Calculate ∫_(−2) ^2 (∣x∣+x)e^(−∣x∣) dx


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Question Number 167372 by LEKOUMA last updated on 14/Mar/22

$C a l c u l a t e$ $\int_{- 2}^{2} (∣ x ∣ + x) e^{- ∣ x ∣} d x$
	Answered by Mathspace last updated on 14/Mar/22
	$I=∫_(−2) ^2 ∣x∣e^(−∣x∣) dx(→paire) +∫_(−2) ^2 xe^(−∣x∣) dx(→impaire) =2∫_0 ^2 xe^(−x) dx +o =2{ [−xe^(−x) ]_0 ^2 −∫_0 ^2 (−e^(−x) )dx} =2{−2e^(−2) +[−e^(−x) ]_0 ^2 } =2{−2e^(−2) +1−e^(−2) } =2{−3e^(−2) +1}=2−(6/e^2 )$
	$I = \int_{- 2}^{2} ∣ x ∣ e^{- ∣ x ∣} d x (\to p a i r e)$ $+ \int_{- 2}^{2} {x e}^{- ∣ x ∣} d x (\to i m p a i r e)$ $= 2 \int_{0}^{2} {x e}^{- x} d x + o$ $= 2 {{[- {x e}^{- x}]}_{0}^{2} - \int_{0}^{2} (- e^{- x}) d x}$ $= 2 {- 2 e^{- 2} + {[- e^{- x}]}_{0}^{2}}$ $= 2 {- 2 e^{- 2} + 1 - e^{- 2}}$ $= 2 {- 3 e^{- 2} + 1} = 2 - \frac{6}{e^{2}}$
	Commented by LEKOUMA last updated on 15/Mar/22

	$T h a n k s$
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