What is the integral of e^{-0.02t} ?

The integral of e^{-0.02t} is -50e^{-0.02t}+C

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integral of e^{-0.02t}

\int e^{- 0.02 t} d t

- 50 e^{- 0.02 t} + C

Solution steps

\int e^{- 0.02 t} d t

Apply u-substitution

= \int - \frac{1}{0.02} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.02} \cdot \int e^{u} d u

Simplify

= - 50 \cdot \int e^{u} d u

Use the common integral:

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

= - 50 e^{u}

Substitute back

u = - 0.02 t

= - 50 e^{- 0.02 t}

Add a constant to the solution

= - 50 e^{- 0.02 t} + C

What is the integral of e^{-0.02t} ?
The integral of e^{-0.02t} is -50e^{-0.02t}+C