Question Number 40152 by maxmathsup by imad last updated on 16/Jul/18
$${let}\:\:{f}\left({x}\right)\:=\:\:\int_{−\mathrm{1}} ^{{x}} \:\:\:\:\frac{{e}^{{t}} }{\:\sqrt{\mathrm{1}−{e}^{{t}} }}{dt}\:\:\:{with}\:{x}<\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\int_{−\mathrm{1}} ^{\mathrm{0}} \:\:\frac{{e}^{{t}} }{\:\sqrt{\mathrm{1}−{e}^{{t}} }}{dt} \\ $$
Commented by maxmathsup by imad last updated on 16/Jul/18
![changement e^t =u give t=ln(u) ⇒ f(x)= ∫_e^(−1) ^e^x (u/( (√(1−u)))) (du/u) = ∫_e^(−1) ^e^x (du/( (√(1−u)))) =[−2(√(1−u))]_e^(−1) ^e^x f(x)=−2{(√(1−e^x )) −(√(1−e^(−1) ))} 2) ∫_(−1) ^0 (e^t /( (√(1−e^t )))) dt =lim_(x→0) f(x)= 2(√(1−e^(−1) ))](https://www.tinkutara.com/question/Q40196.png)
$${changement}\:{e}^{{t}} \:\:={u}\:{give}\:\:{t}={ln}\left({u}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\int_{{e}^{−\mathrm{1}} } ^{{e}^{{x}} } \:\:\:\:\:\frac{{u}}{\:\sqrt{\mathrm{1}−{u}}}\:\frac{{du}}{{u}}\:=\:\int_{{e}^{−\mathrm{1}} } ^{{e}^{{x}} } \:\:\:\:\frac{{du}}{\:\sqrt{\mathrm{1}−{u}}}\:=\left[−\mathrm{2}\sqrt{\mathrm{1}−{u}}\right]_{{e}^{−\mathrm{1}} } ^{{e}^{{x}} } \\ $$$${f}\left({x}\right)=−\mathrm{2}\left\{\sqrt{\mathrm{1}−{e}^{{x}} \:}\:\:−\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }\right\} \\ $$$$\left.\mathrm{2}\right)\:\int_{−\mathrm{1}} ^{\mathrm{0}} \:\:\:\frac{{e}^{{t}} }{\:\sqrt{\mathrm{1}−{e}^{{t}} }}\:{dt}\:={lim}_{{x}\rightarrow\mathrm{0}} \:\:\:{f}\left({x}\right)=\:\mathrm{2}\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} } \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jul/18
$$\int_{−\mathrm{1}} ^{{x}} \frac{{e}^{{t}} }{\:\sqrt{\mathrm{1}−{e}^{{t}} }}{dt} \\ $$$$=−\mathrm{1}\int_{−\mathrm{1}} ^{{x}} \frac{{d}\left(\mathrm{1}−{e}^{{t}} \right)}{\:\sqrt{\mathrm{1}−{e}^{{t}} }} \\ $$$$=−\mathrm{1}×\mid\frac{\sqrt{\mathrm{1}−{e}^{{t}} }}{\frac{\mathrm{1}}{\mathrm{2}}}\mid_{−\mathrm{1}} ^{{x}} \\ $$$$=−\mathrm{2}\left\{\sqrt{\mathrm{1}−{e}^{{x}} \:}\:\:−\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }\:\:\right\} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{0}} \frac{{e}^{{t}} }{\:\sqrt{\mathrm{1}−{e}^{{t}} }\:}{dt} \\ $$$$=−\mathrm{2}\left\{\sqrt{\mathrm{1}−{e}^{\mathrm{0}} }\:\:−\sqrt{\left.\mathrm{1}−{e}_{} ^{−\mathrm{1}} \right\}}\right. \\ $$$$\left.=\mathrm{2}\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }\:\right\} \\ $$$$ \\ $$