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calculate-e-3x-2-2x-dx-




Question Number 73491 by abdomathmax last updated on 13/Nov/19
calculate ∫_(−∞) ^(+∞)   e^(−3x^2 −2x)  dx
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}} \:{dx} \\ $$
Commented by mathmax by abdo last updated on 14/Nov/19
let I =∫_(−∞) ^(+∞)  e^(−3x^2 −2x)  dx ⇒ I=∫_(−∞) ^(+∞)   e^(−(3x^2  +2x)) dx  =∫_(−∞) ^(+∞)  e^(−{ ((√3)x)^2  +2(((√3)x)/( (√3)))  +(1/3)−(1/3))) dx  =∫_(−∞) ^(+∞)   e^(−{  ((√3)x+(1/( (√3) )))^2 −(1/3)}) dx =e^(1/3)  ∫_(−∞) ^(+∞)  e^(−((√3)x+(1/( (√3))))^2 ) dx  =_((√3)x +(1/( (√3) ))=u)    e^(1/3)  ∫_(−∞) ^(+∞)  e^(−u^2 )  (du/( (√3))) =(((^3 (√e)))/( (√3))) ∫_(−∞) ^(+∞)  e^(−u^2 ) du  =((√π)/( (√3))) ×(^3 (√e)).
$${let}\:{I}\:=\int_{−\infty} ^{+\infty} \:{e}^{−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}} \:{dx}\:\Rightarrow\:{I}=\int_{−\infty} ^{+\infty} \:\:{e}^{−\left(\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{2}{x}\right)} {dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:{e}^{−\left\{\:\left(\sqrt{\mathrm{3}}{x}\right)^{\mathrm{2}} \:+\mathrm{2}\frac{\sqrt{\mathrm{3}}{x}}{\:\sqrt{\mathrm{3}}}\:\:+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\right)} {dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:{e}^{−\left\{\:\:\left(\sqrt{\mathrm{3}}{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}\right\}} {dx}\:={e}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\int_{−\infty} ^{+\infty} \:{e}^{−\left(\sqrt{\mathrm{3}}{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} } {dx} \\ $$$$=_{\sqrt{\mathrm{3}}{x}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:}={u}} \:\:\:{e}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\int_{−\infty} ^{+\infty} \:{e}^{−{u}^{\mathrm{2}} } \:\frac{{du}}{\:\sqrt{\mathrm{3}}}\:=\frac{\left(^{\mathrm{3}} \sqrt{{e}}\right)}{\:\sqrt{\mathrm{3}}}\:\int_{−\infty} ^{+\infty} \:{e}^{−{u}^{\mathrm{2}} } {du} \\ $$$$=\frac{\sqrt{\pi}}{\:\sqrt{\mathrm{3}}}\:×\left(^{\mathrm{3}} \sqrt{{e}}\right). \\ $$

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