The area (in sq. units) of the region bounded by the curve $y = 2x^3$, $x$ - axis and ordinates $x = -1$ and $x = 1$ is:

The region is bounded by the curve $y = 2x^3$, the $x$-axis, and the vertical lines (ordinates) at $x = -1$ and $x = 1$.

The function $y = 2x^3$ is an odd function. For negative values of $x$, the curve lies below the $x$-axis, and for positive values of $x$, it lies above the $x$-axis.

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Since $y = 2x^3$ is an odd function, it is symmetric about the origin. The area from $x = -1$ to $x = 0$ equals the area from $x = 0$ to $x = 1$.

The total area can be found by calculating the area from $x = 0$ to $x = 1$ and multiplying by 2.

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For $x \in [0, 1]$, the curve is above the $x$-axis:

$\text{Area} = \int_{0}^{1} 2x^3 \, dx$

$= \left[\frac{x^4}{2}\right]_{0}^{1}$

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

$= \frac{1}{2}$

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Using symmetry, the total area is:

$\text{Total Area} = 2 \times \frac{1}{2}$

$= 1$

Note: Direct integration $\int_{-1}^{1} 2x^3 \, dx = 0$ because the negative area below the $x$-axis cancels the positive area above it. For the bounded region area, absolute values must be used.

Therefore, the area of the region is $1$ square unit.