Tolong Penyelesaiannya Kak

1.

∫(8x - 12)(x² - 3x + 8)³ dx = ∫(8x - 12)(x² - 3x + 8)³ d(x² - 3x + 8)/(2x - 3)
= ∫4(2x - 3)(x² - 3x + 8)³ d(x² - 3x + 8)/(2x - 3)
= ∫4(x² - 3x + 8)³ d(x² - 3x + 8)
= 4/4 (x² - 3x + 8)⁴ + C
= (x² - 3x + 8)⁴ + C

2.

∫2x.sin(2x - 1) dx

u = 2x
du = 2 dx
dv = sin(2x - 1) dx
v = -(1/2)cos(2x - 1)

∫2x.sin(2x - 1) dx = uv - ∫v du
= 2x.-(1/2)cos(2x - 1) - ∫-(1/2)cos(2x - 1).2 dx
= -x.cos(2x - 1) + ∫cos(2x - 1) dx
= -x.cos(2x - 1) + (1/2)sin(2x-1) + C

3.

y = y
6 - x = x²
x² + x - 6 = 0
(x + 3)(x - 2) = 0
x = -3 atau x = 2

pilih x = 2 (karena memotong di kuadran 1), maka

Luas daerah = ∫(6 - x) dx (batas 2 sampai 6) + ∫x² dx (batas 0 sampai 2)
= 6x - (1/2)x² ]₂⁶ + (1/3)x³]₀²
= (6.6 - (1/2).6²) - (6.2 - (1/2)2²) + (1/3)(2)³ - 0
= 18 - 10 + 8/3
= 10 2/3 satuan luas