Question

\(\int_{1}^{4} \log_e [x] dx\) equals

• A. \(\log_e 2\)
• B. \(\log_e 3\)
• C. \(\log_e 6\)
• D. None of the above

Hint:

For \(1 \le x<2\), \([x]=1\); for \(2 \le x<3\), \([x]=2\); for \(3 \le x<4\), \([x]=3\).

Answer 1

The Correct Option is C

To solve the integral \(\int_{1}^{4} \log_e [x] \, dx\), we first need to understand the behavior of the floor function \([x]\), which represents the greatest integer less than or equal to \(x\). The given integral can thus be broken down into segments where \([x]\) has a constant value over each interval:

From \(x = 1\) to \(x < 2\), \([x] = 1\).

From \(x = 2\) to \(x < 3\), \([x] = 2\).

From \(x = 3\) to \(x < 4\), \([x] = 3\).

Now, let's compute the integral over these intervals:

1. The integral from 1 to 2: \(\int_{1}^{2} \log_e 1 \, dx = 0\) 
   Since \(\log_e 1 = 0\), the entire contribution from this segment is 0.
2. The integral from 2 to 3: \(\int_{2}^{3} \log_e 2 \, dx = \log_e 2 \cdot 1 = \log_e 2\) 
   Here, we integrate a constant over a length of 1 unit.
3. The integral from 3 to 4: \(\int_{3}^{4} \log_e 3 \, dx = \log_e 3 \cdot 1 = \log_e 3\) 
   Again, we integrate a constant over a length of 1 unit.

Adding these results together, the total sum of the integrals is:

\(\int_{1}^{4} \log_e [x] \, dx = 0 + \log_e 2 + \log_e 3 = \log_e 2 + \log_e 3\)

Using the property of logarithms that states \(\log_e a + \log_e b = \log_e (a \times b)\), we get:

\(\log_e 2 + \log_e 3 = \log_e (2 \times 3) = \log_e 6\).

Thus, the value of the integral \(\int_{1}^{4} \log_e [x] \, dx\) equals \(\log_e 6\), which matches the given correct answer.