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Question Number 143193 by mohammad17 last updated on 11/Jun/21

$$provethatthefunctionf\left(x\right)=x^2,x\epsilon [1,4]$$$$isRiemannianintegral?$$

Answered by mathmax by abdo last updated on 11/Jun/21

$${\int }_1^4\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_1^4{\mathrm{x}}^2\mathrm{dx}={\left[\frac{{\mathrm{x}}^3}{3}\right]}_1^4=\frac{1}{3}(4^3-1)=\frac{1}{3}\left(63\right)=\frac{63}{3}=21$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{4-1}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{f}(1+(4-1)\times \frac{\mathrm{k}}{\mathrm{n}})$$$$={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{3}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{f}(1+\frac{3\mathrm{k}}{\mathrm{n}})$$$$={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{3}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}{(1+\frac{3\mathrm{k}}{\mathrm{n}})}^2={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{3}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}(1+\frac{6\mathrm{k}}{\mathrm{n}}+\frac{{9\mathrm{k}}^2}{{\mathrm{n}}^2})$$$$=3+{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{18}{{\mathrm{n}}^2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}+{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{27}{{\mathrm{n}}^3}\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2$$$$=3+{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{18}{{\mathrm{n}}^2}\frac{\mathrm{n}(\mathrm{n}+1)}{2}+{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{27}{{\mathrm{n}}^3}\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}$$$$=3+9+9=21\Rightarrow \mathrm{f}\mathrm{is}\mathrm{Rieman}\mathrm{integrable}.$$

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