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

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

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integral of e^{-0.06t}

\int e^{- 0.06 t} d t

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

Solution steps

\int e^{- 0.06 t} d t

Apply u-substitution

= \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 t

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

Add a constant to the solution

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

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