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Find the area common to the regions defined by 0 >= y >= x² + 2x and x - y >= 0.

1. 1/3
2. 2/3
3. 1/6
4. 1/2

Correct answer: 1/6

Solution

Conditions: y <= 0 (from 0 >= y), y >= x² + 2x, and y <= x (from x - y >= 0). For the region y >= x² + 2x and y <= 0: need x² + 2x <= 0, so x(x+2) <= 0, meaning x in [-2, 0]. In this range, x² + 2x <= y <= 0. Now intersect with y <= x: the upper bound on y becomes min(0, x). For x in [-2, 0]: x <= 0, so min(0, x) = x. The valid region requires x² + 2x <= y <= x. This requires x² + 2x <= x, i.e., x² + x <= 0, i.e., x(x+1) <= 0, so x in [-1, 0]. Area = integral from -1 to 0 of [x - (x² + 2x)] dx = integral from -1 to 0 of (-x - x²) dx = [-x²/2 - x³/3] from -1 to 0 = (0) - (-1/2 + 1/3) = -(-3/6 + 2/6) = -(-1/6) = 1/6.

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