The area (in square units) of the region bounded by the curves $$y + 2x^2 = 0$$ and $$y + 3x^2 = 1$$, is equal to

To find the area bounded by the curves $$y + 2x^2 = 0$$ and $$y + 3x^2 = 1$$, we first rewrite the equations in terms of $$y$$. The first curve gives $$y = -2x^2$$, and the second curve gives $$y = 1 - 3x^2$$.

The area between two curves can be found by integrating the difference of the upper curve and the lower curve over the interval where they intersect. So, we need to find the points of intersection by setting the $$y$$-values equal: $$-2x^2 = 1 - 3x^2$$.

Solving for $$x$$:

Add $$3x^2$$ to both sides: $$-2x^2 + 3x^2 = 1$$, which simplifies to $$x^2 = 1$$. Taking square roots, $$x = \pm 1$$. So, the curves intersect at $$x = -1$$ and $$x = 1$$.

Next, we determine which curve is above the other between $$x = -1$$ and $$x = 1$$. Test a point, say $$x = 0$$:

For the first curve, $$y = -2(0)^2 = 0$$.

For the second curve, $$y = 1 - 3(0)^2 = 1$$.

Since $$1 > 0$$, the second curve $$y = 1 - 3x^2$$ is above the first curve $$y = -2x^2$$.

The area $$A$$ is given by the integral from $$-1$$ to $$1$$ of the upper curve minus the lower curve:

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

Simplify the integrand:

$$ (1 - 3x^2) - (-2x^2) = 1 - 3x^2 + 2x^2 = 1 - x^2 $$

So,

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

Now, compute the integral. The antiderivative of $$1 - x^2$$ is $$x - \frac{x^3}{3}$$. Evaluate this from $$-1$$ to $$1$$:

At $$x = 1$$: $$1 - \frac{1^3}{3} = 1 - \frac{1}{3} = \frac{2}{3}$$.

At $$x = -1$$: $$(-1) - \frac{(-1)^3}{3} = -1 - \frac{-1}{3} = -1 + \frac{1}{3} = -\frac{2}{3}$$.

Subtract the value at the lower limit from the value at the upper limit:

$$ A = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} $$

Hence, the area is $$\frac{4}{3}$$ square units. Comparing with the options, this corresponds to Option D.

So, the answer is $$\frac{4}{3}$$.