Question 76
The area (in sq. units) of the region described by $$\{(x, y) : y^2 \leq 2x, y \geq 4x - 1\}$$ is
Solution
y²≤2x and y≥4x-1. Intersection: y²=2x and y=4x-1. y²=2(y+1)/4=(y+1)/2. 2y²=y+1. 2y²-y-1=0. y=1 or y=-1/2.
Area=∫_{-1/2}^{1}[(y+1)/4-y²/2]dy=∫[(y+1)/4-y²/2]dy.
=[y²/8+y/4-y³/6]_{-1/2}^{1}=(1/8+1/4-1/6)-(1/32-1/8+1/48).
=(3/24+6/24-4/24)-(3/96-12/96+2/96)=5/24-(-7/96)=5/24+7/96=20/96+7/96=27/96=9/32.
The answer is Option (4): 9/32.
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