What is the integral of e^{-0.12x} ?

The integral of e^{-0.12x} is -8.33333…e^{-0.12x}+C

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integral of e^{-0.12x}

\int e^{- 0.12 x} d x

- 8.33333 \dots e^{- 0.12 x} + C

Solution steps

\int e^{- 0.12 x} d x

Apply u-substitution

= \int - \frac{1}{0.12} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.12} \cdot \int e^{u} d u

Simplify

= - 8.33333 \dots \cdot \int e^{u} d u

Use the common integral:

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

= - 8.33333 \dots e^{u}

Substitute back

u = - 0.12 x

= - 8.33333 \dots e^{- 0.12 x}

Add a constant to the solution

= - 8.33333 \dots e^{- 0.12 x} + C

\int \frac{1}{z ^{2}} d z

d er i v a t i v e f (x) = - 5 e^{- 5 x + 2}

d er i v a t i v e f (x) = \frac{x}{7 - x}

\frac{d}{d x} (arctan (a x))

\frac{d}{d x} (\frac{x ^{3}}{6} + \frac{1}{2 x})

What is the integral of e^{-0.12x} ?
The integral of e^{-0.12x} is -8.33333…e^{-0.12x}+C