What is the integral of λe^{-λx} ?

The integral of λe^{-λx} is -e^{-λx}+C

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integral of λe^{-λx}

\int λ e^{- λ x} d x

- e^{- λ x} + C

Solution steps

\int λ 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 e^{- λ x} d x

Apply u-substitution

: \int - \frac{e ^{u}}{λ} d u

= λ \cdot \int - \frac{e ^{u}}{λ} d u

Take the constant out:

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

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

Use the common integral:

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

= λ (- \frac{1}{λ} e^{u})

Substitute back

u = - λ x

= λ (- \frac{1}{λ} e^{- λ x})

Simplify

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

= - e^{- λ x}

Add a constant to the solution

= - e^{- λ x} + C

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