Question

The area in sq. units inside the parabola $5 x^{2}-y=0$ but outside the parabola $2 x^{2}-y+9=0$ is A. Find $\frac{A^{2}}{12} .$

Answer 1

Solving $y=5 x^{2}$ and
$y=2 x^{2}+9, $ we get $x=\pm \sqrt{3}$
Area, $A=\int\limits_{0}^{\sqrt{3}}\left(\left(2 x^{2}+9\right)-5 x^{2}\right) d x$
$=2 \int\limits_{0}^{\sqrt{3}}\left(9-3 x^{2}\right) d x$
$=2\left[9 x-x^{3}\right]_{0}^{\sqrt{3}}$
$=2(9 \sqrt{3}-3 \sqrt{3})=12 \sqrt{3}$
$\Rightarrow \frac{A^{2}}{12}=\frac{(12 \sqrt{3})^{2}}{12}=36$