What is the integral of λxe^{-λx} ?

The integral of λxe^{-λx} is -e^{-λx}x-(e^{-λx})/λ+C

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integral of λxe^{-λx}

\int λ x e^{- λ x} d x

- e^{- λ x} x - \frac{e ^{- λ x}}{λ} + C

Solution steps

\int λ x e^{- λ x} d x

Take the constant out:

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

= λ \cdot \int x e^{- λ x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{λ ^{2}} d u

= λ \cdot \int \frac{e ^{u} u}{λ ^{2}} d u

Take the constant out:

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

= λ \frac{1}{λ ^{2}} \cdot \int e^{u} u d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = - λ x

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

Simplify

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

= - e^{- λ x} x - \frac{e ^{- λ x}}{λ}

Add a constant to the solution

= - e^{- λ x} x - \frac{e ^{- λ x}}{λ} + C

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What is the integral of λxe^{-λx} ?
The integral of λxe^{-λx} is -e^{-λx}x-(e^{-λx})/λ+C