What is the integral of 1/(2k^2)*e^{-x/(2k^2)} ?

The integral of 1/(2k^2)*e^{-x/(2k^2)} is -e^{-x/(2k^2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2k^2)*e^{-x/(2k^2)}

\int \frac{1}{2 k ^{2}} \cdot e^{- \frac{x}{2 k ^{2}}} d x

- e^{- \frac{x}{2 k ^{2}}} + C

Solution steps

\int \frac{1}{2 k ^{2}} e^{- \frac{x}{2 k ^{2}}} d x

Take the constant out:

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

= \frac{1}{2 k ^{2}} \cdot \int e^{- \frac{x}{2 k ^{2}}} d x

Apply u-substitution

: \int e^{- u} \cdot 2 k^{2} d u

= \frac{1}{2 k ^{2}} \cdot \int e^{- u} \cdot 2 k^{2} d u

Take the constant out:

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

= \frac{1}{2 k ^{2}} \cdot 2 k^{2} \cdot \int e^{- u} d u

Apply u-substitution

: \int - e^{v} d v

= \frac{1}{2 k ^{2}} \cdot 2 k^{2} \cdot \int - e^{v} d v

Take the constant out:

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

= \frac{1}{2 k ^{2}} \cdot 2 k^{2} (- \int e^{v} d v)

Use the common integral:

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

= \frac{1}{2 k ^{2}} \cdot 2 k^{2} (- e^{v})

Substitute back

= \frac{1}{2 k ^{2}} \cdot 2 k^{2} (- e^{- \frac{x}{2 k ^{2}}})

Simplify

\frac{1}{2 k ^{2}} \cdot 2 k^{2} (- e^{- \frac{x}{2 k ^{2}}}) : - e^{- \frac{x}{2 k ^{2}}}

= - e^{- \frac{x}{2 k ^{2}}}

Add a constant to the solution

= - e^{- \frac{x}{2 k ^{2}}} + C

\int \frac{1 - u}{1 + u ^{2}} d u

\int x^{2} (x^{3} + 5)^{8} cos ((x^{3} + 5)^{9}) d x

\frac{\partial}{\partial x} (y + x^{2})

x y^{^{'}} + 2 y = - 4 x

\frac{d}{d x} ((x^{3})^{5})

What is the integral of 1/(2k^2)*e^{-x/(2k^2)} ?
The integral of 1/(2k^2)*e^{-x/(2k^2)} is -e^{-x/(2k^2)}+C