What is the integral of le^{-lx} ?

The integral of le^{-lx} is -e^{-lx}+C

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integral of le^{-lx}

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

- e^{- l x} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= l \cdot \int - \frac{e ^{u}}{l} d u

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - l x

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

Simplify

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

= - e^{- l x}

Add a constant to the solution

= - e^{- l x} + C

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