calculate ∫_(−∞) ^(+∞) e^(−3x^2 −2x) dx


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Question Number 73491 by abdomathmax last updated on 13/Nov/19

$c a l c u l a t e \int_{- \infty}^{+ \infty} e^{- 3 x^{2} - 2 x} d x$
Commented by mathmax by abdo last updated on 14/Nov/19
$let I =∫_(−∞) ^(+∞) e^(−3x^2 −2x) dx ⇒ I=∫_(−∞) ^(+∞) e^(−(3x^2 +2x)) dx =∫_(−∞) ^(+∞) e^(−{ ((√3)x)^2 +2(((√3)x)/(√3)) +(1/3)−(1/3))) dx =∫_(−∞) ^(+∞) e^(−{ ((√3)x+(1/((√3) )))^2 −(1/3)}) dx =e^(1/3) ∫_(−∞) ^(+∞) e^(−((√3)x+(1/(√3)))^2 ) dx =_((√3)x +(1/((√3) ))=u) e^(1/3) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√3)) =(((^3 (√e)))/(√3)) ∫_(−∞) ^(+∞) e^(−u^2 ) du =((√π)/(√3)) ×(^3 (√e)).$
$l e t I = \int_{- \infty}^{+ \infty} e^{- 3 x^{2} - 2 x} d x \Rightarrow I = \int_{- \infty}^{+ \infty} e^{- (3 x^{2} + 2 x)} d x$ $= \int_{- \infty}^{+ \infty} e^{- {{(\sqrt{3} x)}^{2} + 2 \frac{\sqrt{3} x}{\sqrt{3}} + \frac{1}{3} - \frac{1}{3})} d x$ $= \int_{- \infty}^{+ \infty} e^{- {{(\sqrt{3} x + \frac{1}{\sqrt{3}})}^{2} - \frac{1}{3}}} d x = e^{\frac{1}{3}} \int_{- \infty}^{+ \infty} e^{- {(\sqrt{3} x + \frac{1}{\sqrt{3}})}^{2}} d x$ $=_{\sqrt{3} x + \frac{1}{\sqrt{3}} = u} e^{\frac{1}{3}} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{3}} = \frac{(^{3} \sqrt{e})}{\sqrt{3}} \int_{- \infty}^{+ \infty} e^{- u^{2}} d u$ $= \frac{\sqrt{π}}{\sqrt{3}} \times (^{3} \sqrt{e}) .$
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