What is the integral of 1 (e^{-3x}) ?

The integral of 1 (e^{-3x}) is -1/3 de^{-3x}+C

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integral of 1 (e^{-3x})

\int \frac{d}{1} (e^{- 3 x}) d x

- \frac{1}{3} d e^{- 3 x} + C

Solution steps

\int \frac{d}{1} e^{- 3 x} d x

Take the constant out:

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

= d \cdot \int e^{- 3 x} d x

Apply u-substitution

: \int - \frac{1}{3} e^{u} d u

= d \cdot \int - \frac{1}{3} e^{u} d u

Take the constant out:

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

= d (- \frac{1}{3} \cdot \int e^{u} d u)

Use the common integral:

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

= d (- \frac{1}{3} e^{u})

Substitute back

u = - 3 x

= d (- \frac{1}{3} e^{- 3 x})

Simplify

= - \frac{1}{3} d e^{- 3 x}

Add a constant to the solution

= - \frac{1}{3} d e^{- 3 x} + C

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What is the integral of 1 (e^{-3x}) ?
The integral of 1 (e^{-3x}) is -1/3 de^{-3x}+C