Question:
The value of
\(\displaystyle\sum_{n=1}^{100}\int_{n-1}^{n}e^{x-[x]}\,dx\)
, where [ x ] is the greatest integer \(\le\) x, is :
The value of
\(\displaystyle\sum_{n=1}^{100}\int_{n-1}^{n}e^{x-[x]}\,dx\)
, where [ x ] is the greatest integer \(\le\) x, is :
\(\displaystyle\sum_{n=1}^{100}\int_{n-1}^{n}e^{x-[x]}\,dx\)
, where [ x ] is the greatest integer \(\le\) x, is :
Show Hint
\(e^{x-[x]}\) has period 1, so each integral over a unit interval equals \(\int_0^1e^x\,dx\).
Updated On: Nov 4, 2026
- \(100e\)
- \(100(e-1)\)
- \(100(1+e)\)
- \(100(1-e)\)
Show Solution
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The Correct Option is B
Solution and Explanation
The function \(e^{x-[x]}=e^{\{x\}}\) is periodic with period 1, so \(\int_{n-1}^{n}e^{x-[x]}dx=\int_0^1e^{x}dx=e-1\) for every \(n\).
The sum has 100 equal terms: \(100(e-1)\).
Answer: (2) \(100(e-1)\)
The sum has 100 equal terms: \(100(e-1)\).
Answer: (2) \(100(e-1)\)
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