The value of $$\int_{-1}^{1} x^2 e^{[x^3]} dx$$, where $$[t]$$ denotes the greatest integer $$\leq t$$, is:

We need to evaluate $$\int_{-1}^{1} x^2 e^{[x^3]}\, dx$$, where $$[t]$$ denotes the greatest integer function.

We split the integral based on the behavior of $$[x^3]$$. For $$x \in [-1, 0)$$: $$x^3 \in [-1, 0)$$, so $$[x^3] = -1$$. For $$x \in [0, 1)$$: $$x^3 \in [0, 1)$$, so $$[x^3] = 0$$. At $$x = 1$$: $$x^3 = 1$$, but a single point does not affect the integral.

Therefore: $$\int_{-1}^{1} x^2 e^{[x^3]}\, dx = \int_{-1}^{0} x^2 e^{-1}\, dx + \int_{0}^{1} x^2 e^{0}\, dx$$.

Computing the first integral: $$e^{-1} \int_{-1}^{0} x^2\, dx = e^{-1} \left[\frac{x^3}{3}\right]_{-1}^{0} = e^{-1}\left(0 - \left(-\frac{1}{3}\right)\right) = \frac{1}{3e}$$.

Computing the second integral: $$\int_{0}^{1} x^2\, dx = \left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1}{3}$$.

Adding the two parts: $$\frac{1}{3e} + \frac{1}{3} = \frac{1 + e}{3e} = \frac{e + 1}{3e}$$.

Therefore, the value of the integral is $$\frac{e + 1}{3e}$$.