JEE Main 2025 Jan 24 Shift 1, Mathematics Q13: Area as Definite Integral

Substitute $u=x+2$. Then $x^2+4x+2=(x+2{)}^2-2=u^2-2$ and $|x+2|=|u|$. The area equals the area of $\{(u,y):u^2-2\le y\le |u|\}$, which is symmetric about $u=0$.

Intersections: $|u|=u^2-2$. For $u\ge 0$: $u=u^2-2\Rightarrow u^2-u-2=0\Rightarrow u=2$. By symmetry $u=\pm 2$.

Using symmetry, compute for $u\in [0,2]$ and double:

Area $=2{\int }_0^2\left[u-(u^2-2)\right]du=2{\left[\frac{u^2}{2}-\frac{u^3}{3}+2u\right]}_0^2$

$=2\left[2-\frac{8}{3}+4\right]=2\left[6-\frac{8}{3}\right]=2\cdot \frac{10}{3}=\frac{20}{3}$.

$\therefore$ Option (3).