Question Number 74889 by abdomathmax last updated on 03/Dec/19
$${find}\:\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{+\infty} \:\:{e}^{−{x}} \sqrt{\mathrm{2}{x}+\mathrm{1}}{dx} \\ $$
Commented by mathmax by abdo last updated on 06/Dec/19
![let I =∫_(−(1/2)) ^∞ e^(−x) (√(2x+1))dx changement (√(2x+1))=t give 2x+1=t^2 ⇒2x=t^2 −1 ⇒x=((t^2 −1)/2) ⇒ I =∫_0 ^∞ e^(−((t^2 −1)/2)) t tdt =(√e) ∫_0 ^∞ t^2 e^(−(t^2 /2)) dt =_((t/( (√2)))=u) (√e)∫_0 ^∞ 2u^2 e^(−u^2 ) (√2)du =2(√(2e))∫_0 ^∞ u^2 e^(−u^2 ) du but ∫_0 ^∞ u e^(−u^2 ) udu =_(bypsrts) [−(1/2)ue^(−u^2 ) ]_0 ^(+∞) −∫_0 ^(+∞) (−(1/2)e^(−u^2 ) ) du =(1/2)∫_0 ^∞ e^(−u^2 ) du =(1/2)×((√π)/2) =((√π)/4) ⇒I =2(√(2e)) ((√π)/4) =(((√2)(√e)(√π))/2) =(√((πe)/2)) ★ I=(√((πe)/2))★](https://www.tinkutara.com/question/Q75049.png)
$${let}\:{I}\:=\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\infty} \:{e}^{−{x}} \sqrt{\mathrm{2}{x}+\mathrm{1}}{dx}\:\:{changement}\:\sqrt{\mathrm{2}{x}+\mathrm{1}}={t}\:{give} \\ $$$$\mathrm{2}{x}+\mathrm{1}={t}^{\mathrm{2}} \:\Rightarrow\mathrm{2}{x}={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow{x}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}} {t}\:\:{tdt}\:=\sqrt{{e}}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:{t}^{\mathrm{2}} \:{e}^{−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} \:{dt} \\ $$$$=_{\frac{{t}}{\:\sqrt{\mathrm{2}}}={u}} \:\:\:\sqrt{{e}}\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{2}{u}^{\mathrm{2}} \:{e}^{−{u}^{\mathrm{2}} } \sqrt{\mathrm{2}}{du}\:=\mathrm{2}\sqrt{\mathrm{2}{e}}\int_{\mathrm{0}} ^{\infty} \:\:{u}^{\mathrm{2}} \:{e}^{−{u}^{\mathrm{2}} } {du}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{u}\:{e}^{−{u}^{\mathrm{2}} } {udu}\:\:=_{{bypsrts}} \:\:\:\left[−\frac{\mathrm{1}}{\mathrm{2}}{ue}^{−{u}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{+\infty} −\int_{\mathrm{0}} ^{+\infty} \left(−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{u}^{\mathrm{2}} } \right)\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{\pi}}{\mathrm{2}}\:=\frac{\sqrt{\pi}}{\mathrm{4}}\:\Rightarrow{I}\:=\mathrm{2}\sqrt{\mathrm{2}{e}}\:\frac{\sqrt{\pi}}{\mathrm{4}}\:=\frac{\sqrt{\mathrm{2}}\sqrt{{e}}\sqrt{\pi}}{\mathrm{2}} \\ $$$$=\sqrt{\frac{\pi{e}}{\mathrm{2}}} \\ $$$$\bigstar\:{I}=\sqrt{\frac{\pi{e}}{\mathrm{2}}}\bigstar \\ $$
Answered by mind is power last updated on 04/Dec/19
![u=(√(2x+1))⇒x=((u^2 −1)/2)⇒dx=udu =∫_0 ^(+∞) e^(−((u^2 −1)/2)) u^2 du =(√e)∫_0 ^(+∞) −u.(−ue^(−(u^2 /2)) )du by part ∫_0 ^(+∞) (−u).(−u.e^(−(u^2 /2)) )du=[−u.e^(−(u^2 /2)) ]+∫e^(−(u^2 /2)) du =∫_0 ^(+∞) e^(−(u^2 /2)) du=(√2)∫_0 ^(+∞) e^(−u^2 ) du=((√π)/( (√2))) we get (√e).(√(π/2))=(√((π.e)/2))](https://www.tinkutara.com/question/Q74960.png)
$$\mathrm{u}=\sqrt{\mathrm{2x}+\mathrm{1}}\Rightarrow\mathrm{x}=\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{dx}=\mathrm{udu} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \mathrm{e}^{−\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}} \mathrm{u}^{\mathrm{2}} \mathrm{du} \\ $$$$=\sqrt{\mathrm{e}}\int_{\mathrm{0}} ^{+\infty} −\mathrm{u}.\left(−\mathrm{ue}^{−\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}} \right)\mathrm{du}\:\mathrm{by}\:\mathrm{part} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \left(−\mathrm{u}\right).\left(−\mathrm{u}.\mathrm{e}^{−\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}} \right)\mathrm{du}=\left[−\mathrm{u}.\mathrm{e}^{−\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}} \right]+\int\mathrm{e}^{−\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}} \mathrm{du} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \mathrm{e}^{−\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}} \mathrm{du}=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{+\infty} \mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du}=\frac{\sqrt{\pi}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{we}\:\mathrm{get}\: \\ $$$$\sqrt{\mathrm{e}}.\sqrt{\frac{\pi}{\mathrm{2}}}=\sqrt{\frac{\pi.\mathrm{e}}{\mathrm{2}}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 04/Dec/19
$${thank}\:{you}\:{sir}. \\ $$