Question

Area enclosed between the curves $y^2=x$ and $x^2=y$ is equal to

• A. $2{\int }_0^1(x-x^2)dx$
• B. $\frac{1}{3}$
• C. area of region $\{(x,y):x^2\le y\le \left|x\right|\}$
• D. $\frac{2}{3}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $2{\int }_0^1(x-x^2)dx$
B $\frac{1}{3}$
C area of region $\{(x,y):x^2\le y\le \left|x\right|\}$

The $y^2=x$ and $x^2=y$ intersect each other at point $(0,0)$ and $(1,1)$ as shown below.

$\therefore$ required area is given by $A={\int }_0^1(\sqrt{x}-x^2)dx=2A_2$

Where $A_2=$area between $y=x$ and $y=x^2$ between

$x=0$ and $x=1=\{(x,y):x^2\le y\le \left|x\right|\}$

$=2{\int }_0^1(x-x^2)dx=2{[\frac{x^2}{2}-\frac{x^3}{3}]}_0^1$

$=2[\frac{1}{2}-\frac{1}{3}]=\frac{1}{3}$ square units.