What is the integral of 4xe^{-2x^2} ?

The integral of 4xe^{-2x^2} is -e^{-2x^2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4xe^{-2x^2}

\int 4 x e^{- 2 x^{2}} d x

- e^{- 2 x^{2}} + C

Solution steps

\int 4 x e^{- 2 x^{2}} d x

Take the constant out:

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

= 4 \cdot \int x e^{- 2 x^{2}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - 2 x^{2}

= 4 (- \frac{1}{4} e^{- 2 x^{2}})

Simplify

4 (- \frac{1}{4} e^{- 2 x^{2}}) : - e^{- 2 x^{2}}

= - e^{- 2 x^{2}}

Add a constant to the solution

= - e^{- 2 x^{2}} + C

\frac{d}{d x} ((cos (x) + 2)^{3})

\int \frac{4 x}{( x ^{2} + 6 ) ^{2}} d x

\frac{d y}{d x} = (x + y)^{2} - 1

\frac{d}{d x} (- \frac{2}{9 x ^{\frac{5}{3}}})

\int 2 x^{2} + 1 d x

What is the integral of 4xe^{-2x^2} ?
The integral of 4xe^{-2x^2} is -e^{-2x^2}+C