The value of `int_(-1)^(1)x^(2)e^([x^(3)])dx`, where [t] denotes the greatest integer `le t`, is :

Options

A. `(e-1)/(3e)`
B. `(e+1)/(3)`
C. `(e+1)/(3e)`
D. `(1)/(3e)`

Answer

To solve the integral \( I = \int_{-1}^{1} x^2 e^{[\text{x}^3]} \, dx \), where \([\cdot]\) denotes the greatest integer function, we will break the integral into two parts based on the behavior of the greatest integer function over the interval \([-1, 1]\).

### Step 1: Break the Integral
The integral can be split into two parts:
\[ I = \int_{-1}^{0} x^2 e^{[\text{x}^3]} \, dx + \int_{0}^{1} x^2 e^{[\text{x}^3]} \, dx \]

### Step 2: Evaluate the First Integral \(\int_{-1}^{0} x^2 e^{[\text{x}^3]} \, dx\)
For \( x \) in the interval \([-1, 0]\), \( x^3 \) will also lie in \([-1, 0]\). The greatest integer function \([\text{x}^3]\) will yield \(-1\) since the values of \( x^3 \) will be less than \( 0 \) but greater than or equal to \(-1\). Therefore, we have:
\[ I_1 = \int_{-1}^{0} x^2 e^{-1} \, dx = e^{-1} \int_{-1}^{0} x^2 \, dx \]

### Step 3: Calculate \(\int_{-1}^{0} x^2 \, dx\)
The integral can be computed as follows:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from \(-1\) to \(0\):
\[ \int_{-1}^{0} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{0} = \frac{0^3}{3} - \frac{(-1)^3}{3} = 0 - \left(-\frac{1}{3}\right) = \frac{1}{3} \]
Thus,
\[ I_1 = e^{-1} \cdot \frac{1}{3} = \frac{1}{3e} \]

### Step 4: Evaluate the Second Integral \(\int_{0}^{1} x^2 e^{[\text{x}^3]} \, dx\)
For \( x \) in the interval \([0, 1]\), \( x^3 \) will also lie in \([0, 1]\). The greatest integer function \([\text{x}^3]\) will yield \(0\) since \(x^3\) will be less than \(1\) but greater than or equal to \(0\). Therefore, we have:
\[ I_2 = \int_{0}^{1} x^2 e^{0} \, dx = \int_{0}^{1} x^2 \, dx \]

### Step 5: Calculate \(\int_{0}^{1} x^2 \, dx\)
Using the same integral formula:
\[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]
Thus,
\[ I_2 = \frac{1}{3} \]

### Step 6: Combine the Results
Now, we can combine both parts of the integral:
\[ I = I_1 + I_2 = \frac{1}{3e} + \frac{1}{3} = \frac{1}{3} \left( \frac{1}{e} + 1 \right) \]

### Final Result
Thus, the final value of the integral is:
\[ I = \frac{1 + e}{3e} \]