What is the integral of xe^{-0.2x} ?

The integral of xe^{-0.2x} is 25(-0.2e^{-0.2x}x-e^{-0.2x})+C

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integral of xe^{-0.2x}

\int x e^{- 0.2 x} d x

25 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x}) + C

Solution steps

\int x e^{- 0.2 x} d x

Apply u-substitution

: \int \frac{5}{0.2} e^{u} u d u

= \int \frac{5}{0.2} e^{u} u d u

Take the constant out:

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

= \frac{5}{0.2} \cdot \int e^{u} u d u

Simplify

= 25 \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 25 (e^{u} u - \int e^{u} d u)

\int e^{u} d u = e^{u}

= 25 (e^{u} u - e^{u})

Substitute back

u = - 0.2 x

= 25 (e^{- 0.2 x} (- 0.2 x) - e^{- 0.2 x})

Simplify

= 25 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x})

Add a constant to the solution

= 25 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x}) + C

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What is the integral of xe^{-0.2x} ?
The integral of xe^{-0.2x} is 25(-0.2e^{-0.2x}x-e^{-0.2x})+C