What is the integral of (xe^{-x})/2 ?

The integral of (xe^{-x})/2 is 1/2 (-e^{-x}x-e^{-x})+C

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integral of (xe^{-x})/2

\int \frac{x e ^{- x}}{2} d x

\frac{1}{2} (- e^{- x} x - e^{- x}) + C

Solution steps

\int \frac{x e ^{- x}}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int x e^{- x} d x

Apply Integration By Parts

: - e^{- x} x - \int - e^{- x} d x

= \frac{1}{2} (- e^{- x} x - \int - e^{- x} d x)

\int - e^{- x} d x = e^{- x}

= \frac{1}{2} (- e^{- x} x - e^{- x})

Add a constant to the solution

= \frac{1}{2} (- e^{- x} x - e^{- x}) + C

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What is the integral of (xe^{-x})/2 ?
The integral of (xe^{-x})/2 is 1/2 (-e^{-x}x-e^{-x})+C