What is the integral of 15e^{-5t} ?

The integral of 15e^{-5t} is -3e^{-5t}+C

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integral of 15e^{-5t}

\int 15 e^{- 5 t} d t

- 3 e^{- 5 t} + C

Solution steps

\int 15 e^{- 5 t} d t

Take the constant out:

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

= 15 \cdot \int e^{- 5 t} d t

Apply u-substitution

: \int - \frac{1}{5} e^{u} d u

= 15 \cdot \int - \frac{1}{5} e^{u} d u

Take the constant out:

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

= 15 (- \frac{1}{5} \cdot \int e^{u} d u)

Use the common integral:

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

= 15 (- \frac{1}{5} e^{u})

Substitute back

u = - 5 t

= 15 (- \frac{1}{5} e^{- 5 t})

Simplify

15 (- \frac{1}{5} e^{- 5 t}) : - 3 e^{- 5 t}

= - 3 e^{- 5 t}

Add a constant to the solution

= - 3 e^{- 5 t} + C

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