Question

Find the area enclosed by the curves $y=4x^2$ and $y^2=2x$.

Answer 1

Point of intersection

$y=4x^2$

$⟹(4x{)}^2=4x$

$⟹16x^4-2x=0$

$⟹2x(8x^3-1)=0$

$x=0,x=1/2$

$y=0,y=1$

Area$={\int }_0^{1/2}(\sqrt{2}{x}^{1/2}-4x^2)dx$

$=\frac{(2\sqrt{2}{x}^{3/2}-4x^2)}{3}-\frac{4x^3}{3}{]}_0^{1/2}$

$=\frac{2\sqrt{2}}{3}.\frac{1}{2\sqrt{2}}-\frac{4}{3}.\frac{1}{8}$

$=\frac{1}{6}$ sq units.