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Let the region in the plane be defined by {(x, y) in R²: x² <= y <= 3 - 2x}. If the area of this region is k square units, find the value of 3k.

1. 16
2. 24
3. 32
4. 48

Correct answer: 32

Solution

The bounding curves are y = x² and y = 3 - 2x. They intersect where x² = 3 - 2x, i.e. x² + 2x - 3 = 0 -> (x + 3)(x - 1) = 0, giving x = -3 and x = 1. Between these the line lies above the parabola, so the area is the integral of (3 - 2x - x²). Evaluating gives k = 32/3, hence 3k = 32.

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