integral batas atas a batas bawah 0 (x+2) dx = 6

∫(x + 2) dx = 6
x²/2 + 2x [a...0] = 6
a²/2 + 2a - 0 = 6
a²/2 + 2a - 6 = 0 kalikn 2
a² + 4a - 12 = 0
(a - 2)(a + 6) = 0
a = 2 atau a = -6

Krn a batas atas, maka yg memenuhi : a = 2