`int_(3)^(10)[log[x]]dx` is equal to (where [.] represents the greatest integer function)

Options

A. 9
B. `16-e`
C. `10`
D. `10+e`

Explanation

To solve the integral \( \int_{3}^{10} [\log x] \, dx \), where \([\cdot]\) represents the greatest integer function, we will first analyze the behavior of the function \([\log x]\) over the interval from 3 to 10.

### Step 1: Determine the intervals for \([\log x]\)

1. **Calculate \(\log x\) at the endpoints:**
- \(\log 3 \approx 1.0986\)
- \(\log 10 = 1\)

2. **Identify the range of \(\log x\) from \(x = 3\) to \(x = 10\):**
- At \(x = 3\), \([\log 3] = 1\)
- At \(x = 10\), \([\log 10] = 2\)

3. **Identify critical points where \([\log x]\) changes:**
- \(\log 4 \approx 1.386\) → \([\log 4] = 1\)
- \(\log 5 \approx 1.609\) → \([\log 5] = 1\)
- \(\log 6 \approx 1.792\) → \([\log 6] = 1\)
- \(\log 7 \approx 1.946\) → \([\log 7] = 1\)
- \(\log 8 \approx 2.079\) → \([\log 8] = 2\)
- \(\log 9 \approx 2.197\) → \([\log 9] = 2\)

### Step 2: Break the integral into intervals

From the analysis, we can break the integral into two parts based on the values of \([\log x]\):

- From \(x = 3\) to \(x = 8\), \([\log x] = 1\)
- From \(x = 8\) to \(x = 10\), \([\log x] = 2\)

Thus, we can write:
\[ \int_{3}^{10} [\log x] \, dx = \int_{3}^{8} 1 \, dx + \int_{8}^{10} 2 \, dx \]

### Step 3: Evaluate the integrals

1. **Evaluate \(\int_{3}^{8} 1 \, dx\):**
\[ \int_{3}^{8} 1 \, dx = [x]_{3}^{8} = 8 - 3 = 5 \]

2. **Evaluate \(\int_{8}^{10} 2 \, dx\):**
\[ \int_{8}^{10} 2 \, dx = 2[x]_{8}^{10} = 2(10 - 8) = 2 \times 2 = 4 \]

### Step 4: Combine the results

Now, we combine the results of the two integrals:
\[ \int_{3}^{10} [\log x] \, dx = 5 + 4 = 9 \]

### Final Answer
Thus, the value of the integral \( \int_{3}^{10} [\log x] \, dx \) is \( \boxed{9} \).