What is the integral of 17xe^{-x^2} ?

The integral of 17xe^{-x^2} is -17/2 e^{-x^2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 17xe^{-x^2}

\int 17 x e^{- x^{2}} d x

- \frac{17}{2} e^{- x^{2}} + C

Solution steps

\int 17 x e^{- x^{2}} d x

Take the constant out:

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

= 17 \cdot \int x e^{- x^{2}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{2} d u

= 17 \cdot \int - \frac{e ^{u}}{2} d u

Take the constant out:

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

= 17 (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= 17 (- \frac{1}{2} e^{u})

Substitute back

u = - x^{2}

= 17 (- \frac{1}{2} e^{- x^{2}})

Simplify

17 (- \frac{1}{2} e^{- x^{2}}) : - \frac{17}{2} e^{- x^{2}}

= - \frac{17}{2} e^{- x^{2}}

Add a constant to the solution

= - \frac{17}{2} e^{- x^{2}} + C

\frac{d y}{d x} = \frac{cos ( 13 x )}{e ^{13 y}}

\int 7^{- 2 x} d x

\frac{d}{d x} (5 x - 2)

x \to \infty lim ((2)^{x})

\frac{d y}{d x} - 4 y = 2 x y^{2}

What is the integral of 17xe^{-x^2} ?
The integral of 17xe^{-x^2} is -17/2 e^{-x^2}+C