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

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

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integral of 1/3 e^{-x/3}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - \frac{x}{3}

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

Simplify

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

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

Add a constant to the solution

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

x \to - 4 - lim (x^{2} + \frac{2}{x + 4})

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θ \to - π + lim (csc (θ))

\int - \frac{1}{4} cos (4 x) d x

\int \frac{1 + x}{x x} d x

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