Question

The area inside the parabola $5x^2-y=0$ but outside the parabola $2x^2-y+9=0$ is

Answer 1

Inside parabola $5x^2-y=0$

outside parabola $2x^2-y+9=0$

for intersection point $y=5x^2$

$2x^2-5x^2+9=0$

$-3x^2+9=0$

$x=\pm \sqrt{3}$

$y=5x^2=15$

$Area={\int }_{\sqrt{-}3}^{\sqrt{3}}(2x^2+9)-\left(5x^2\right)dx={\int }_{\sqrt{-}3}^{\sqrt{3}}9-3x^{2}dx$

$=2{\int }_{\sqrt{0}}^{\sqrt{3}}9-3x^2=2[9x-x^3{]}_0^{\sqrt{3}}=2(9\sqrt{3}3\sqrt{3})$

$=12\sqrt{3}$ sq. unit