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Find the area (in square units) of the region {(x, y) in R²: 4x² <= y <= 8x + 12}.

1. 127/3
2. 125/3
3. 124/3
4. 128/3

Correct answer: 128/3

Solution

The curves meet at x = -1 and x = 3; the area is integral from -1 to 3 of (8x + 12 - 4x²) dx = 128/3 square units.

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