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

Popular >

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}}

\int_{0}^{4} (x - 3) (x - 1) d x

\int_{0}^{1} z e^{- z^{2}} d z

\int_{- 1}^{3} \frac{1}{x ^{3}} d x

\int_{0}^{0.5} sin (x^{2}) d x

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