What is the integral of 1/4 e^{(-x)/4} ?

The integral of 1/4 e^{(-x)/4} is -e^{-x/4}+C

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integral of 1/4 e^{(-x)/4}

\int \frac{1}{4} e^{\frac{- x}{4}} d x

- e^{- \frac{x}{4}} + C

Solution steps

\int \frac{1}{4} e^{\frac{- x}{4}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{\frac{- x}{4}} d x

Simplify

= \frac{1}{4} \cdot \int e^{- \frac{x}{4}} d x

Apply u-substitution

= \frac{1}{4} \cdot \int - 4 e^{u} d u

Take the constant out:

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

= \frac{1}{4} (- 4 \cdot \int e^{u} d u)

Use the common integral:

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

= \frac{1}{4} (- 4 e^{u})

Substitute back

u = - \frac{x}{4}

= \frac{1}{4} (- 4 e^{- \frac{x}{4}})

Simplify

\frac{1}{4} (- 4 e^{- \frac{x}{4}}) : - e^{- \frac{x}{4}}

= - e^{- \frac{x}{4}}

Add a constant to the solution

= - e^{- \frac{x}{4}} + C

What is the integral of 1/4 e^{(-x)/4} ?
The integral of 1/4 e^{(-x)/4} is -e^{-x/4}+C