Question Number 143193 by mohammad17 last updated on 11/Jun/21
![prove that the function f(x)=x^2 ,xε[1,4] is Riemannian integral ?](https://www.tinkutara.com/question/Q143193.png)
$${prove}\:{that}\:{the}\:{function}\:{f}\left({x}\right)={x}^{\mathrm{2}} \:\:,{x}\varepsilon\left[\mathrm{1},\mathrm{4}\right] \\ $$$${is}\:{Riemannian}\:{integral}\:? \\ $$
Answered by mathmax by abdo last updated on 11/Jun/21
![∫_1 ^4 f(x)dx=∫_1 ^4 x^2 dx =[(x^3 /3)]_1 ^4 =(1/3)(4^3 −1)=(1/3)(63)=((63)/3)=21 lim_(n→+∞) ((4−1)/n)Σ_(k=1) ^n f(1+(4−1)×(k/n)) =lim_(n→+∞) (3/n)Σ_(k=1) ^n f(1+((3k)/n)) =lim_(n→+∞) (3/n)Σ_(k=1) ^n (1+((3k)/n))^2 =lim_(n→+∞) (3/n)Σ_(k=1) ^n (1+((6k)/n)+((9k^2 )/n^2 )) =3 +lim_(n→+∞) ((18)/n^2 )Σ_(k=1) ^n k +lim_(n→+∞) ((27)/n^3 )Σ_(k=1) ^n k^2 =3+lim_(n→+∞) ((18)/n^2 )((n(n+1))/2) +lim_(n→+∞) ((27)/n^3 )((n(n+1)(2n+1))/6) =3+9 +9 =21 ⇒f is Rieman integrable.](https://www.tinkutara.com/question/Q143242.png)
$$\int_{\mathrm{1}} ^{\mathrm{4}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{1}} ^{\mathrm{4}} \:\mathrm{x}^{\mathrm{2}} \mathrm{dx}\:=\left[\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{1}} ^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{4}^{\mathrm{3}} −\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{63}\right)=\frac{\mathrm{63}}{\mathrm{3}}=\mathrm{21} \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{4}−\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{f}\left(\mathrm{1}+\left(\mathrm{4}−\mathrm{1}\right)×\frac{\mathrm{k}}{\mathrm{n}}\right) \\ $$$$=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{3}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{f}\left(\mathrm{1}+\frac{\mathrm{3k}}{\mathrm{n}}\right) \\ $$$$=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{3}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\mathrm{1}+\frac{\mathrm{3k}}{\mathrm{n}}\right)^{\mathrm{2}} \:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{3}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\mathrm{1}+\frac{\mathrm{6k}}{\mathrm{n}}+\frac{\mathrm{9k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\right) \\ $$$$=\mathrm{3}\:+\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{18}}{\mathrm{n}^{\mathrm{2}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{k}\:+\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{27}}{\mathrm{n}^{\mathrm{3}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{k}^{\mathrm{2}} \\ $$$$=\mathrm{3}+\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{18}}{\mathrm{n}^{\mathrm{2}} }\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\:+\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{27}}{\mathrm{n}^{\mathrm{3}} }\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$=\mathrm{3}+\mathrm{9}\:+\mathrm{9}\:=\mathrm{21}\:\Rightarrow\mathrm{f}\:\mathrm{is}\:\mathrm{Rieman}\:\mathrm{integrable}. \\ $$