question_answer

Evaluate \[\int_{0}^{2}{{{e}^{x-[x]}}dx.}\]

Answer:

Let        \[l=\int_{0}^{2}{{{e}^{x-[x]}}dx}\] \[=\int_{0}^{1}{{{e}^{x\,-\,[x]}}dx}+\int_{1}^{2}{{{e}^{x\,-\,[x]}}dx}\] \[=\int_{0}^{1}{{{e}^{x}}dx}+\int_{1}^{2}{{{e}^{(x\,\,-\,\,1)}}dx}\] \[[\because [x]=0,\,\,0<x<1\,\,and\,\,[x]=1,\,\,1<x<2]\] \[=[{{e}^{x}}]_{0}^{1}+[{{e}^{x\,\,-1\,\,}}]_{1}^{2}=[e-e{}^\circ ]+[e-e{}^\circ ]\] \[=(e-1)+(e-1)=2e-2=2\,(e-1)\]