JEE Main 2025 Jan 28 Shift 1, Mathematics Q20: Area as Definite Integral

For each $x$, the height of the region is $\min (2|x|+1,x^2+1)$. Find where they meet: $2|x|+1=x^2+1\Rightarrow x^2=2|x|\Rightarrow |x|(|x|-2)=0$, so the curves intersect at $|x|=0$ and $|x|=2$.

Between $|x|=0$ and $|x|=2$: $x^2+1\le 2|x|+1$, so the parabola is lower.
For $|x|>2$: $2|x|+1\le x^2+1$, so the V-curve is lower.

By symmetry in $x\to -x$, compute over $x\in [0,3]$ and double:

$\text{Area}=2\left[{\int }_0^2(x^2+1)dx+{\int }_2^3(2x+1)dx\right]$

${\int }_0^2(x^2+1)dx=\frac{8}{3}+2=\frac{14}{3}$
${\int }_2^3(2x+1)dx=(9+3)-(4+2)=6$

$\text{Area}=2\left(\frac{14}{3}+6\right)=2\cdot \frac{32}{3}=\frac{64}{3}$.