The area inside the parabola y=5x2 but outside y =2x2+9 is

The intersection of two curves is given by

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

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

the curves are symmetrical about $y$ -axis.

Reqd. area $= 2 \int_{0}^{\sqrt{3}} ((2 x^{2} + 9) - 5 x^{2}) d x$

$\begin{array}{l} = 2 \int_{0}^{\sqrt{3}} (9 - 3 x^{2}) d x = 2 [9 \sqrt{3} - {x^{3}|}_{0}^{\sqrt{3}}) \\ = 12 \sqrt{3} \end{array}$

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