The area enclosed between the curves y2=x and y=∣x∣ is
(A)2/3
(B)1
(C)1/6✓ Correct
(D)1/3
Answer: (C)
Solution
The curves y2=x and y=∣x∣ meet in the first quadrant where y=x equals y=x, that is at x=0 and x=1. Between these x lies above x, so the enclosed area is ∫01(x−x)dx=[32x3/2−2x2]01=32−21=61.
JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option