Mathematics - Definite Integration Question with Solution | TestHub

$\begin{array}{rl}& x^2=2y\&y=2x+6\\ & x^2=4x+12\end{array}$
$x^2-4x-12=0\Rightarrow \begin{array}{c}x=6\\ y=18\end{array}|\begin{array}{c}\text{ if }x=-2\\ y=2\end{array}$
$\therefore (6,18)\&(-2,2)$
Here $(6,18)$ Rejected because $(a,b)$ lies in $2^{\text{nd }}$ quadrant
$\begin{array}{rl}& \therefore \mathrm{a}=-2\&\mathrm{b}=2\\ & \therefore \mathrm{I}={\int }_{-2}^2\frac{9{\mathrm{x}}^2}{1+5^{\mathrm{x}}}\mathrm{dx}={\int }_{-2}^2\frac{9\cdot 5^{\mathrm{x}}\cdot {\mathrm{x}}^2}{1+5^{\mathrm{x}}}\mathrm{dx}\\ & \therefore 2\mathrm{I}={\int }_{-2}^{2}9{\mathrm{x}}^2\mathrm{dx}=18{\int }_0^2{\mathrm{x}}^2\mathrm{dx}=18{\left(\frac{{\mathrm{x}}^3}{3}\right)}_0^2\\ & 2\mathrm{I}=48\\ & \therefore \mathrm{I}=24\end{array}$