Let `f(x)=inte^x (x - 1)(x-2) dx`. Then f decreases in the interval

Options

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

Answer

To find the interval in which 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
To determine where the function is decreasing, we first differentiate \( f(x) \). By the Fundamental Theorem of Calculus, the derivative of the integral is simply the integrand:

\[ f'(x) = e^x (x - 1)(x - 2) \]

### Step 2: Set the Derivative Less Than or Equal to Zero
The function \( f(x) \) is decreasing where its derivative is less than or equal to zero:

\[ f'(x) \leq 0 \implies e^x (x - 1)(x - 2) \leq 0 \]

### Step 3: Analyze the Exponential Function
The term \( e^x \) is always positive for all \( x \). Therefore, we can simplify our inequality to:

\[ (x - 1)(x - 2) \leq 0 \]

### Step 4: Solve the Inequality
Next, we need to find the values of \( x \) that satisfy this inequality. The critical points occur when \( (x - 1)(x - 2) = 0 \), which gives us:

\[ x = 1 \quad \text{and} \quad x = 2 \]

### Step 5: Test Intervals
We will test the intervals determined by the critical points \( x = 1 \) and \( x = 2 \):

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

2. **Interval \( (1, 2) \)**:
- Choose \( x = 1.5 \):
\[ (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 \quad (\text{negative}) \]

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

### Step 6: Conclusion on Intervals
From our tests, we find that \( (x - 1)(x - 2) \leq 0 \) is satisfied in the interval \( [1, 2] \). Therefore, the function \( f(x) \) is decreasing in the interval:

\[ \boxed{[1, 2]} \]