If `I= underset(1)overset(2)(int)log_(11)(x^3 - x^2 + 6x -5)` dx, then:

A. `0 lt I lt 1/2`
B. `-1ltIlt1/2`
C. `-1ltIlt0`
D. `0lt Ilt 1`

Options

A. `0 lt I lt 1/2`
B. `-1ltIlt1/2`
C. `-1ltIlt0`
D. `0lt Ilt 1`

Explanation

To solve the integral \( I = \int_{1}^{2} \log_{11}(x^3 - x^2 + 6x - 5) \, dx \), we will follow these steps:

### Step 1: Define the function
Let \( f(x) = x^3 - x^2 + 6x - 5 \).

### Step 2: Find the derivative
We need to find the derivative \( f'(x) \) to analyze the behavior of the function:
\[ f'(x) = 3x^2 - 2x + 6 \]

### Step 3: Analyze the derivative
To determine if \( f'(x) \) is always positive, we can check the discriminant of the quadratic:
\[ D = (-2)^2 - 4 \cdot 3 \cdot 6 = 4 - 72 = -68 \]
Since the discriminant is negative, \( f'(x) \) has no real roots and is always positive. Therefore, \( f(x) \) is an increasing function.

### Step 4: Evaluate \( f(x) \) at the endpoints
Now, we will evaluate \( f(x) \) at the bounds of the integral:
- At \( x = 1 \):
\[ f(1) = 1^3 - 1^2 + 6 \cdot 1 - 5 = 1 - 1 + 6 - 5 = 1 \]
- At \( x = 2 \):
\[ f(2) = 2^3 - 2^2 + 6 \cdot 2 - 5 = 8 - 4 + 12 - 5 = 11 \]

### Step 5: Evaluate the logarithm at the endpoints
Now we can evaluate \( \log_{11}(f(x)) \) at the endpoints:
- At \( x = 1 \):
\[ \log_{11}(f(1)) = \log_{11}(1) = 0 \]
- At \( x = 2 \):
\[ \log_{11}(f(2)) = \log_{11}(11) = 1 \]

### Step 6: Analyze the integral
Since \( f(x) \) is increasing from 1 to 11 as \( x \) goes from 1 to 2, \( \log_{11}(f(x)) \) increases from 0 to 1. Therefore, the integral \( I \) represents the area under the curve of \( \log_{11}(f(x)) \) from \( x = 1 \) to \( x = 2 \).

### Step 7: Estimate the value of the integral
The area under the curve will be between the area of a triangle with base 1 and height 1 (which gives \( \frac{1}{2} \)) and the area of a rectangle with height 1 and base 1 (which gives 1). Thus, we can conclude:
\[ \frac{1}{2} < I < 1 \]

### Conclusion
Based on our analysis, we find that:
\[ 0 < I < 1 \]
Thus, the correct option is **D. \( 0 < I < 1 \)**.