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Question Number 65665 by mathmax by abdo last updated on 01/Aug/19
calculate ∫_(−2) ^(+∞) (e^(−x) /( (√(x+2)))) dx
$${calculate}\:\int_{−\mathrm{2}} ^{+\infty} \:\:\frac{{e}^{−{x}} }{\:\sqrt{{x}+\mathrm{2}}}\:{dx} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 02/Aug/19
let I =∫_(−2) ^(+∞) (e^(−x) /( (√(x+2))))dx changement (√(x+2))=t give x+2=t^2 ⇒ I = ∫_0 ^(+∞) (e^(−(t^2 −2)) /t)(2t)dt =2 ∫_0 ^(+∞) e^(−t^2 +2 ) dt =e^2 ∫_(−∞) ^(+∞) e^(−t^2 ) dt =e^2 (√π)
$${let}\:{I}\:=\int_{−\mathrm{2}} ^{+\infty} \:\:\frac{{e}^{−{x}} }{\:\sqrt{{x}+\mathrm{2}}}{dx}\:\:{changement}\:\sqrt{{x}+\mathrm{2}}={t}\:{give}\:{x}+\mathrm{2}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{e}^{−\left({t}^{\mathrm{2}} −\mathrm{2}\right)} }{{t}}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{+\infty} \:{e}^{−{t}^{\mathrm{2}} +\mathrm{2}\:} {dt}\:\:={e}^{\mathrm{2}} \int_{−\infty} ^{+\infty} \:{e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$={e}^{\mathrm{2}} \sqrt{\pi} \\ $$

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