If `I=int_(3)^(4) (1)/(3sqrt(logx))dx`then

Options

A. `I lt 1`
B. `I gt 1`
C. `I lt 0.92`
D. none of these

Answer

To solve the integral \( I = \int_{3}^{4} \frac{1}{\sqrt[3]{\log x}} \, dx \), we will follow these steps:

### Step 1: Understand the behavior of the logarithm
Since \( x \) is in the interval \( [3, 4] \), we can analyze \( \log x \):
- For \( x = 3 \), \( \log 3 \) is approximately \( 1.0986 \).
- For \( x = 4 \), \( \log 4 \) is approximately \( 1.3863 \).

Thus, \( \log x \) is always greater than \( 1 \) in the interval \( [3, 4] \).

### Step 2: Analyze \( \sqrt[3]{\log x} \)
Since \( \log x > 1 \), we have:
\[ \sqrt[3]{\log x} > 1 \]
This implies:
\[ \frac{1}{\sqrt[3]{\log x}} < 1 \]

### Step 3: Set up the inequality for the integral
From the above analysis, we can establish the following inequality:
\[ \int_{3}^{4} \frac{1}{\sqrt[3]{\log x}} \, dx < \int_{3}^{4} 1 \, dx \]

### Step 4: Evaluate the right-hand side integral
Now we compute the integral on the right:
\[ \int_{3}^{4} 1 \, dx = x \bigg|_{3}^{4} = 4 - 3 = 1 \]

### Step 5: Combine results
Thus, we have:
\[ I < 1 \]

### Conclusion
Therefore, we conclude that:
\[ I = \int_{3}^{4} \frac{1}{\sqrt[3]{\log x}} \, dx < 1 \]