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 equal to:

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

Hint:

When integrating a step function like \( \lfloor x \rfloor \), break the integral into subintervals where the function is constant and apply the integral rule \( \int_a^b k \, dx = k(b - a) \).

Answer 1

The Correct Option is A

We are given: \[ \int_1^4 \log(\lfloor x \rfloor) \, dx \] The greatest integer function \( \lfloor x \rfloor \) is constant on the intervals: - \( [1, 2) \Rightarrow \lfloor x \rfloor = 1 \) - \( [2, 3) \Rightarrow \lfloor x \rfloor = 2 \) - \( [3, 4) \Rightarrow \lfloor x \rfloor = 3 \) Since the integral is from 1 to 4, and 4 is not included due to the floor function being constant in half-open intervals, we break the integral as: \[ = \int_1^2 \log(1) \, dx + \int_2^3 \log(2) \, dx + \int_3^4 \log(3) \, dx \] Evaluate each: - \( \int_1^2 \log(1) \, dx = 0 \) because \( \log 1 = 0 \) - \( \int_2^3 \log(2) \, dx = \log 2 \cdot (3 - 2) = \log 2 \) - \( \int_3^4 \log(3) \, dx = \log 3 \cdot (4 - 3) = \log 3 \) \[ \Rightarrow \int_1^4 \log(\lfloor x \rfloor) \, dx = \log 2 + \log 3 = \log(2 \cdot 3) = \log 6 \] \[ \boxed{\text{Correct answer: } \log 6} \]