What is \int_{n}^{n+1}(x-[x])\,dx, where [x] is the greatest integer function and n is a natural number?
- A. \frac{4n+1}{2}
- B. \frac{2n+1}{2}
- C. \frac{1}{2} ✓
- D. 1
Correct Answer: C. \frac{1}{2}
Explanation
The integrand x - [x] represents the fractional part function \{x\}, which is periodic with a period of 1. The integral of a periodic function over one complete period is constant regardless of the starting point n. Thus, \int_{n}^{n+1} \{x\} \,dx = \int_{0}^{1} x \,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}.
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