What is the integral of ae^{-bx} ?

The integral of ae^{-bx} is -ae^{-bx} 1/b+C

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integral of ae^{-bx}

\int a e^{- b x} d x

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

Solution steps

\int a e^{- b x} d x

Take the constant out:

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

= a \cdot \int e^{- b x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - b x

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

Simplify

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

Add a constant to the solution

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

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