If `I=int_(0)^(1) (1)/(1+x^(pi//2))dx`then

Options

A. `"In" 2 lt1lt(pi)/(4)`
B. ` I lt"In "2`
C. ` "In "gt(pi)/(4)`
D. none of these

Explanation

To solve the integral \( I = \int_{0}^{1} \frac{1}{1 + x^{\frac{\pi}{2}}} \, dx \), we will use the method of comparison.

### Step 1: Establish inequalities
We know that \( \frac{\pi}{2} \) is approximately between 1 and 2. Therefore, we can compare \( x^{\frac{\pi}{2}} \) with \( x^1 \) and \( x^2 \) for \( x \in [0, 1] \):

\[ 1 < x^{\frac{\pi}{2}} < 2 \quad \text{for } x \in [0, 1] \]

This gives us the following inequalities:

\[ \frac{1}{1 + x^2} < \frac{1}{1 + x^{\frac{\pi}{2}}} < \frac{1}{1 + x} \]

### Step 2: Integrate the inequalities
Now we will integrate all parts of the inequality from 0 to 1:

\[ \int_{0}^{1} \frac{1}{1 + x^2} \, dx < \int_{0}^{1} \frac{1}{1 + x^{\frac{\pi}{2}}} \, dx < \int_{0}^{1} \frac{1}{1 + x} \, dx \]

### Step 3: Calculate the integrals
1. **Left Integral**:
\[ \int_{0}^{1} \frac{1}{1 + x^2} \, dx = \tan^{-1}(x) \Big|_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

2. **Right Integral**:
\[ \int_{0}^{1} \frac{1}{1 + x} \, dx = \ln(1 + x) \Big|_{0}^{1} = \ln(2) - \ln(1) = \ln(2) - 0 = \ln(2) \]

### Step 4: Combine results
From the inequalities and the calculated integrals, we have:

\[ \frac{\pi}{4} > I > \ln(2) \]

### Conclusion
Thus, we conclude that:

\[ \ln(2) < I < \frac{\pi}{4} \]

### Final Answer
The answer is that \( I \) lies between \( \ln(2) \) and \( \frac{\pi}{4} \).

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