Question

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

• A. $\frac{1}{3}$.sq.unit
• B. $2{\int }_0^1(x-x^2)dx$
• C. area of the region $\{(x,y):x^2\le y\le \left|x\right|\}$
• D. none of the above

Answer 1

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

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

$\because$ Curves $x^2=y$ and $y^2=x$ intersect at $(0,0)$ and $(1,1)$.
$\therefore$Required Area$={\int }_0^1(\sqrt{x}-x^2)dx$
$={[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{x^3}{3}]}_0^1$
$=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$.sq.unit.
Also, both curves $x^2=y$ and $y^2=x$ are symmetrical about $y=x$
$\therefore$ Required Area$=2{\int }_0^1(x-x^2)dx$
Option$\left(c\right):x^2\le y\le \left|x\right|$
$y=x^2,y=\left|x\right|$ point of intersection is $(0,0)$ and $(1,1)$.
$\therefore$ Required Area$=2{\int }_0^1(x-x^2)dx$
$=2{[\frac{x^2}{2}-\frac{x^3}{3}]}_0^1$
$=2(\frac{1}{2}-\frac{1}{3})$
$=1-\frac{2}{3}=\frac{1}{3}$.sq.unit