Question

Area lying between the curves $y^2=2x$ and $y=x$ is

• A. $\frac{2}{3}$ sq. units
• B. $\frac{1}{3}$ sq. units
• C. $\frac{1}{4}$ sq. units
• D. $\frac{3}{4}$ sq. units

Answer 1

The correct option is A $\frac{2}{3}$ sq. units


Point of intersection of the curves :
$x^2=2x$
$\Rightarrow x=0,2$
Therefore, the curves intersect at $(0,0)$ and $(2,2).$
$\text{Required area}=\underset{0}{\overset{2}{\int }}(\sqrt{2x}-x)dx$
$=\frac{2}{3}$ sq. units