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

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

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

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

\frac{1}{b ^{2}}

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

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