Question

The expression \(\frac{\int_{0}^{n}[x]dx}{\int_{0}^{n}\left \{ x\right \}dx}\), where [x] and {x} are respectively integral and fractional part of x and n∈N, is equal to

• A. \(\frac{1}{n-1}\)
• B. \(\frac{1}{n}\)
• C. n
• D. n-1

Answer 1

The Correct Option is D

Given:
\[ \int_{0}^{n}[x]\,dx \] Since \([x]\) is the greatest integer function, we break the interval \([0, n]\) into parts:
\[ = \int_{0}^{1} 0\,dx + \int_{1}^{2} 1\,dx + \int_{2}^{3} 2\,dx + \cdots + \int_{n-1}^{n} (n-1)\,dx \] \[ = 0 + 1(2-1) + 2(3-2) + \cdots + (n-1)(n - (n-1)) = 1 + 2 + \cdots + (n-1) \] Using the formula:
\[ 1 + 2 + \cdots + (n-1) = \frac{n(n-1)}{2} \tag{i} \] Also, \[ \int_{0}^{n} \{x\}\,dx = \int_{0}^{n}(x - [x])\,dx = \int_{0}^{n} x\,dx - \int_{0}^{n} [x]\,dx \] \[ = \frac{n^2}{2} - \frac{n(n-1)}{2} = \frac{n}{2} \tag{ii} \] Now: 
\[ \frac{\int_{0}^{n}[x]\,dx}{\int_{0}^{n} \{x\}\,dx} = \frac{\frac{n(n-1)}{2}}{\frac{n}{2}} = n - 1 \] Final Answer: Option (D): \(n - 1\)