The area of the region bounded by the curves $y = x$ and $y = x^3$ is:

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

The correct answer is Option (3) → $\frac{1}{2}$ sq. units

The area bounded by the curves $y = x$ and $y = x^3$ is found by integrating the difference between the two functions over the interval where they intersect.

First, find the points of intersection:

$x = x^3 \Rightarrow x^3 - x = 0 \Rightarrow x(x^2 - 1) = 0 \Rightarrow x = -1, 0, 1$

So, the area is:

$A = \int_{-1}^{0} (x - x^3)\,dx + \int_{0}^{1} (x - x^3)\,dx$

Since the integrand is odd (symmetric), we can double the integral from 0 to 1:

$A = 2 \int_{0}^{1} (x - x^3)\,dx$

$= 2\left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1$

$= 2\left(\frac{1}{2} - \frac{1}{4}\right) = 2 \cdot \frac{1}{4} = \frac{1}{2}$