What is the integral from 0 to infinity of t*e^{-st} ?

The integral from 0 to infinity of t*e^{-st} is 1/(s^2)

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integral from 0 to infinity of t*e^{-st}

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

\frac{1}{s ^{2}}

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

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

\frac{d y}{d x}, y = x^{2} + 2

\int_{7}^{8} \frac{x}{x ^{2} + 4 x + 13} d x

\frac{1}{s} e^{- 2 s}

f (x) = 3 ln (x) + 4 e^{x} - 5 x^{2}

\frac{d y}{d x} = y^{2} (e^{x} cos (3 x))

What is the integral from 0 to infinity of t*e^{-st} ?
The integral from 0 to infinity of t*e^{-st} is 1/(s^2)