To solve the given problem, we need to evaluate the integral \( I = \int_{-2}^{2} f(x) \, dx \) where the function \( f(x) \) is defined piecewise. We will break it down into three parts based on the intervals provided.
### Step 1: Define the Function
The function \( f(x) \) is defined as follows:
- For \( x \leq -1 \): \( f(x) = x |x| = x^2 \) (since \( |x| = -x \))
- For \( -1 < x < 1 \): \( f(x) = [x + 1] + [1 - x] \)
- For \( x \geq 1 \): \( f(x) = -x |x| = -x^2 \)
### Step 2: Break Down the Integral
We can split the integral into three parts according to the definition of \( f(x) \):
\[
I = \int_{-2}^{-1} f(x) \, dx + \int_{-1}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx
\]
### Step 3: Evaluate Each Integral
#### Part 1: \( \int_{-2}^{-1} f(x) \, dx \)
For \( x \leq -1 \), \( f(x) = x^2 \):
\[
\int_{-2}^{-1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-2}^{-1} = \left( \frac{(-1)^3}{3} - \frac{(-2)^3}{3} \right) = \left( -\frac{1}{3} + \frac{8}{3} \right) = \frac{7}{3}
\]
#### Part 2: \( \int_{-1}^{1} f(x) \, dx \)
For \( -1 < x < 1 \), we need to evaluate \( f(x) = [x + 1] + [1 - x] \):
- For \( -1 < x < 0 \): \( [x + 1] = 0 \) and \( [1 - x] = 1 \) → \( f(x) = 0 + 1 = 1 \)
- For \( 0 < x < 1 \): \( [x + 1] = 1 \) and \( [1 - x] = 1 \) → \( f(x) = 1 + 1 = 2 \)
Thus, we can break this integral into two parts:
\[
\int_{-1}^{0} 1 \, dx + \int_{0}^{1} 2 \, dx = [x]_{-1}^{0} + [2x]_{0}^{1} = (0 - (-1)) + (2 \cdot 1 - 0) = 1 + 2 = 3
\]
#### Part 3: \( \int_{1}^{2} f(x) \, dx \)
For \( x \geq 1 \), \( f(x) = -x^2 \):
\[
\int_{1}^{2} -x^2 \, dx = -\left[ \frac{x^3}{3} \right]_{1}^{2} = -\left( \frac{2^3}{3} - \frac{1^3}{3} \right) = -\left( \frac{8}{3} - \frac{1}{3} \right) = -\left( \frac{7}{3} \right) = -\frac{7}{3}
\]
### Step 4: Combine the Results
Now, we can combine the results from all three parts:
\[
I = \frac{7}{3} + 3 - \frac{7}{3} = 3
\]
### Step 5: Calculate \( |3I| \)
Finally, we need to find \( |3I| \):
\[
|3I| = |3 \cdot 3| = |9| = 9
\]
### Final Answer
Thus, the final answer is:
\[
\boxed{9}
\]
