To solve the integral \( I = \int_{-2}^{2} |3x^2 - 3x - 6| \, dx \), we will follow these steps:
### Step 1: Factor the expression inside the absolute value
First, we need to factor the quadratic expression \( 3x^2 - 3x - 6 \).
\[
3x^2 - 3x - 6 = 3(x^2 - x - 2)
\]
Next, we factor \( x^2 - x - 2 \):
\[
x^2 - x - 2 = (x - 2)(x + 1)
\]
Thus, we have:
\[
3x^2 - 3x - 6 = 3(x - 2)(x + 1)
\]
### Step 2: Find the roots of the quadratic
The roots of \( 3(x - 2)(x + 1) = 0 \) are \( x = 2 \) and \( x = -1 \). These points will help us determine the intervals where the expression is positive or negative.
### Step 3: Determine the sign of the expression in each interval
We will evaluate the sign of \( 3(x - 2)(x + 1) \) in the intervals defined by the roots:
1. For \( x < -1 \) (e.g., \( x = -2 \)):
\[
3(-2 - 2)(-2 + 1) = 3(-4)(-1) = 12 > 0
\]
So, \( 3x^2 - 3x - 6 > 0 \).
2. For \( -1 < x < 2 \) (e.g., \( x = 0 \)):
\[
3(0 - 2)(0 + 1) = 3(-2)(1) = -6 < 0
\]
So, \( 3x^2 - 3x - 6 < 0 \).
3. For \( x > 2 \) (e.g., \( x = 3 \)):
\[
3(3 - 2)(3 + 1) = 3(1)(4) = 12 > 0
\]
So, \( 3x^2 - 3x - 6 > 0 \).
### Step 4: Rewrite the integral using the sign information
Now we can rewrite the integral \( I \) as follows:
\[
I = \int_{-2}^{-1} (3(x - 2)(x + 1)) \, dx + \int_{-1}^{2} -(3(x - 2)(x + 1)) \, dx + \int_{2}^{2} (3(x - 2)(x + 1)) \, dx
\]
### Step 5: Calculate the integrals
1. **First integral**:
\[
\int_{-2}^{-1} 3(x - 2)(x + 1) \, dx
\]
2. **Second integral**:
\[
\int_{-1}^{2} -3(x - 2)(x + 1) \, dx
\]
3. **Third integral**:
\[
\int_{2}^{2} 3(x - 2)(x + 1) \, dx = 0 \quad \text{(since the limits are the same)}
\]
### Step 6: Evaluate the first integral
Calculating the first integral:
\[
\int_{-2}^{-1} 3(x^2 - x - 2) \, dx
\]
Calculating:
\[
= 3 \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-2}^{-1}
\]
Evaluating at the limits:
\[
= 3 \left[ \left( \frac{(-1)^3}{3} - \frac{(-1)^2}{2} + 2 \right) - \left( \frac{(-2)^3}{3} - \frac{(-2)^2}{2} + 4 \right) \right]
\]
Calculating:
\[
= 3 \left[ \left( -\frac{1}{3} - \frac{1}{2} + 2 \right) - \left( -\frac{8}{3} - 2 + 4 \right) \right]
\]
### Step 7: Evaluate the second integral
Calculating the second integral:
\[
\int_{-1}^{2} -3(x^2 - x - 2) \, dx
\]
Calculating:
\[
= -3 \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-1}^{2}
\]
Evaluating at the limits:
\[
= -3 \left[ \left( \frac{(2)^3}{3} - \frac{(2)^2}{2} - 4 \right) - \left( \frac{(-1)^3}{3} - \frac{(-1)^2}{2} + 2 \right) \right]
\]
### Final Calculation
Combine the results from both integrals and simplify to find the final value of \( I \).
### Final Result
After evaluating all integrals and simplifying, we find:
\[
I = 19
\]
