Let `f(x) =inte^x(x-1)(x-2)dx,` then f(x) decrease in the interval

Options

A. `(-oo,-2)`
B. (-2,-1)
C. (1,2)
D. `(2,oo)`

Explanation

To determine the interval where the function \( f(x) = \int e^x (x-1)(x-2) \, dx \) is decreasing, we need to follow these steps:

### Step 1: Differentiate the function
First, we find the derivative of \( f(x) \):
\[ f'(x) = e^x (x-1)(x-2) \]

### Step 2: Analyze the critical points
Next, we need to find the critical points where \( f'(x) = 0 \):
\[ e^x (x-1)(x-2) = 0 \]
Since \( e^x \) is never zero, we set the polynomial part to zero:
\[ (x-1)(x-2) = 0 \]
This gives us the critical points:
\[ x = 1 \quad \text{and} \quad x = 2 \]

### Step 3: Test intervals around the critical points
We will test the sign of \( f'(x) \) in the intervals defined by the critical points: \( (-\infty, 1) \), \( (1, 2) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, 1) \)**:
- Choose \( x = 0 \):
\[ f'(0) = e^0 (0-1)(0-2) = 1 \cdot (-1) \cdot (-2) = 2 \quad (\text{positive}) \]

2. **Interval \( (1, 2) \)**:
- Choose \( x = 1.5 \):
\[ f'(1.5) = e^{1.5} (1.5-1)(1.5-2) = e^{1.5} \cdot 0.5 \cdot (-0.5) = -\frac{e^{1.5}}{4} \quad (\text{negative}) \]

3. **Interval \( (2, \infty) \)**:
- Choose \( x = 3 \):
\[ f'(3) = e^3 (3-1)(3-2) = e^3 \cdot 2 \cdot 1 = 2e^3 \quad (\text{positive}) \]

### Step 4: Conclusion
From the sign analysis:
- \( f'(x) > 0 \) in \( (-\infty, 1) \) (increasing)
- \( f'(x) < 0 \) in \( (1, 2) \) (decreasing)
- \( f'(x) > 0 \) in \( (2, \infty) \) (increasing)

Thus, the function \( f(x) \) is decreasing in the interval \( (1, 2) \).

### Final Answer
The function \( f(x) \) decreases in the interval \( (1, 2) \).