Question

The value of \[ \int_1^4 \log(\lfloor x \rfloor) \, dx \] where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \), is:

• A. \( \log 2 \)
• B. \( \log 5 \)
• C. \( \log 6 \)
• D. \( \log 3 \)

Hint:

For integrals involving the greatest integer function \( \lfloor x \rfloor \), break the integral into intervals between integers and treat the function as constant over each subinterval.

Answer 1

The Correct Option is C

We are given the definite integral: \[ \int_1^4 \log(\lfloor x \rfloor) \, dx \] The greatest integer function \( \lfloor x \rfloor \) is constant within each interval from \( n \) to \( n+1 \), where \( n \in \mathbb{Z} \). So, break the integral at integer points: \[ = \int_1^2 \log(1) \, dx + \int_2^3 \log(2) \, dx + \int_3^4 \log(3) \, dx \] Evaluate each part: - \( \int_1^2 \log(1) \, dx = \log(1)(2 - 1) = 0 \cdot 1 = 0 \) - \( \int_2^3 \log(2) \, dx = \log(2)(3 - 2) = \log 2 \) - \( \int_3^4 \log(3) \, dx = \log(3)(4 - 3) = \log 3 \) So the total integral is: \[ 0 + \log 2 + \log 3 = \log(2 \cdot 3) = \log 6 \] \[ \boxed{\int_1^4 \log(\lfloor x \rfloor) \, dx = \log 6} \]