Integration Questions

Question Number 43145 by rahul 19 last updated on 07/Sep/18

$$\int_{\mathrm{0}} ^{\infty} \:\left[\:\mathrm{2e}^{−{x}} \right]{dx}\:=\:?\: \\$$ $${where}\:\left[.\right]=\:{gif}. \\$$

Commented bymaxmathsup by imad last updated on 07/Sep/18

$${changement}\:\mathrm{2}\:{e}^{−{x}} \:={t}\:{give}\:{e}^{−{x}} =\frac{{t}}{\mathrm{2}}\:\Rightarrow−{x}={ln}\left(\frac{{t}}{\mathrm{2}}\right)\:\Rightarrow{x}=−{ln}\left(\frac{{t}}{\mathrm{2}}\right)\:{and} \\$$ $${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\left[\:\mathrm{2}\:{e}^{−{x}} \right]{dx}\:=−\int_{\mathrm{0}} ^{\mathrm{2}} \:\left[{t}\right]\left(−\frac{\mathrm{1}}{{t}}\right){dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{\left[{t}\right]}{{t}}{dt} \\$$ $$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{0}}{{t}}{dt}\:+\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{1}}{{t}}{dt}\:=\:\left[{ln}\mid{t}\mid\right]_{\mathrm{1}} ^{\mathrm{2}} \:\:={ln}\left(\mathrm{2}\right)\:. \\$$

Commented bytanmay.chaudhury50@gmail.com last updated on 07/Sep/18

Commented byrahul 19 last updated on 08/Sep/18

$$\mathrm{thanks}\:\mathrm{prof}\:\mathrm{abdo}. \\$$

Commented bymath khazana by abdo last updated on 08/Sep/18

$${look}\:{sir}\:{raul}\:{x}=−{ln}\left({t}\right)+{ln}\left(\mathrm{2}\right)\:\Rightarrow{dx}\:=−\frac{{dt}}{{t}} \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Sep/18

$$\int_{\mathrm{0}} ^{\mathrm{0}.\mathrm{7}} \left[\mathrm{2}{e}^{−{x}} \right]{dx}+\int_{\mathrm{0}.\mathrm{7}} ^{\infty} \left[\mathrm{2}{e}^{−{x}} \right]{dx} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{0}.\mathrm{7}} \mathrm{1}×{dx}+\int_{\mathrm{0}.\mathrm{7}} ^{\infty} \mathrm{0}×{dx} \\$$ $$=\mathrm{0}.\mathrm{7} \\$$ $${ln}\mathrm{2}\approx\mathrm{0}.\mathrm{7} \\$$

Commented bytanmay.chaudhury50@gmail.com last updated on 07/Sep/18

Commented byrahul 19 last updated on 08/Sep/18

$$\mathrm{thanks}\:\mathrm{sir}. \\$$

Commented byabdo.msup.com last updated on 08/Sep/18

$${nevermind}\:{sir}... \\$$