The area of the plane region bounded by the curves x+2y2=0 and x+3y2=1 is equal to
(A)35
(B)31
(C)32
(D)34✓ Correct
Answer: (D)
Solution
The curves are x=−2y2 and x=1−3y2; setting them equal, −2y2=1−3y2 gives y2=1, so they meet at y=±1. Between these the rightmost curve minus the leftmost is (1−3y2)−(−2y2)=1−y2, and by symmetry the area is 2∫01(1−y2)dy=2[y−3y3]01=2⋅32=34.
JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option