JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of \(\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]} d x,\) where \([x]\) is the greatest integer \(\leq x ,\) is
- A \(100( e -1)\)
- B \(100(1- e )\)
- C \(100 e\)
- D \(100(1+ e )\)
Answer & Solution
Correct Answer
(A) \(100( e -1)\)
Step-by-step Solution
Detailed explanation
\(\sum_{n=1}^{100} \int_{n-1}^{n} e^{[x\}} d x,\) period of \(\{x\}=1\) \(\sum_{n=1}^{100} \int_{0}^{1} e^{\mid x\}} d x=\sum_{n=1}^{100} \int_{0}^{1} e^{x} d x\) \(\sum_{n=1}^{100}(e-1)=100(e-1)\)
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