The integral `I=int_(e)^(e+1)(1+x^(2))/(1+x^(3))dx` satisfies

Options

A. `I gt 2`
B. `I gt e`
C. `I lt 0`
D. `I lt1`

Answer

To solve the integral \( I = \int_{e}^{e+1} \frac{1 + x^2}{1 + x^3} \, dx \) and determine the inequalities it satisfies, we can follow these steps:

### Step 1: Analyze the integrand
We start with the integrand \( \frac{1 + x^2}{1 + x^3} \). We want to compare this expression to 1.

### Step 2: Establish the inequality
We know that for \( x \) in the interval \( [e, e+1] \):
- \( x^2 < x^3 \) (since \( x > 1 \) in this interval)
- Therefore, \( 1 + x^2 < 1 + x^3 \).

This implies:
\[ \frac{1 + x^2}{1 + x^3} < 1 \]

### Step 3: Integrate both sides
Now, we can integrate both sides over the interval \( [e, e+1] \):
\[ \int_{e}^{e+1} \frac{1 + x^2}{1 + x^3} \, dx < \int_{e}^{e+1} 1 \, dx \]

### Step 4: Calculate the right side
The integral on the right side is straightforward:
\[ \int_{e}^{e+1} 1 \, dx = [x]_{e}^{e+1} = (e + 1) - e = 1 \]

### Step 5: Conclude the inequality
Thus, we have:
\[ I < 1 \]

### Step 6: Verify the options
Now, let's check the options given in the problem:
- \( I > 2 \) (False)
- \( I > e \) (False, since \( e \approx 2.718 \))
- \( I < 0 \) (False, since \( I \) is positive)
- \( I > 1 \) (False, since \( I < 1 \))

The only valid conclusion we can draw is:
\[ I < 1 \]

### Final Answer
Thus, the integral \( I \) satisfies \( I < 1 \).