JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Area Under Curves

Real $y$ needs $(4-x)(x-2)\ge 0$, so the closed loop of the curve lies between $x=2$ and $x=4$, with $x=0$ contributing only the isolated point $(0,0)$. On $2\le x\le 4$,

and the curve is symmetric about the $x$-axis, so the enclosed area is

Substituting $x=3+\sin \theta$ makes $(4-x)(x-2)=1-{\sin }^2\theta ={\cos }^2\theta$ and $dx=\cos \theta d\theta$, with $\theta$ running over $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ where $\cos \theta \ge 0$:

the $\sin \theta {\cos }^2\theta$ part being odd over a symmetric interval. The area bounded by the curve is $\frac{3\pi }{2}$.