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Find the area of the region satisfying both 12x - y² >= 0 and y >= 2x.
- (a) 3
- (b) 4
- (c) 6
- (d) 9
Correct answer: (a) 3
Solution
The region 12x - y² >= 0 is the interior of the rightward-opening parabola y² = 12x. The region y >= 2x is above the line through the origin. Intersections: substitute y = 2x into y² = 12x: 4x² = 12x => x = 0 or x = 3. Points: (0,0) and (3,6). For x in [0,3], the bounded region lies between y = 2x (lower) and y = sqrt(12x) (upper). Area = integral from 0 to 3 of [sqrt(12x) - 2x] dx = [2*sqrt(12)/3 * x^(3/2) - x²] from 0 to 3 = (4*sqrt(3)/3 * 3*sqrt(3)) - 9 = 12 - 9 = 3.
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