Question

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

• A. $\frac{2}{3}\mathrm{square}\mathrm{units}$
• B. $\frac{1}{2}\mathrm{square}\mathrm{units}$
• C. $\frac{1}{3}\mathrm{square}\mathrm{units}$
• D. $\frac{1}{6}\mathrm{square}\mathrm{units}$

Answer 1

The correct option is D

$\frac{1}{6}\mathrm{square}\mathrm{units}$

Explanation for the correct option

Given curves are $y^2=x$ and $y=x$

The intersecting points of two curves are $O\left(0,0\right)$ and $A\left(1,1\right)$

The graph of the given curves is shown below,

Now the required area is given by,

$$\begin{gathered} \mathrm{Area}={\int }_0^1\left(\sqrt{\mathrm{x}}-\mathrm{x}\right)d\mathrm{x}\left[\because \mathrm{y}=\sqrt{\mathrm{x}},\mathrm{y}=\mathrm{x}\right] \\ ={\left[\frac{2}{3}{\mathrm{x}}^{\frac{3}{2}}-\frac{{\mathrm{x}}^2}{2}\right]}_0^1 \\ =\left[\frac{2}{3}-\frac{1}{2}-\left(0-0\right)\right] \\ =\left[\frac{2}{3}-\frac{1}{2}\right] \\ =\frac{1}{6}\mathrm{square}\mathrm{units} \end{gathered}$$

Therefore the area enclosed between the given curves is equal to $\frac{1}{6}\mathrm{square}\mathrm{units}$.

Hence, option (D) is correct.