What is the integral from 0 to infinity of te^{-pt} ?

The integral from 0 to infinity of te^{-pt} is 1/(p^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to infinity of te^{-pt}

\int_{0}^{\infty} t e^{- pt} d t

\frac{1}{p ^{2}}

Solution steps

\int_{0}^{\infty} t e^{- pt} d t

Compute the indefinite integral:

\int t e^{- pt} d t = \frac{1}{p ^{2}} (- p e^{- pt} t - e^{- pt}) + C

Compute the boundaries:

\int_{0}^{\infty} t e^{- pt} d t = 0 - (- \frac{1}{p ^{2}})

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

Simplify

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

What is the integral from 0 to infinity of te^{-pt} ?
The integral from 0 to infinity of te^{-pt} is 1/(p^2)