JEE Main 2025 — 24 January, Morning Shift — Question 13

$\text{Region:\; } \{(x,y):\;x^2+4x+2\le y\le |x+2|\}.$

$\text{Solve\; intersections\; }x^2+4x+2=|x+2|.$

$\text{Put\; }t=x+2\Rightarrow t^2-2=|t|\Rightarrow t^2-|t|-2=0.$

$\text{For\; }t\ge0:\;t^2-t-2=0\Rightarrow t=2\ (\Rightarrow x=0).$

$\text{For\; }t<0:\;t^2+t-2=0\Rightarrow t=-2\ (\Rightarrow x=-4).$

$\text{Thus\; the\; interval\; of\; interest\; is\; }x\in[-4,0].$

$\text{Split\; at\; }x=-2\ (\text{where\; }x+2=0).$

$\text{For\; }x\in[-4,-2]:\;|x+2|=-(x+2)=-x-2.$

$\text{For\; }x\in[-2,0]:\;|x+2|=x+2.$

$\text{Area\; }A=\int_{-4}^{-2}\big[(-x-2)-(x^2+4x+2)\big]dx+\int_{-2}^{0}\big[(x+2)-(x^2+4x+2)\big]dx.$

$\text{Simplify\; integrands:\; }(-x-2)-(x^2+4x+2)=-x^2-5x-4,\quad (x+2)-(x^2+4x+2)=-x^2-3x.$

$\displaystyle \int_{-4}^{-2}(-x^2-5x-4)\,dx=\Big[-\frac{x^3}{3}-\frac{5x^2}{2}-4x\Big]_{-4}^{-2}=\frac{10}{3}.$

$\displaystyle \int_{-2}^{0}(-x^2-3x)\,dx=\Big[-\frac{x^3}{3}-\frac{3x^2}{2}\Big]_{-2}^{0}=\frac{10}{3}.$

$\therefore A=\frac{10}{3}+\frac{10}{3}=\frac{20}{3}.$

$\boxed{\dfrac{20}{3}}$