Question-98687

Question Number 98687 by abony1303 last updated on 15/Jun/20

Commented by abony1303 last updated on 15/Jun/20

$$\mathrm{pls}\:\mathrm{help} \\ $$

Commented by MWSuSon last updated on 15/Jun/20

i am not sure, but i think sir used integration as the sum of infinite series where lim_(n→∞) (1/n)Σ_(k=1) ^(k=n) 𝛗((k/n))=∫_0 ^1 𝛗(x)dx

$$\boldsymbol{{i}}\:\boldsymbol{{am}}\:\boldsymbol{{not}}\:\boldsymbol{{sure}},\:\boldsymbol{{but}}\:\boldsymbol{{i}}\:\boldsymbol{{think}}\:\boldsymbol{{sir}}\:\boldsymbol{{used}} \\ $$$$\boldsymbol{{integration}}\:\boldsymbol{{as}}\:\boldsymbol{{the}}\:\boldsymbol{{sum}}\:\boldsymbol{{of}}\:\boldsymbol{{infinite}} \\ $$$$\boldsymbol{{series}}\:\boldsymbol{{where}}\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\boldsymbol{\mathrm{lim}}}\:\frac{\mathrm{1}}{\boldsymbol{{n}}}\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{k}}=\boldsymbol{{n}}} {\sum}}\boldsymbol{\phi}\left(\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}}\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\boldsymbol{\phi}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}} \\ $$

Commented by PRITHWISH SEN 2 last updated on 15/Jun/20

lim_(n→∞) (π/n)Σ_(k=1) ^n ((πk)/n)sin (((πk)/n))=π^2 ∫_0 ^1 xsin πxdx ={−πxcos πx+sin πx}_0 ^1 =𝛑 please check

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\pi}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\pi\mathrm{k}}{\mathrm{n}}\mathrm{sin}\:\left(\frac{\pi\mathrm{k}}{\mathrm{n}}\right)=\pi^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{xsin}\:\pi\mathrm{xdx} \\ $$$$=\left\{−\pi\mathrm{xcos}\:\pi\mathrm{x}+\mathrm{sin}\:\pi\mathrm{x}\right\}_{\mathrm{0}} ^{\mathrm{1}} =\boldsymbol{\pi}\:\:\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$

Commented by abony1303 last updated on 15/Jun/20

sorry sir, could you please explain more detailed?

$$\mathrm{sorry}\:\mathrm{sir},\:\mathrm{could}\:\mathrm{you}\:\mathrm{please}\:\mathrm{explain}\:\mathrm{more} \\ $$$$\mathrm{detailed}? \\ $$

Answered by mathmax by abdo last updated on 15/Jun/20

let S_n =(π/n) Σ_(k=1) ^n x_k sin(x_k ) ⇒S_n =(π/n)Σ_(k=1) ^n ((kπ)/n)sin(((kπ)/n)) so S_n is a Rieman sum snd lim_(n→+∞) S_n =∫_0 ^π xsinx dx by parts ∫_0 ^π xsinx dx =[−xcosx]_0 ^π +∫_0 ^π cosx dx =π +[sinx]_0 ^π =π ⇒ lim_(n→+∞) S_n =π

$$\mathrm{let}\:\mathrm{S}_{\mathrm{n}} =\frac{\pi}{\mathrm{n}}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{x}_{\mathrm{k}} \mathrm{sin}\left(\mathrm{x}_{\mathrm{k}} \right)\:\Rightarrow\mathrm{S}_{\mathrm{n}} =\frac{\pi}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}\pi}{\mathrm{n}}\mathrm{sin}\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right)\:\mathrm{so}\:\mathrm{S}_{\mathrm{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{Rieman}\:\mathrm{sum} \\ $$$$\mathrm{snd}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{S}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\pi} \:\mathrm{xsinx}\:\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\mathrm{xsinx}\:\mathrm{dx}\:=\left[−\mathrm{xcosx}\right]_{\mathrm{0}} ^{\pi} \:+\int_{\mathrm{0}} ^{\pi} \mathrm{cosx}\:\mathrm{dx}\:=\pi\:\:+\left[\mathrm{sinx}\right]_{\mathrm{0}} ^{\pi} \:=\pi\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{S}_{\mathrm{n}} =\pi \\ $$$$ \\ $$

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