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Find the area (in square units) enclosed between the curve a² * y = x² * (x + a) and the x-axis.

1. a²/3 sq. units
2. a²/4 sq. units
3. 3a²/4 sq. units
4. a²/12 sq. units

Correct answer: a²/12 sq. units

Solution

The curve is y = x²*(x+a)/a². It crosses x-axis at x = 0 (double root) and x = -a. For a > 0, on (-a, 0): x² > 0, and (x+a) > 0, so y > 0. Area = integral from -a to 0 of [x²*(x+a)/a²] dx = (1/a²) * integral(-a to 0) [x³ + a*x²] dx = (1/a²)*[x⁴/4 + a*x³/3] from -a to 0 = (1/a²)*{0 - (a⁴/4 - a⁴/3)} = (1/a²)*{-(3a⁴ - 4a⁴)/12} = (1/a²)*(a⁴/12) = a²/12.

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