JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Area Under Curves

The region lies inside the square $|x|+|y|\le 1$ and outside the parabola, where $2y^2\ge |x|$; by symmetry compute the first-quadrant part and multiply by four. There the bounds are $x+y\le 1$ and $x\le 2y^2$, and the curves $x=2y^2$ and $x+y=1$ meet where $2y^2+y-1=0$, i.e. $y=\frac{1}{2}$. For $0\le y\le \frac{1}{2}$ the parabola $x=2y^2$ is the tighter cap, contributing ${\int }_0^{1/2}2y^{2}dy=\frac{1}{12}$, and for $\frac{1}{2}\le y\le 1$ the line $x=1-y$ bounds a triangle of area ${\int }_{1/2}^1(1-y)dy=\frac{1}{8}$. The first-quadrant area is $\frac{1}{12}+\frac{1}{8}=\frac{5}{24}$, so the whole region has area $4\times \frac{5}{24}=\frac{5}{6}$.