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Question: Answered & Verified by Expert
$\int_{0}^{100} e^{x-[x]} d x$ is equal to
MathematicsDefinite IntegrationWBJEEWBJEE 2017
Options:
  • A $\frac{e^{100}-1}{100}$
  • B $\frac{e^{100}-1}{e-1}$
  • C $100(e-1)$
  • D $\frac{e-1}{100}$
Solution:
2970 Upvotes Verified Answer
The correct answer is: $100(e-1)$
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 $0 < x < 1]$
$=100\left[e^{x}\right]_{0}^{1}$
$=100\left[e^{1}-e^{0}\right]$
$=100(e-1)$

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