Key Concepts and Formulas

• Greatest Integer Function: The greatest integer function, $[x]$, gives the largest integer less than or equal to $x$.
• Properties of Definite Integrals: When evaluating a definite integral of a function involving the greatest integer function, the interval of integration must be split into sub-intervals where the greatest integer function is constant.
• Integral of $x^n$: The indefinite integral of $x^n$ (where $n \neq -1$) is $\frac{x^{n+1}}{n+1} + C$.

Step-by-Step Solution

Step 1: Analyze the argument of the greatest integer function. The integrand is $x^2 e^{[x^3]}$. We need to understand the behavior of $[x^3]$ over the interval of integration $[-1, 1]$. As $x$ varies from $-1$ to $1$, $x^3$ varies from $(-1)^3 = -1$ to $(1)^3 = 1$. Thus, we have $-1 \le x^3 \le 1$. The possible integer values for $[x^3]$ in this range are $-1$, $0$, and $1$.

Step 2: Determine the intervals for each value of the greatest integer function. We need to find the values of $x$ for which $[x^3]$ takes on these integer values:

• $[x^3] = -1$ when $-1 \le x^3 < 0$. Taking the cube root, this corresponds to $-1 \le x < 0$.
• $[x^3] = 0$ when $0 \le x^3 < 1$. Taking the cube root, this corresponds to $0 \le x < 1$.
• $[x^3] = 1$ when $x^3 = 1$. Taking the cube root, this corresponds to $x = 1$.

Step 3: Split the definite integral based on the intervals found. The interval of integration is $[-1, 1]$. Based on the analysis in Step 2, the value of $[x^3]$ changes at $x=0$. Therefore, we split the integral into two parts: from $-1$ to $0$ and from $0$ to $1$. $\int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}} dx = \int\limits_{ - 1}^0 {{x^2}{e^{[{x^3}]}}} dx + \int\limits_0^1 {{x^2}{e^{[{x^3}]}}} dx$

Step 4: Evaluate the first sub-integral. For the interval $[-1, 0)$, we have $[x^3] = -1$. So, the first integral becomes: $\int\limits_{ - 1}^0 {{x^2}{e^{ - 1}}} dx$ Since $e^{-1}$ is a constant, we can pull it out of the integral: ${e^{ - 1}} \int\limits_{ - 1}^0 {{x^2}} dx$ Now, we integrate $x^2$: ${e^{ - 1}} \left[ {{{x^3}} \over 3} \right]_{ - 1}^0$ Applying the limits of integration: {e^{ - 1}} \left( {{{0^3}} \over 3} - {{{(-1)}^3}} \over 3} \right) = {e^{ - 1}} \left( 0 - \left( {{ - 1} \over 3} \right) \right) = {e^{ - 1}} \left( {{1 \over 3}} \right) = {1 \over {3e}}

Step 5: Evaluate the second sub-integral. For the interval $[0, 1)$, we have $[x^3] = 0$. The value of the integrand at a single point ($x=1$) does not affect the value of the definite integral. Thus, for integration purposes over $[0, 1]$, we can consider $[x^3] = 0$. The second integral becomes: $\int\limits_0^1 {{x^2}{e^0}} dx$ Since $e^0 = 1$, the integral simplifies to: $\int\limits_0^1 {{x^2}} dx$ Now, we integrate $x^2$: $\left[ {{{x^3}} \over 3} \right]_0^1$ Applying the limits of integration: $\left( {{{1^3}} \over 3} - {{{0^3}} \over 3} \right) = {{1 \over 3} - 0} = {{1 \over 3}}$

Step 6: Combine the results of the sub-integrals. Add the values of the two sub-integrals to find the total value of the original integral: $\int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}} dx = {1 \over {3e}} + {1 \over 3}$ To combine these terms, find a common denominator, which is $3e$: ${{1} \over {3e}} + {{e} \over {3e}} = {{1 + e} \over {3e}}$

Common Mistakes & Tips

• Incorrectly splitting the interval: Ensure the integral is split at every point where the argument of the greatest integer function crosses an integer value.
• Misinterpreting the greatest integer function: Remember that $[x]$ is the greatest integer less than or equal to $x$. For example, $[ -0.5 ] = -1$.
• Handling the endpoint: For definite integrals, the value of the integrand at a single point does not affect the integral's value. This is crucial when the argument of the greatest integer function reaches an integer at an endpoint of a sub-interval.

Summary To evaluate the given definite integral involving the greatest integer function, we first analyzed the behavior of $x^3$ over the interval $[-1, 1]$ to determine the values of $[x^3]$. We then split the integral into sub-intervals where $[x^3]$ was constant. Each sub-integral was evaluated separately and then the results were summed to obtain the final answer. The first sub-integral from $-1$ to $0$ yielded $\frac{1}{3e}$, and the second sub-integral from $0$ to $1$ yielded $\frac{1}{3}$. Their sum is $\frac{1+e}{3e}$.

The final answer is \boxed{\frac{e + 1}{3e}} which corresponds to option (D).