Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 48174 by Abdo msup. last updated on 20/Nov/18

find lim_(n→+∞)    ∫_0 ^n    sin(((πx)/n))dx .

$${find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{{n}} \:\:\:{sin}\left(\frac{\pi{x}}{{n}}\right){dx}\:. \\ $$

Commented by Abdo msup. last updated on 20/Nov/18

let A_n =∫_0 ^n  sin(((πx)/n))dx ⇒  A_n =[−(n/π)cos(((πx)/n))]_0 ^n  =−(n/π){cos(π)−1}  =((2n)/π) ⇒ lim_(n→+∞) A_n =+∞

$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{{n}} \:{sin}\left(\frac{\pi{x}}{{n}}\right){dx}\:\Rightarrow \\ $$$${A}_{{n}} =\left[−\frac{{n}}{\pi}{cos}\left(\frac{\pi{x}}{{n}}\right)\right]_{\mathrm{0}} ^{{n}} \:=−\frac{{n}}{\pi}\left\{{cos}\left(\pi\right)−\mathrm{1}\right\} \\ $$$$=\frac{\mathrm{2}{n}}{\pi}\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} {A}_{{n}} =+\infty\: \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Nov/18

∫_0 ^n sin(((πx)/n))dx  =[((−cos(((πx)/n)))/(π/n))]_0 ^n =((−n)/π)[cosπ−cos0]  =((−n)/π)[−2]=((2n)/π)  when n→∞ limit also→∞

$$\int_{\mathrm{0}} ^{{n}} {sin}\left(\frac{\pi{x}}{{n}}\right){dx} \\ $$$$=\left[\frac{−{cos}\left(\frac{\pi{x}}{{n}}\right)}{\frac{\pi}{{n}}}\right]_{\mathrm{0}} ^{{n}} =\frac{−{n}}{\pi}\left[{cos}\pi−{cos}\mathrm{0}\right] \\ $$$$=\frac{−{n}}{\pi}\left[−\mathrm{2}\right]=\frac{\mathrm{2}{n}}{\pi} \\ $$$${when}\:{n}\rightarrow\infty\:{limit}\:{also}\rightarrow\infty \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com