JEE Main · 2005 · 08 May 2005 · Offline · Area Under Curves

The parabolas $y^2=4x$ and $x^2=4y$ are reflections of each other in the line $y=x$, meeting at $(0,0)$ and $(4,4)$. The middle strip $S_2$, lying between the two curves, has area that by symmetry about $y=x$ equals $2{\int }_0^4(2\sqrt{x}-x)dx$; since ${\int }_0^4(2\sqrt{x}-x)dx=\frac{8}{3}$, this makes $S_2=\frac{16}{3}$. The whole square has area $16$, so the two outer pieces together have area $16-\frac{16}{3}=\frac{32}{3}$, and by the same symmetry they are equal, each $\frac{16}{3}$. Hence $S_1:S_2:S_3=1:1:1$.