The area (in sq units) of region described by and is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
Find intersections of y² = 2x and y = 4x - 1. Substitute x = (y+1)/4 into y² = 2x: y² = (y+1)/2, so 2y² - y - 1 = 0, giving (2y+1)(y-1) = 0, so y = -1/2 and y = 1.
The region requires y² ≤ 2x (right of parabola) and y ≥ 4x-1 (above/left of line), i.e., x ≤ (y+1)/4.
Area = ∫ from y = -1/2 to 1 of [(y+1)/4 - y²/2] dy
= [(y+1)²/8 - y³/6] from -1/2 to 1
At y=1: (4/8) - (1/6) = 1/2 - 1/6 = 1/3
At y=-1/2: (1/2)²/8 - (-1/8)/6 = (1/4)/8 + 1/48 = 1/32 + 1/48 = 3/96 + 2/96 = 5/96
Area = 1/3 - 5/96 = 32/96 - 5/96 = 27/96 = 9/32
The region requires y² ≤ 2x (right of parabola) and y ≥ 4x-1 (above/left of line), i.e., x ≤ (y+1)/4.
Area = ∫ from y = -1/2 to 1 of [(y+1)/4 - y²/2] dy
= [(y+1)²/8 - y³/6] from -1/2 to 1
At y=1: (4/8) - (1/6) = 1/2 - 1/6 = 1/3
At y=-1/2: (1/2)²/8 - (-1/8)/6 = (1/4)/8 + 1/48 = 1/32 + 1/48 = 3/96 + 2/96 = 5/96
Area = 1/3 - 5/96 = 32/96 - 5/96 = 27/96 = 9/32
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