Jika integral 0 2(ax-b) dx=4 dan integral 1 3(x^2+2b)

3a + 6b = 9.

Pembahasan

Kita diberikan dua informasi integral:
1. ∫₀² (ax - b) dx = 4
2. ∫₁³ (x² + 2b) dx = 10

Mari kita selesaikan integral pertama:
∫₀² (ax - b) dx = [ax²/2 - bx]₀²
= (a(2)²/2 - b(2)) - (a(0)²/2 - b(0))
= (4a/2 - 2b) - (0)
= 2a - 2b

Jadi, 2a - 2b = 4. Kita bisa menyederhanakannya menjadi a - b = 2. (Persamaan 1)

Sekarang, mari kita selesaikan integral kedua:
∫₁³ (x² + 2b) dx = [x³/3 + 2bx]₁³
= (3³/3 + 2b(3)) - (1³/3 + 2b(1))
= (27/3 + 6b) - (1/3 + 2b)
= (9 + 6b) - (1/3 + 2b)
= 9 + 6b - 1/3 - 2b
= (9 - 1/3) + (6b - 2b)
= (27/3 - 1/3) + 4b
= 26/3 + 4b

Jadi, 26/3 + 4b = 10.
4b = 10 - 26/3
4b = 30/3 - 26/3
4b = 4/3
b = (4/3) / 4
b = 1/3

Sekarang kita substitusikan nilai b ke Persamaan 1 (a - b = 2):
a - (1/3) = 2
a = 2 + 1/3
a = 6/3 + 1/3
a = 7/3

Terakhir, kita hitung 3a + 6b:
3a + 6b = 3(7/3) + 6(1/3)
= 7 + 2
= 9

Jadi, 3a + 6b = 9.