How do you integrate $\int x^{2} e^{- x} d x$ $int x^2 e^(-x) dx$ using integration by parts?

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How do you integrate $\int x^{2} e^{- x} d x$ $int x^2 e^(-x) dx$ using integration by parts?

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Answer 1 · 2016-01-09T01:34:44

$I = - e^{- x} (x^{2} + 2 x + 2) + c$ $I = -e^(-x)(x^2 + 2x + 2) + c$

Explanation:

$I = \int x^{2} e^{- x} d x$ $I = intx^2e^(-x)dx$

Say $u = x^{2}$ $u = x^2$ so $d u = 2 x$ $du = 2x$ and $d v = e^{- x}$ $dv = e^(-x)$ so $v = - e^{- x}$ $v = -e^(-x)$

$I = - x^{2} e^{- x} + 2 \int x e^{- x} d x$ $I = -x^2e^(-x) + 2intxe^(-x)dx$

Say $u = x$ $u = x$ so $d u = 1$ $du = 1$ and $d v = e^{- x}$ $dv = e^(-x)$ so $v = - e^{- x}$ $v = -e^(-x)$

$I = - x^{2} e^{- x} + 2 (- x e^{- x} + \int e^{- x} d x)$ $I = -x^2e^(-x) + 2(-xe^(-x) + inte^(-x)dx)$
$I = - x^{2} e^{- x} - 2 x e^{- x} + 2 \int e^{- x} d x$ $I = -x^2e^(-x) -2xe^(-x) + 2inte^(-x)dx$
$I = - x^{2} e^{- x} - 2 x e^{- x} - 2 e^{- x} + c$ $I = -x^2e^(-x)-2xe^(-x)-2e^(-x)+c$
$I = - e^{- x} (x^{2} + 2 x + 2) + c$ $I = -e^(-x)(x^2 + 2x + 2) + c$

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How do you integrate $\int x^{2} e^{- x} d x$ $int x^2 e^(-x) dx$ using integration by parts?

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How do you integrate ∫x2e−xdxint x^2 e^(-x) dx using integration by parts?

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How do you integrate $\int x^{2} e^{- x} d x$ $int x^2 e^(-x) dx$ using integration by parts?