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

Apply u-substitution

: \int - 4 e^{u} d u

= \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

y = f (x), at x = 2

x \to - 2 - lim (\frac{- 3}{x ^{2} - 4})

n = 1 \sum \infty \frac{1}{n ^{3} + 2}

f (x) = x^{2} (1 - 7 x)

\int \frac{2}{x ( x ^{2} + 25 )} d x

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