What is the integral of 1/500 e^{-1/500 t} ?

The integral of 1/500 e^{-1/500 t} is -e^{-t/(500)}+C

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integral of 1/500 e^{-1/500 t}

\int \frac{1}{500} e^{- \frac{1}{500} t} d t

- e^{- \frac{t}{500}} + C

Solution steps

\int \frac{1}{500} e^{- \frac{1}{500} t} d t

Take the constant out:

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

= \frac{1}{500} \cdot \int e^{- \frac{1}{500} t} d t

Simplify

- \frac{1}{500} t : - \frac{t}{500}

= \frac{1}{500} \cdot \int e^{- \frac{t}{500}} d t

Apply u-substitution

: \int - 500 e^{u} d u

= \frac{1}{500} \cdot \int - 500 e^{u} d u

Take the constant out:

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

= \frac{1}{500} (- 500 \cdot \int e^{u} d u)

Use the common integral:

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

= \frac{1}{500} (- 500 e^{u})

Substitute back

u = - \frac{t}{500}

= \frac{1}{500} (- 500 e^{- \frac{t}{500}})

Simplify

\frac{1}{500} (- 500 e^{- \frac{t}{500}}) : - e^{- \frac{t}{500}}

= - e^{- \frac{t}{500}}

Add a constant to the solution

= - e^{- \frac{t}{500}} + C

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What is the integral of 1/500 e^{-1/500 t} ?
The integral of 1/500 e^{-1/500 t} is -e^{-t/(500)}+C