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The area (in sq. units) of the region described by $$\{(x, y) : y^2 \leq 2x, y \geq 4x - 1\}$$ is
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=$$\int_{-1/2}^1[(y+1)/4-y^2/2]dy$$=$$\int[(y+1)/4-y^2/2]dy.$$
=$$[y^2/8+y/4-y^3/6]_{-1/2}^1$$=(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.
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