If `I= int_(0)^(1)(x)/(8+x^(3))dx` then the smallest interval is which I less is

Options

A. (0,1/8)
B. (0,1/9)
C. (0,1/10)
D. (0,1/7)

Explanation

To solve the integral \( I = \int_{0}^{1} \frac{x}{8 + x^3} \, dx \) and find the smallest interval in which \( I \) lies, we can follow these steps:

### Step 1: Define the function
Let \( f(x) = \frac{x}{8 + x^3} \).

### Step 2: Find the derivative
To analyze the behavior of \( f(x) \), we need to find its derivative \( f'(x) \):
\[ f'(x) = \frac{(8 + x^3)(1) - x(3x^2)}{(8 + x^3)^2} \]
This simplifies to:
\[ f'(x) = \frac{8 + x^3 - 3x^3}{(8 + x^3)^2} = \frac{8 - 2x^3}{(8 + x^3)^2} \]

### Step 3: Determine the sign of the derivative
We need to analyze the sign of \( f'(x) \) over the interval \( [0, 1] \):
- The numerator \( 8 - 2x^3 \) is positive for \( x \in [0, 1] \) because \( 2x^3 \leq 2 \) when \( x \leq 1 \).
- Therefore, \( f'(x) > 0 \) for \( x \in [0, 1] \).

This implies that \( f(x) \) is an increasing function on the interval \( [0, 1] \).

### Step 4: Evaluate the function at the endpoints
Now we evaluate \( f(x) \) at the endpoints of the interval:
- At \( x = 0 \):
\[ f(0) = \frac{0}{8 + 0^3} = 0 \]
- At \( x = 1 \):
\[ f(1) = \frac{1}{8 + 1^3} = \frac{1}{9} \]

### Step 5: Establish the bounds for the integral
Since \( f(x) \) is increasing on \( [0, 1] \), we have:
\[ f(0) \leq f(x) \leq f(1) \]
Thus,
\[ 0 \leq I \leq \frac{1}{9} \]

### Conclusion
The integral \( I \) lies in the interval:
\[ I \in [0, \frac{1}{9}] \]

### Final Answer
The smallest interval in which \( I \) lies is \( (0, \frac{1}{9}) \).

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