Let $(a, b)$ be the point of intersection of the curve $x^2=2 y$ and the straight line $y-2 x-6=0$ in the second quadrant. Then the integral $\mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}} \frac{9 x^2}{1+5^x} \mathrm{~d} x$ is equal to :

<p>$$\begin{aligned} & x^2=2 y \text { and } y-2 x-6=0 \\ & \frac{x^2}{2}-2 x-6=0 \\ & x^2-4 x-12=0 \\ & x^2-6 x+2 x-12=0 \\ & x(x-6)+2(x-6)=0 \\ & (x-6)(x+2)=0 \end{aligned}$$</p> <p>Point of intersection are $(6,18)$ and $(-2,2)$</p> <p>$(-2,2)$ is in second quadrant</p> <p>$$\begin{aligned} & a=-2, b=2 \\ & I=\int_{-2}^2 \frac{9 x^2}{1+5^x} d x\quad\text{..... (i)} \end{aligned}$$</p> <p>$I=\int_{-2}^2 \frac{9 x^2}{1+5^{-x}} d x\quad\text{..... (ii)}$</p> <p>$$\begin{aligned} &\text { Adding (i) and (ii) }\\ &\begin{aligned} & 2 I=\int_{-2}^2 9 x^2 d x \\ & I=9 \int_0^2 x^2 d x \\ & I=9\left(\frac{x^3}{3}\right)_0^2 \Rightarrow I=24 \end{aligned} \end{aligned}$$</p>