WBJEE · Maths · Definite Integration
\(\int_{0}^{100} e^{x-[x]} d x\) is equal to
- A \(\frac{e^{100}-1}{100}\)
- B \(\frac{e^{100}-1}{e-1}\)
- C \(100(e-1)\)
- D \(\frac{e-1}{100}\)
Answer & Solution
Correct Answer
(C) \(100(e-1)\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_{0}^{100} e^{x-[x]} d x\) \(=100 \int_{0}^{1} e^{x-[x]} d x\) \([\because x-[x]\) is a periodic function of period 1 and \(\int_{0}^{mT} f(x) d x=m \int_{0}^{T} f(x) d x,\) where \(T\) is period of \(f(x)]\) \(=100 \int_{0}^{1} e^{x} d x[\because x-[x]=x\) for…
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