What is the integral of e^{-0.09x} ?

The integral of e^{-0.09x} is -11.11111…e^{-0.09x}+C

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integral of e^{-0.09x}

\int e^{- 0.09 x} d x

- 11.11111 \dots e^{- 0.09 x} + C

Solution steps

\int e^{- 0.09 x} d x

Apply u-substitution

: \int - \frac{1}{0.09} e^{u} d u

= \int - \frac{1}{0.09} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.09} \cdot \int e^{u} d u

Simplify

= - 11.11111 \dots \cdot \int e^{u} d u

Use the common integral:

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

= - 11.11111 \dots e^{u}

Substitute back

u = - 0.09 x

= - 11.11111 \dots e^{- 0.09 x}

Add a constant to the solution

= - 11.11111 \dots e^{- 0.09 x} + C

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What is the integral of e^{-0.09x} ?
The integral of e^{-0.09x} is -11.11111…e^{-0.09x}+C