What is $F (x) = \int x - x e^{- x} d x$ $F(x) = int x-xe^(-x) dx$ if $F (0) = 2$ $F(0) = 2$ ?

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What is $F (x) = \int x - x e^{- x} d x$ $F(x) = int x-xe^(-x) dx$ if $F (0) = 2$ $F(0) = 2$ ?

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Answer 1 · 2017-11-29T07:10:52

$F (x) = \frac{1}{2} x^{2} + x e^{- x} + e^{- x} + 1$ $F(x) = 1/2x^2 +xe^-x + e^-x + 1$

Explanation:

$F (x) = \int x - x e^{- x} d x = \frac{1}{2} x^{2} - \int x e^{- x} d x$ $F(x) = intx - xe^-xdx = 1/2x^2 - intxe^-xdx$
Finding $\int x e^{- x} d x$ $intxe^-xdx$ is tricky, we'll have to use integration of parts.
$\int x e^{- x} d x = - x e^{- x} - e^{- x} + C$ $intxe^-xdx = -xe^-x - e^-x + C$
Therefore, $F (x) = \frac{1}{2} x^{2} + x e^{- x} + e^{- x} + C$ $F(x) = 1/2x^2 +xe^-x + e^-x + C$ .
Now we have to solve for $C$ $C$ .
$F (0) = \frac{1}{2} \cdot 0^{2} + 0 e^{0} + e^{0} + C$ $F(0) = 1/2 * 0^2 + 0e^0 + e^0 + C$
$= 1 + C = 2$ $=1+C = 2$ (since $F (0) = 2$ $F(0) = 2$ )
Therefore, $C = 1$ $C = 1$ .
$F (x) = \frac{1}{2} x^{2} + x e^{- x} + e^{- x} + 1$ $F(x) = 1/2x^2 +xe^-x + e^-x + 1$ .

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What is $F (x) = \int x - x e^{- x} d x$ $F(x) = int x-xe^(-x) dx$ if $F (0) = 2$ $F(0) = 2$ ?

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What is F(x) = int x-xe^(-x) dxF(x)=∫x−xe−xdxF(x) = int x-xe^(-x) dx if F(0) = 2 F(0)=2F(0) = 2 ?

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Explanation:

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What is $F (x) = \int x - x e^{- x} d x$ $F(x) = int x-xe^(-x) dx$ if $F (0) = 2$ $F(0) = 2$ ?