What is the integral of (e^{-ax}) ?

The integral of (e^{-ax}) is -1/a e^{-ax}+C

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integral of (e^{-ax})

\int (e^{- a x}) d x

- \frac{1}{a} e^{- a x} + C

Solution steps

\int e^{- a x} d x

Apply u-substitution

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

= \int - \frac{e ^{u}}{a} d u

Take the constant out:

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

= - \frac{1}{a} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{a} e^{u}

Substitute back

u = - a x

= - \frac{1}{a} e^{- a x}

Add a constant to the solution

= - \frac{1}{a} e^{- a x} + C

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