tolong di jawab yaaa

batas integral : titik potong kedua kurva
x² = 2x + 3
x² - 2x - 3 = 0
(x + 1)(x - 3) = 0
x = -1  dan  x = 3

Luas = ∫ (2x + 3 - x²) dx  {-1 ... 3}
= x² + 3x - x³/3  {-1 ... 3}
= (3)² + 3(3) - (3)³/3 - (-1)² - 3(-1) + (-1)³/3
= 9 + 9 - 9 - 1 + 3 - 1/3
= 33/3 - 1/3
= 32/3