Integration Questions

Question Number 119681 by TANMAY PANACEA last updated on 26/Oct/20

$$\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}}\:{dx} \\$$$$\int_{\mathrm{0}} ^{\infty} \left[{ne}^{−{x}} \right]{dx} \\$$

Answered by bemath last updated on 26/Oct/20

$$\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}\:}\:=\:\sqrt{\frac{\mathrm{1}+\mathrm{2cos}\:^{\mathrm{2}} {x}−\mathrm{1}}{\mathrm{2}}} \\$$$$\:=\:\mid\:\mathrm{cos}\:{x}\:\mid\:\rightarrow\begin{cases}{\mathrm{cos}\:{x}\:{for}\:\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}}\\{−\mathrm{cos}\:{x}\:{for}\:\frac{\pi}{\mathrm{2}}\leqslant{x}\leqslant\pi}\end{cases} \\$$$${I}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mid\mathrm{cos}\:{x}\mid\:{dx}\:+\:\underset{\frac{\pi}{\mathrm{2}}} {\overset{\pi} {\int}}\:\mid\mathrm{cos}\:{x}\mid\:{dx} \\$$$$\left.{I}\left.\:=\:\mathrm{sin}\:{x}\:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\left(\mathrm{sin}\:{x}\right)\right]_{\frac{\pi}{\mathrm{2}}} ^{\pi} \\$$$${I}=\mathrm{1}−\left(\mathrm{0}−\mathrm{1}\right)=\mathrm{2} \\$$

Commented by TANMAY PANACEA last updated on 26/Oct/20

$${excellent}\:{sir} \\$$

Answered by mindispower last updated on 26/Oct/20

$$\int_{\mathrm{0}} ^{\infty} \left[{ne}^{−{x}} \right]{dx}={u}_{{n}} \\$$$${t}={ne}^{−{x}} \Rightarrow{x}=−{ln}\left(\frac{{t}}{{n}}\right)\Rightarrow{dx}=−\frac{\mathrm{1}}{{t}}{dt} \\$$$$=\int_{\mathrm{0}} ^{{n}} \frac{\left[{t}\right]}{{t}}{dt} \\$$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \frac{\left[{t}\right]{dt}}{{t}}=\underset{{k}\leqslant{n}−\mathrm{1}} {\sum}\int_{{k}} ^{{k}+\mathrm{1}} \frac{{k}}{{t}}{dt} \\$$$$=\underset{{k}\leqslant{n}−\mathrm{1}} {\sum}{kln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)=\underset{{k}\leqslant{n}−\mathrm{1}} {\sum}\left({kln}\left({k}+\mathrm{1}\right)−{kln}\left({k}\right)\right) \\$$$$=\underset{{k}\leqslant{n}−\mathrm{1}} {\sum}\left(\left({k}+\mathrm{1}\right){ln}\left({k}+\mathrm{1}\right)−{kln}\left({k}\right)−{ln}\left({k}+\mathrm{1}\right)\right)={u}_{{n}} \\$$$${n}=\mathrm{0},{u}_{\mathrm{0}} =\mathrm{0} \\$$$${n}\geqslant\mathrm{1} \\$$$${u}_{{n}} ={nln}\left({n}\right)−{ln}\left[\underset{{k}\leqslant{n}−\mathrm{1}} {\prod}\left({k}+\mathrm{1}\right)\right]={nln}\left({n}\right)−{ln}\left({n}!\right) \\$$$${u}_{{n}} =\begin{cases}{\mathrm{0}\:\:,{n}=\mathrm{0}}\\{{nln}\left({n}\right)−{ln}\left({n}!\right)\:,{n}\geqslant\mathrm{1}}\end{cases} \\$$$$\\$$$$\\$$$$\\$$$$\\$$$$\\$$$$\\$$$$\\$$

Commented by TANMAY PANACEA last updated on 26/Oct/20

$${thank}\:{you}\:{sir} \\$$

Commented by mindispower last updated on 26/Oct/20

$${withe}\:{pleasur}\:{have}\:{a}\:{nice}\:{day} \\$$

Answered by mathmax by abdo last updated on 26/Oct/20

$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}\mathrm{dx}\:\Rightarrow\mathrm{A}\:=\int_{\mathrm{0}} ^{\pi} \mid\mathrm{cosx}\mid\mathrm{dx} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cosx}\:\mathrm{dx}\:−\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{cosx}\:\mathrm{dx}\:=\left[\mathrm{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\left[\mathrm{sinx}\right]_{\frac{\pi}{\mathrm{2}}} ^{\pi} \\$$$$=\mathrm{1}−\left(−\mathrm{1}\right)\:=\mathrm{2}\:\Rightarrow\:\mathrm{A}\:=\mathrm{2} \\$$

Answered by mathmax by abdo last updated on 26/Oct/20

$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \left[\mathrm{ne}^{−\mathrm{x}} \right]\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{ne}^{−\mathrm{x}} \:=\mathrm{t}\:\Rightarrow\mathrm{e}^{−\mathrm{x}} \:=\frac{\mathrm{t}}{\mathrm{n}} \\$$$$\Rightarrow−\mathrm{x}=\mathrm{ln}\left(\frac{\mathrm{t}}{\mathrm{n}}\right)\:\Rightarrow\mathrm{x}\:=−\mathrm{ln}\left(\frac{\mathrm{t}}{\mathrm{n}}\right)=\mathrm{ln}\left(\mathrm{n}\right)−\mathrm{ln}\left(\mathrm{t}\right)\:\left(\mathrm{we}\:\mathrm{suppose}\:\mathrm{n}\neq\mathrm{0}\right) \\$$$$\Rightarrow\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{n}} ^{\mathrm{0}} \left[\mathrm{t}\right]\left(−\frac{\mathrm{dt}}{\mathrm{t}}\right)\:=\int_{\mathrm{0}} ^{\mathrm{n}} \:\frac{\left[\mathrm{t}\right]}{\mathrm{t}}\mathrm{dt}\:\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{k}}{\mathrm{t}}\mathrm{dt} \\$$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \mathrm{k}\left\{\mathrm{ln}\left(\mathrm{k}+\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{k}\right)\right\} \\$$$$=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \mathrm{kln}\left(\frac{\mathrm{k}+\mathrm{1}}{\mathrm{k}}\right)\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \mathrm{k}\:\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{k}}\right) \\$$$$=\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{3ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)+...+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}\right) \\$$$$\mathrm{ifn}=\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{I}_{\mathrm{n}} =\mathrm{0} \\$$$$\\$$