Question

The area enclosed between $y^2=x$ and $y=x$

• A. $\frac{2}{3}$ sq unit
• B. $\frac{1}{2}$ sq unit
• C. $\frac{1}{3}$ sq unit
• D. $\frac{1}{6}$ sq unit

Answer 1

The correct option is D $\frac{1}{6}$ sq unit

Intersection point of two curves are $O(0,0)$ and $A(1,1)$
$\therefore$ Required area $={\int }_0^4(y_2-y_1)dx$
${\int }_0^1(\sqrt{x}-x)dx$
$={[\frac{x^{3/2}}{3/2}-\frac{x^2}{2}]}_0^1$
$=\frac{2}{3}[1-0]-\frac{1}{2}(1-0)$
$=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$ unit