What is the integral from 0 to infinity of xe^{-sx} ?

The integral from 0 to infinity of xe^{-sx} is 1/(s^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to infinity of xe^{-sx}

\int_{0}^{\infty} x e^{- s x} d x

\frac{1}{s ^{2}}

Solution steps

\int_{0}^{\infty} x e^{- s x} d x

Compute the indefinite integral:

\int x e^{- s x} d x = \frac{1}{s ^{2}} (- s e^{- s x} x - e^{- s x}) + C

Compute the boundaries:

\int_{0}^{\infty} x e^{- s x} d x = 0 - (- \frac{1}{s ^{2}})

= 0 - (- \frac{1}{s ^{2}})

Simplify

= \frac{1}{s ^{2}}

\int tan^{5} (x) sec^{6} (x) d x

(2 x - 2)^{4} (x^{2} + x + 1)^{4}

7 sec^{5} (π t - 8)

\int \frac{1}{( 225 - x ^{2} ) ^{\frac{3}{2}}} d x

(y^{2} sin (x)) d x + (\frac{1}{x} - \frac{y}{x}) d y = 0

What is the integral from 0 to infinity of xe^{-sx} ?
The integral from 0 to infinity of xe^{-sx} is 1/(s^2)