What is the integral of e^{-0.06x} ?

The integral of e^{-0.06x} is -16.66666…e^{-0.06x}+C

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integral of e^{-0.06x}

\int e^{- 0.06 x} d x

- 16.66666 \dots e^{- 0.06 x} + C

Solution steps

\int e^{- 0.06 x} d x

Apply u-substitution

: \int - \frac{1}{0.06} e^{u} d u

= \int - \frac{1}{0.06} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.06} \cdot \int e^{u} d u

Simplify

= - 16.66666 \dots \cdot \int e^{u} d u

Use the common integral:

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

= - 16.66666 \dots e^{u}

Substitute back

u = - 0.06 x

= - 16.66666 \dots e^{- 0.06 x}

Add a constant to the solution

= - 16.66666 \dots e^{- 0.06 x} + C

\frac{\partial}{\partial y} (sin (x - y) + (x y))

\frac{\partial}{\partial x} (- 4 x^{2} y - 4 x)

\frac{d}{d x} (\frac{x ^{2} + 4 x + 8}{( x + 2 ) ^{2}})

f (x) = - x^{\frac{4}{3}}

\int_{- 1}^{1} (8 - 8 x^{2})^{2} d x

What is the integral of e^{-0.06x} ?
The integral of e^{-0.06x} is -16.66666…e^{-0.06x}+C