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
- A \(\frac{1}{n-1}\)
- B \(\frac{1}{n}\)
- C \(n\)
- D \(n-1\)
Answer & Solution
Correct Answer
(D) \(n-1\)
Step-by-step Solution
Detailed explanation
We have, \(I_{1}=\int_{0}^{n}[x] d x\) \(=\int_{0}^{1}[x] d x+\int_{1}^{2}[x] d x\) \(\quad+\int_{2}^{3}[x] d x+\ldots+\int_{n-1}^{n}[x] d x\) \(=\int_{0}^{1} 0 d x+\int_{2}^{3} 1 d x+\int_{2}^{3} 2 d x\) \(\quad+\ldots+\int_{n-1}^{n}(n-1) d x\)…
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