Question Number 59637 by mathtype last updated on 12/May/19
$$\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{x}} \mid\mathrm{sin}\:{x}\mid \\ $$
Commented by Mr X pcx last updated on 13/May/19
![there is a problem here let take x=nπ and find lim_(n→+∞) (1/(nπ)) ∫_0 ^(nπ) ∣sint∣dt wehave ∫_0 ^(nπ) ∣sint∣dt =Σ_(k=0) ^(n−1) ∫_(kπ) ^()k+1)π) ∣sint∣dt =_(t=kπ +u) Σ_(k=0) ^(n−1) ∫_0 ^π ∣(−1)^k sinu∣du =Σ_(k=0) ^(n−1) ∫_0 ^π sinu du =Σ_(k=0) ^(n−1) [−cosu]_0 ^π =2 Σ_(k=0) ^(n−1) )1) =2n ⇒ lim_(n→+∞) (1/(nπ)) ∫_0 ^(nπ) ∣sinx∣dx =(2/π) ...!](Q59712.png)
$${there}\:{is}\:{a}\:{problem}\:{here}\:{let}\:{take}\:{x}={n}\pi \\ $$$${and}\:{find}\:{lim}_{{n}\rightarrow+\infty} \frac{\mathrm{1}}{{n}\pi}\:\int_{\mathrm{0}} ^{{n}\pi} \mid{sint}\mid{dt}\:{wehave} \\ $$$$\int_{\mathrm{0}} ^{{n}\pi} \mid{sint}\mid{dt}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}\pi} ^{\left.\right)\left.{k}+\mathrm{1}\right)\pi} \mid{sint}\mid{dt} \\ $$$$=_{{t}={k}\pi\:+{u}} \:\:\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\int_{\mathrm{0}} ^{\pi} \mid\left(−\mathrm{1}\right)^{{k}} {sinu}\mid{du} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\pi} \:{sinu}\:{du}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left[−{cosu}\right]_{\mathrm{0}} ^{\pi} \\ $$$$\left.=\left.\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \right)\mathrm{1}\right)\:=\mathrm{2}{n}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:\frac{\mathrm{1}}{{n}\pi}\:\int_{\mathrm{0}} ^{{n}\pi} \:\mid{sinx}\mid{dx}\:=\frac{\mathrm{2}}{\pi}\:...! \\ $$$$ \\ $$
Answered by tanmay last updated on 13/May/19
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\int_{\mathrm{0}} ^{{x}} \mid{sinx}\mid\:{dx}}{{x}} \\ $$$${now}\:\int_{\mathrm{0}} ^{{x}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{x}} {f}\left({t}\right){dt} \\ $$$$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{g}\left({x}\right)}{{x}} \\ $$$${g}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \mid{sint}\mid{dt} \\ $$$$\frac{{dg}}{{dx}}=\int_{\mathrm{0}} ^{{x}} \frac{\partial\mid{sint}\mid}{\partial{x}}{dt}+\mid{sinx}\mid×\frac{{dx}}{{dx}} \\ $$$$\frac{{dg}}{{dx}}=\mid{sinx}\mid \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{{g}\left({x}\right)}{{x}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{g}'\left({x}\right)}{\mathrm{1}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mid{sinx}\mid \\ $$$$\:\:\mathrm{1}\geqslant{sinx}\geqslant−\mathrm{1} \\ $$$$\mathrm{1}\geqslant\mid{sinx}\mid\geqslant\mathrm{0} \\ $$$$ \\ $$$${so}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mid{sinx}\mid\: \\ $$$${limit}\:{does}\:{not}\:{exist}\:{since}\:\mid{sinx}\mid\:{oscillate} \\ $$$${between}\:\mathrm{0}\:{and}\:\mathrm{1} \\ $$$${am}\:{i}\:{correct}...{pls}\:{check} \\ $$