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Find the area (in square units) bounded between the curves |x| + y = 7 and 4y = |4 - x²|.

1. 4
2. 8
3. 16
4. 32

Correct answer: 32

Solution

Curve 1: y = 7 - |x| (line going down from vertex (0,7)) Curve 2: y = |4 - x²|/4 For |x| <= 2: 4 - x² >= 0, so y2 = (4 - x²)/4 For |x| > 2: y2 = (x² - 4)/4 Find intersections: For x in [0, 2]: 7 - x = (4 - x²)/4 => 28 - 4x = 4 - x² => x² - 4x + 24 = 0. Discriminant = 16 - 96 < 0. No intersection here. For x in [2, 7]: 7 - x = (x² - 4)/4 => 28 - 4x = x² - 4 => x² + 4x - 32 = 0 => x = (-4 +- sqrt(16 + 128))/2 = (-4 +- 12)/2. So x = 4 or x = -8. x = 4 is in range. By symmetry, intersection at x = -4 as well. Bounded region (by symmetry, compute for x >= 0 and double): From x = 0 to x = 2: upper curve y1 = 7 - x, lower curve y2 = (4-x²)/4. From x = 2 to x = 4: upper curve y1 = 7 - x, lower curve y2 = (x²-4)/4. Area for x in [0,4]: A1 = integral from 0 to 2 of [(7-x) - (4-x²)/4] dx = integral [7 - x - 1 + x²/4] dx = integral [6 - x + x²/4] dx = [6x - x²/2 + x³/12] from 0 to 2 = 12 - 2 + 8/12 = 10 + 2/3 = 32/3 A2 = integral from 2 to 4 of [(7-x) - (x²-4)/4] dx = integral [7 - x - x²/4 + 1] dx = integral [8 - x - x²/4] dx = [8x - x²/2 - x³/12] from 2 to 4 At x=4: 32 - 8 - 64/12 = 24 - 16/3 = 56/3 At x=2: 16 - 2 - 8/12 = 14 - 2/3 = 40/3 A2 = 56/3 - 40/3 = 16/3 Total for x in [0,4] = 32/3 + 16/3 = 48/3 = 16. Total area (both sides) = 2 * 16 = 32 sq units.

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