Find the area enclosed between the curve $y = x^2 - 4$ and the x-axis

The curve crosses the x-axis at x = -2 and x = 2. The area = |∫₋₂² (x² - 4) dx| = |[x³/3 - 4x]₋₂²| = |(8/3 - 8) - (-8/3 + 8)| = |16/3 - 16| = |-32/3| = 32/3 square units. Since the curve is below the x-axis between x = -2 and x = 2, the absolute value is taken.