What is the integral of λe^{-λt} ?

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

Popular >

integral of λe^{-λt}

\int λ e^{- λ t} d t

- e^{- λ t} + C

Solution steps

\int λ e^{- λ t} d t

Take the constant out:

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

= λ \cdot \int e^{- λ t} d t

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 t

= λ (- \frac{1}{λ} e^{- λ t})

Simplify

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

= - e^{- λ t}

Add a constant to the solution

= - e^{- λ t} + C

n = 0 \sum \infty (\frac{1}{3})^{n - 1}

\frac{\partial}{\partial x} (9 e^{x} x)

\frac{d}{d x} ((2 x - 1) (x + 3))

\int \frac{x}{7 x ^{2} + 6} d x

\int 25 e^{5 θ} sin (5 θ) d θ