JEE Main 2025 Jan 29 Shift 1, Mathematics Q7: Area as Definite Integral

The region is bounded above by the V-shaped envelope $y+|x|=3$, below by $y=|x-1|$ on one side and by the parabola $2y=x^2+3$ (i.e. $y=(x^2+3)/2$) on the other.

Key vertices: $A(1,0)$ (apex of lower V), $B(2,1)$, $D(0,3)$, $E(-1,2)$, with the parabola cutting through at $C(1,2)$.

It is convenient to compute $A$ as the rectangular/trapezoidal hull $ABDE$ (which has area $4$) minus the curved region $EDC$ that lies above the parabola but below the upper bound. The region $EDC$ contributes $2{\int }_0^1[(3-x)-(\frac{x^2+3}{2}\left)\right]dx.$

Evaluate: ${\int }_0^1\left(3-x-\frac{x^2}{2}-\frac{3}{2}\right)dx={\left[\frac{3}{2}x-\frac{x^2}{2}-\frac{x^3}{6}\right]}_0^1=\frac{3}{2}-\frac{1}{2}-\frac{1}{6}=\frac{5}{6}.$

So $A=4-2\cdot \frac{5}{6}=4-\frac{5}{3}=\frac{7}{3}.$

Therefore $6A=14.$