find the value of ∫_(−1) ^(+∞) (√(x+1))e^(−x) dx


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Question Number 40043 by abdo mathsup 649 cc last updated on 15/Jul/18

$f i n d t h e v a l u e o f \int_{- 1}^{+ \infty} \sqrt{x + 1} e^{- x} d x$
	Answered by prof Abdo imad last updated on 22/Jul/18
	$changement (√(x+1))=t give x+1=t^2 I = ∫_0 ^∞ t e^(−(t^2 −1)) 2t dt =2e ∫_0 ^∞ t^2 e^(−t^2 ) dt and by partsu^′ =t e^(−t^2 ) v=t I =2e{ [−(1/2) e^(−t^2 ) t]_0 ^(+∞) +∫_0 ^∞ (1/2) e^(−t^2 ) dt} = e ∫_0 ^∞ e^(−t^2 ) dt = e ((√π)/2) ⇒ I = (e/2)(√π).$
	$c h a n g e m e n t \sqrt{x + 1} = t g i v e x + 1 = t^{2}$ $I = \int_{0}^{\infty} t e^{- (t^{2} - 1)} 2 t d t$ $= 2 e \int_{0}^{\infty} t^{2} e^{- t^{2}} d t a n d b y {p a r t s u}^{^{'}} = t e^{- t^{2}} v = t$ $I = 2 e {{[- \frac{1}{2} e^{- t^{2}} t]}_{0}^{+ \infty} + \int_{0}^{\infty} \frac{1}{2} e^{- t^{2}} d t}$ $= e \int_{0}^{\infty} e^{- t^{2}} d t = e \frac{\sqrt{π}}{2} \Rightarrow$ $I = \frac{e}{2} \sqrt{π} .$
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