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WBJEE · Maths · Definite Integration

Let \(I_{1}=\int_{0}^{n}[x] d x\) and \(I_{2}=\int_{0}^{n}\{x\} d x,\) where \([x]\)
and \(\{x\}\) are integral and fractional parts of \(x\) and \(n \in N-\{1\} .\) Then, \(I_{1} / I_{2}\) is equal to

  1. A \(\frac{1}{n-1}\)
  2. B \(\frac{1}{n}\)
  3. C \(n\)
  4. D \(n-1\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(n-1\)

Step-by-step Solution

Detailed explanation

We have, \(I_{1}=\int_{0}^{n}[x] d x\) \(=\int_{0}^{1}[x] d x+\int_{1}^{2}[x] d x\) \(\quad+\int_{2}^{3}[x] d x+\ldots+\int_{n-1}^{n}[x] d x\) \(=\int_{0}^{1} 0 d x+\int_{2}^{3} 1 d x+\int_{2}^{3} 2 d x\) \(\quad+\ldots+\int_{n-1}^{n}(n-1) d x\)…