Question Number 38101 by maxmathsup by imad last updated on 21/Jun/18
$${let}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:{e}^{{x}−\left[{x}\right]} {dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$
Commented by prof Abdo imad last updated on 22/Jun/18
$${A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{e}^{−{k}} \:{e}^{{x}} {dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{−{k}} \:\:\left(\:{e}^{{k}+\mathrm{1}} \:−{e}^{{k}} \right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({e}\:−\mathrm{1}\right)=\left({e}−\mathrm{1}\right)\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{1}\right)={n}\left({e}−\mathrm{1}\right) \\ $$$${A}_{{n}} ={n}\left({e}−\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:{e}−\mathrm{1}>\mathrm{0}\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =+\infty. \\ $$