The area inside the parabola $5 x^2-y=0$ but outside the parabola $2 x^2-y+9=0 \quad$ is equal to: (

Question

The area inside the parabola $5 x^2-y=0$ but outside the parabola $2 x^2-y+9=0 \quad$ is equal to: (sq. unit)

Professor PC Thomas Classes · Answer

$\begin{aligned} & x^2=\frac{y}{5}, x^2=\frac{1}{2}(y-9) \ & 	ext { Area }=2 \int_0^{\sqrt{3}}\left[\left(2 x^2+9ight)-\left(5 x^2ight)ight] d x \ & =2 \int_0^{\sqrt{3}}\left(9-3 x^2ight) d x=18 x-2 x^3=18 \sqrt{3}-6 \sqrt{3}=12 \sqrt{3}\end{aligned}$