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

The integral from 0 to infinity of te^{-t} is 1

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

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

1

Solution steps

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

Compute the indefinite integral:

\int t e^{- t} d t = - e^{- t} t - e^{- t} + C

Compute the boundaries:

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

= 0 - (- 1)

Simplify

= 1

\frac{x}{x ^{2} + 49}

t \to 2 lim (\frac{t ^{2} - 4}{t - 2})

x^{^{'}} + 4 x = t e^{- 4 t}

\frac{d}{d x} (e^{y} sin (x y))

x \to - \infty lim (2 x - x^{2})

What is the integral from 0 to infinity of te^{-t} ?
The integral from 0 to infinity of te^{-t} is 1