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

The integral of-1/4 e^{-4x} is 1/16 e^{-4x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{16} e^{- 4 x} + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = - 4 x

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

Simplify

- \frac{1}{4} (- \frac{1}{4} e^{- 4 x}) : \frac{1}{16} e^{- 4 x}

= \frac{1}{16} e^{- 4 x}

Add a constant to the solution

= \frac{1}{16} e^{- 4 x} + C

\int_{0}^{1} 2 x (6 - x^{2}) d x

y^{^{''}} - 4 y^{^{'}} = 3 t

\int 1 - cos (x) d x

\int t^{4} + 1 d t

\int csc^{2} (9 θ + 2) d θ

What is the integral of-1/4 e^{-4x} ?
The integral of-1/4 e^{-4x} is 1/16 e^{-4x}+C