Question Number 40043 by abdo mathsup 649 cc last updated on 15/Jul/18
$${find}\:\:{the}\:{value}\:{of}\:\:\int_{−\mathrm{1}} ^{+\infty} \:\:\sqrt{{x}+\mathrm{1}}{e}^{−{x}} \:{dx} \\ $$$$ \\ $$$$ \\ $$
Answered by prof Abdo imad last updated on 22/Jul/18
![changement (√(x+1))=t give x+1=t^2 I = ∫_0 ^∞ t e^(−(t^2 −1)) 2t dt =2e ∫_0 ^∞ t^2 e^(−t^2 ) dt and by partsu^′ =t e^(−t^2 ) v=t I =2e{ [−(1/2) e^(−t^2 ) t]_0 ^(+∞) +∫_0 ^∞ (1/2) e^(−t^2 ) dt} = e ∫_0 ^∞ e^(−t^2 ) dt = e ((√π)/2) ⇒ I = (e/2)(√π).](Q40488.png)
$${changement}\:\sqrt{{x}+\mathrm{1}}={t}\:{give}\:{x}+\mathrm{1}={t}^{\mathrm{2}} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{t}\:{e}^{−\left({t}^{\mathrm{2}} −\mathrm{1}\right)} \:\mathrm{2}{t}\:{dt} \\ $$$$=\mathrm{2}{e}\:\:\int_{\mathrm{0}} ^{\infty} \:\:{t}^{\mathrm{2}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:\:{and}\:{by}\:{partsu}^{'} \:={t}\:{e}^{−{t}^{\mathrm{2}} } \:\:\:{v}={t} \\ $$$${I}\:=\mathrm{2}{e}\left\{\:\left[−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{t}^{\mathrm{2}} } \:{t}\right]_{\mathrm{0}} ^{+\infty} \:+\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{t}^{\mathrm{2}} } {dt}\right\} \\ $$$$=\:{e}\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}^{\mathrm{2}} } {dt}\:\:=\:{e}\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\:\Rightarrow \\ $$$${I}\:=\:\frac{{e}}{\mathrm{2}}\sqrt{\pi}. \\ $$$$ \\ $$