The area of the plane region bounded by the curves x+2y2=0 and x+3y2=1 is equal to
43squnits
53squnits
13squnits
23squnits
Given, x+2y2=0 and x+3y2=1
Equate x of both the equation,
1-3y2=-2y2⇒y2=1⇒y=±1
Find the corresponding value of x
y=-1⇒x=-2y=1⇒x=-2
We have two points of intersection,
-2,1and-2,-1
The desired area,
A=2∫011-3y2--2y2dy=2∫011-y2dy=2y-y3301=2×23=43squnits
Hence option(A) is correct.