Consider `I=int_(0)^(1)(dx)/(1+x^(5)).` Then, I satisfies

Options

A. `I gt 1`
B. `I=1`
C. `I lt1`
D. `I+1 lt0`

Answer

To solve the integral \( I = \int_{0}^{1} \frac{dx}{1 + x^5} \) and determine the bounds for \( I \), we can follow these steps:

### Step 1: Identify the bounds of the function
We know that \( 1 + x^5 \) is a continuous and increasing function for \( x \) in the interval \( [0, 1] \).

- At \( x = 0 \):
\[ 1 + 0^5 = 1 \]
- At \( x = 1 \):
\[ 1 + 1^5 = 2 \]

Thus, we have:
\[ 1 < 1 + x^5 < 2 \quad \text{for } x \in [0, 1] \]

### Step 2: Take the reciprocal
Taking the reciprocal of the inequality (and reversing the inequality signs), we get:
\[ \frac{1}{2} < \frac{1}{1 + x^5} < 1 \quad \text{for } x \in [0, 1] \]

### Step 3: Integrate the inequalities
Now, we can integrate the entire inequality from 0 to 1:
\[ \int_{0}^{1} \frac{1}{2} \, dx < \int_{0}^{1} \frac{1}{1 + x^5} \, dx < \int_{0}^{1} 1 \, dx \]

Calculating the integrals:
- The left integral:
\[ \int_{0}^{1} \frac{1}{2} \, dx = \frac{1}{2} \cdot (1 - 0) = \frac{1}{2} \]
- The right integral:
\[ \int_{0}^{1} 1 \, dx = 1 \cdot (1 - 0) = 1 \]

### Step 4: Combine the results
Thus, we have:
\[ \frac{1}{2} < I < 1 \]

### Conclusion
This means that the value of \( I \) satisfies:
\[ \frac{1}{2} < I < 1 \]

Therefore, we conclude that \( I < 1 \).