To solve the integral \( I = \int_{-1}^{1} x^2 e^{\lfloor x^3 \rfloor} \, dx \), we will break it down into two parts based on the behavior of the greatest integer function \( \lfloor x^3 \rfloor \).
### Step 1: Break the Integral
The function \( \lfloor x^3 \rfloor \) changes its value at \( x = 0 \). Therefore, we can split the integral as follows:
\[
I = \int_{-1}^{0} x^2 e^{\lfloor x^3 \rfloor} \, dx + \int_{0}^{1} x^2 e^{\lfloor x^3 \rfloor} \, dx
\]
### Step 2: Evaluate the Integral from -1 to 0
For \( x \) in the interval \([-1, 0]\):
- Here, \( x^3 \) ranges from \(-1\) to \(0\).
- Thus, \( \lfloor x^3 \rfloor = -1 \).
So, we have:
\[
\int_{-1}^{0} x^2 e^{\lfloor x^3 \rfloor} \, dx = \int_{-1}^{0} x^2 e^{-1} \, dx = e^{-1} \int_{-1}^{0} x^2 \, dx
\]
### Step 3: Calculate \( \int_{-1}^{0} x^2 \, dx \)
The integral \( \int x^2 \, dx \) is given by:
\[
\int x^2 \, dx = \frac{x^3}{3}
\]
Evaluating from \(-1\) to \(0\):
\[
\int_{-1}^{0} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{0} = \frac{0^3}{3} - \frac{(-1)^3}{3} = 0 + \frac{1}{3} = \frac{1}{3}
\]
Thus,
\[
\int_{-1}^{0} x^2 e^{-1} \, dx = e^{-1} \cdot \frac{1}{3} = \frac{1}{3e}
\]
### Step 4: Evaluate the Integral from 0 to 1
For \( x \) in the interval \([0, 1]\):
- Here, \( x^3 \) ranges from \(0\) to \(1\).
- Thus, \( \lfloor x^3 \rfloor = 0 \).
So, we have:
\[
\int_{0}^{1} x^2 e^{\lfloor x^3 \rfloor} \, dx = \int_{0}^{1} x^2 e^{0} \, dx = \int_{0}^{1} x^2 \, dx
\]
### Step 5: Calculate \( \int_{0}^{1} x^2 \, dx \)
Evaluating the integral:
\[
\int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}
\]
### Step 6: Combine the Results
Now, we can combine both parts:
\[
I = \frac{1}{3e} + \frac{1}{3} = \frac{1}{3e} + \frac{1}{3} = \frac{1 + e}{3e}
\]
### Final Answer
Thus, the value of the integral is:
\[
I = \frac{1 + e}{3e}
\]
