What is the integral of e^{-rx} ?

The integral of e^{-rx} is -1/r e^{-rx}+C

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integral of e^{-rx}

\int e^{- r x} d x

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

Solution steps

\int e^{- r x} d x

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - r x

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

Add a constant to the solution

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