The area (in square units) of the region bounded by curves $y=x$ and $y=x^3$ is:

1/2

The correct answer is Option (2) → 1/2

$y=x$   ...(1)

$y=x^3$   ...(2)

from (1) and (2),

$x=x^3$

$⇒x(x-1)(x+1)=0$

$⇒x=-1,0,1$

$A=\int\limits_{-1}^1(x-x^3)dx$

$=\left[\int x\,dx-\int x^3\,dx\right]_{-1}^1$

$=\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_{-1}^1$

$=\frac{1}{2}$ sq. units