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

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

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

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

\frac{1}{k ^{2}}

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

\int \frac{x}{x ^{2} + 2 x + 1} d x

\frac{d}{d x} (2 x^{2} - 4)

\frac{d}{d x} (ln x^{2} - 4)

f (x) = - 2 x^{2} - 12 x - 4

\frac{t d y}{d t} = t^{3} + 8 t^{3} y

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