JEE Main 2025 — 2 April, Evening Shift — Question 25

$x^{2}=2 y$ and $y-2 x-6=0$

$$\begin{aligned} & \frac{x^{2}}{2}-2 x-6=0 \\& x^{2}-4 x-12=0 \\& x^{2}-6 x+2 x-12=0 \\& x(x-6)+2(x-6)=0 \\& (x-6)(x+2)=0 \end{aligned}$$

Point of intersection are $(6,18)$ and $(-2,2)$ $(-2,2)$ is in second quadrant $a=-2, b=2$

$I=\int_{-2}^{2} \frac{9 x^{2}}{1+5^{x}} d x …(i)$

$I=\int_{-2}^{2} \frac{9 x^{2}}{1+5^{-x}} d x …(ii)$

Adding (i) and (ii) $2 I=\int_{-2}^{2} 9 x^{2} d x$

$I=9 \int_{0}^{2} x^{2} d x$

$I=9\left(\frac{x^{3}}{3}\right)_{0}^{2} \Rightarrow I=24$