Aspire Faculty ID #13639 · Topic: JAMIA MILLIA ISLAMIA MCA 2023 · Just now
JAMIA MILLIA ISLAMIA MCA 2023

$\displaystyle \int_{0}^{1000} e^{\,x-[x]}\,dx$ is –

Solution

Solution: For $x \in [n,n+1)$, we have $[x]=n$. $\displaystyle \int_n^{n+1} e^{\,x-[x]}dx = e^{-n}\!\int_n^{n+1} e^{x}dx = e^{-n}(e^{n+1}-e^{n})=e-1.$ The interval $[0,1000)$ has $1000$ such unit pieces, so the total integral is $1000(e-1)$.

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