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 \( I = \int_0^1 \frac{dx}{1+x^5} \), we will analyze the behavior of the function in the integrand and apply properties of inequalities.

### Step 1: Determine the bounds of the integrand
The function in the denominator is \( 1 + x^5 \). We need to find its minimum and maximum values in the interval \( [0, 1] \).

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

Thus, \( 1 + x^5 \) ranges from \( 1 \) to \( 2 \) as \( x \) varies from \( 0 \) to \( 1 \).

### Step 2: Establish inequalities for the integrand
Since \( 1 + x^5 \) is between \( 1 \) and \( 2 \), we can write:
\[ 1 < 1 + x^5 < 2 \]

Taking the reciprocal (and reversing the inequalities because the values are positive), we get:
\[ \frac{1}{2} < \frac{1}{1 + x^5} < 1 \]

### Step 3: Integrate the inequalities
Now we will integrate the inequalities over the interval \( [0, 1] \):
\[ \int_0^1 \frac{1}{2} \, dx < \int_0^1 \frac{1}{1 + x^5} \, dx < \int_0^1 1 \, dx \]

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:
\[ \int_0^1 1 \, dx = 1 \cdot (1 - 0) = 1 \]

Thus, we have:
\[ \frac{1}{2} < I < 1 \]

### Conclusion
From the inequality \( \frac{1}{2} < I < 1 \), we can conclude that \( I \) satisfies the condition \( I < 1 \).

### Final Answer
The correct option is \( C: I < 1 \).

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