan

d)
y = 9 - x²
y = x + 3
kurangkan
-x² - x + 6 = 0
-(x+3)(x-2) = 0
x = -3 atau x = 2
L = ₋₃²∫ -x² - x + 6 dx =
L = [- 1/3 x³ - 1/2 x² + 6x ]²₋₃ 
L= - 1/3 (8 +27)  - 1/2 (4 - 9) + 6(2+ 3)
L = - 35/3 + 5/2 + 30
L =  125/6 = 20  5/6