The area inside the parabola $y=5x^2$ but outside $y =$$2x^2+9$ is

The intersection of two curves is given by

$5x^2=2x^2+9\Rightarrow x^2=3\Rightarrow x=\sqrt{3},-\sqrt{3}$

$y=2x^2+9$ is a parabola with focus at $\left(0,\frac{73}{8}\right)$ Both

the curves are symmetrical about $y$-axis.

Reqd. area $=2{\int }_0^{\sqrt{3}} \left(\left(2x^2+9\right)-5x^2\right)dx$

$\begin{array}{l}=2{\int }_0^{\sqrt{3}} \left(9-3x^2\right)dx=2\left[9\sqrt{3}-{\left.x^3\right|}_0^{\sqrt{3}}\right)\\ =12\sqrt{3}\end{array}$