What is the integral of 6e^{-3000t} ?

The integral of 6e^{-3000t} is -1/500 e^{-3000t}+C

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integral of 6e^{-3000t}

\int 6 e^{- 3000 t} d t

- \frac{1}{500} e^{- 3000 t} + C

Solution steps

\int 6 e^{- 3000 t} d t

Take the constant out:

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

= 6 \cdot \int e^{- 3000 t} d t

Apply u-substitution

: \int - \frac{1}{3000} e^{u} d u

= 6 \cdot \int - \frac{1}{3000} e^{u} d u

Take the constant out:

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

= 6 (- \frac{1}{3000} \cdot \int e^{u} d u)

Use the common integral:

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

= 6 (- \frac{1}{3000} e^{u})

Substitute back

u = - 3000 t

= 6 (- \frac{1}{3000} e^{- 3000 t})

Simplify

6 (- \frac{1}{3000} e^{- 3000 t}) : - \frac{1}{500} e^{- 3000 t}

= - \frac{1}{500} e^{- 3000 t}

Add a constant to the solution

= - \frac{1}{500} e^{- 3000 t} + C

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What is the integral of 6e^{-3000t} ?
The integral of 6e^{-3000t} is -1/500 e^{-3000t}+C