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Question Number 93941 by Ar Brandon last updated on 16/May/20
Show that the function f(x)=e^(x ) is of Reimann within x∈[1;5] hence calculate ∫_1 ^5 e^x dx
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{x}\:} \:\mathrm{is}\:\mathrm{of}\:\mathrm{Reimann}\:\mathrm{within} \\ $$$$\mathrm{x}\in\left[\mathrm{1};\mathrm{5}\right]\:\mathrm{hence}\:\mathrm{calculate}\:\int_{\mathrm{1}} ^{\mathrm{5}} \mathrm{e}^{\mathrm{x}} \mathrm{dx} \\ $$
Commented by mathmax by abdo last updated on 16/May/20
f continue we have ∫_1 ^5 e^x dx =lim_(n→+∞) ((5−1)/n)Σ_(k=1) ^n e^(1+((4k)/n)) =lim_(n→+∞) ((4e)/n) Σ_(k=0) ^(n−1) (e^(4/n) )^k =lim_(n→+∞) ((4e)/n)×((1−(e^(4/n) )^n )/(1−e^(4/n) )) =lim_(n→+∞) ((4e)/n)×((1−e^4 )/(1−e^(4/n) )) =lim_(n→+∞) 4e(1−e^4 )×(1/(n(1−e^(4/n) ))) we have e^(4/n) ∼1+(4/n) ⇒1−e^(4/n) ∼−(4/n) ⇒n(1−e^(4/n) ) ∼−4 ⇒ lim_(n→+∞) (1/(n(1−e^(4/n) ))) =−4 ⇒∫_1 ^5 f(x)dx =−(1/4)×4e(1−e^4 ) =e^5 −e .
$${f}\:{continue}\:{we}\:{have}\:\:\int_{\mathrm{1}} ^{\mathrm{5}} \:{e}^{{x}} \:{dx}\:={lim}_{{n}\rightarrow+\infty} \:\:\:\frac{\mathrm{5}−\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \:{e}^{\mathrm{1}+\frac{\mathrm{4}{k}}{{n}}} \\ $$$$={lim}_{{n}\rightarrow+\infty} \frac{\mathrm{4}{e}}{{n}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\left({e}^{\frac{\mathrm{4}}{{n}}} \right)^{{k}} \:\:={lim}_{{n}\rightarrow+\infty} \:\frac{\mathrm{4}{e}}{{n}}×\frac{\mathrm{1}−\left({e}^{\frac{\mathrm{4}}{{n}}} \right)^{{n}} }{\mathrm{1}−{e}^{\frac{\mathrm{4}}{{n}}} } \\ $$$$={lim}_{{n}\rightarrow+\infty} \:\frac{\mathrm{4}{e}}{{n}}×\frac{\mathrm{1}−{e}^{\mathrm{4}} }{\mathrm{1}−{e}^{\frac{\mathrm{4}}{{n}}} }\:\:\:={lim}_{{n}\rightarrow+\infty} \mathrm{4}{e}\left(\mathrm{1}−{e}^{\mathrm{4}} \right)×\frac{\mathrm{1}}{{n}\left(\mathrm{1}−{e}^{\frac{\mathrm{4}}{{n}}} \right)} \\ $$$${we}\:{have}\:{e}^{\frac{\mathrm{4}}{{n}}} \:\sim\mathrm{1}+\frac{\mathrm{4}}{{n}}\:\Rightarrow\mathrm{1}−{e}^{\frac{\mathrm{4}}{{n}}} \:\sim−\frac{\mathrm{4}}{{n}}\:\Rightarrow{n}\left(\mathrm{1}−{e}^{\frac{\mathrm{4}}{{n}}} \right)\:\sim−\mathrm{4}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\frac{\mathrm{1}}{{n}\left(\mathrm{1}−{e}^{\frac{\mathrm{4}}{{n}}} \right)}\:=−\mathrm{4}\:\Rightarrow\int_{\mathrm{1}} ^{\mathrm{5}} {f}\left({x}\right){dx}\:=−\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{4}{e}\left(\mathrm{1}−{e}^{\mathrm{4}} \right) \\ $$$$={e}^{\mathrm{5}} −{e}\:. \\ $$
Commented by Ar Brandon last updated on 16/May/20
Thanks ��

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