How do you find the integral of $(x^{2}) (e^{- x})$ $(x^2)(e^(-x))$ ?

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How do you find the integral of $(x^{2}) (e^{- x})$ $(x^2)(e^(-x))$ ?

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Answer 1 · 2015-09-23T19:43:06

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

Explanation:

$\int x^{2} e^{- x} d x$ $intx^2e^(-x)dx$
if f(x)=x^2e^(-x) let u(x)=x^2 and v(x)=e^(-x)
using integration by parts
$x^{2} e^{- x} d x = x^{2} \frac{e^{- x}}{- 1} - \int 2 x \cdot \frac{e^{- x}}{- 1} = x^{2} e^{- x} + 2 \int x e^{- x} d x = x^{2} e^{- x} + 2 [x \frac{e^{- x}}{- 1} - \frac{e^{- x}}{- 1}] = x^{2} e^{- x} - 2 x e^{- x} + 2 e^{- x} + c$ $x^2e^(-x)dx=x^2e^(-x)/-1-int2x*e^(-x)/-1=x^2e^(-x)+2intxe^(-x)dx=x^2e^(-x)+2[xe^(-x)/-1-e^(-x)/-1]=x^2e^(-x)-2xe^(-x)+2e^(-x)+c$

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How do you find the integral of $(x^{2}) (e^{- x})$ $(x^2)(e^(-x))$ ?

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