JEE Main 2025 Apr 7 Shift 1, Mathematics Q16: Area as Definite Integral

Find intersection points. Set the two expressions for $y$ equal:

$4-\frac{x^2}{4}=\frac{x-4}{2}$

Multiply through by 4: $16-x^2=2(x-4)=2x-8$

$\Rightarrow x^2+2x-24=0\Rightarrow (x+6)(x-4)=0\Rightarrow x=-6,4$.

On the interval $[-6,4]$, the parabola lies above the line, so

$\alpha ={\int }_{-6}^4\left[\left(4-\frac{x^2}{4}\right)-\frac{x-4}{2}\right]dx={\int }_{-6}^4\left(6-\frac{x^2}{4}-\frac{x}{2}\right)dx$

Integrate term-by-term:

${\int }_{-6}^4\left(6-\frac{x^2}{4}-\frac{x}{2}\right)dx={\left[6x-\frac{x^3}{12}-\frac{x^2}{4}\right]}_{-6}^4$

At $x=4$: $24-\frac{64}{12}-4=20-\frac{16}{3}=\frac{60-16}{3}=\frac{44}{3}$.

At $x=-6$: $-36-\frac{-216}{12}-9=-36+18-9=-27$.

$\alpha =\frac{44}{3}-(-27)=\frac{44}{3}+27=\frac{44+81}{3}=\frac{125}{3}$

$6\alpha =6\cdot \frac{125}{3}=250$

$\therefore$ Option (1).