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

The curves intersect where they have the same value:

$x = x^3$

$x - x^3 = 0$

$x(1 - x^2) = 0$

$x(1-x)(1+x) = 0$

The curves intersect at $x = -1, 0, 1$

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For the region from $x = -1$ to $x = 0$:

Testing $x = -0.5$:

• $y = x$ gives $-0.5$
• $y = x^3$ gives $-0.125$

Since $-0.125 > -0.5$, the curve $y = x^3$ is above $y = x$ in this region.

For the region from $x = 0$ to $x = 1$:

Testing $x = 0.5$:

• $y = x$ gives $0.5$
• $y = x^3$ gives $0.125$

Since $0.5 > 0.125$, the curve $y = x$ is above $y = x^3$ in this region.

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The total area is:

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

For the first integral:

$\int_{-1}^{0} (x^3 - x)\,dx = \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^{0}$

$= \left(0 - 0\right) - \left(\frac{1}{4} - \frac{1}{2}\right)$

$= 0 - \left(-\frac{1}{4}\right)$

$= \frac{1}{4}$

For the second integral:

$\int_{0}^{1} (x - x^3)\,dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{0}^{1}$

$= \left(\frac{1}{2} - \frac{1}{4}\right) - (0 - 0)$

$= \frac{1}{4}$

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Total Area $= \frac{1}{4} + \frac{1}{4}$

$= \frac{1}{2}$ square units

Therefore, the area of the region bounded by the curves is $\frac{1}{2}$ square units.