What is the integral from 0 to 1 of (1-t)e^{-st} ?

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

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integral from 0 to 1 of (1-t)e^{-st}

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

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

Solution steps

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

Apply Integration By Parts

: [- e^{- s t} \frac{1}{s} (1 - t) - \int e^{- s t} \frac{1}{s} d t]_{0}^{1}

= [- e^{- s t} \frac{1}{s} (1 - t) - \int e^{- s t} \frac{1}{s} d t]_{0}^{1}

\int e^{- s t} \frac{1}{s} d t = - e^{- s t} \frac{1}{s ^{2}}

= [- e^{- s t} \frac{1}{s} (1 - t) - (- e^{- s t} \frac{1}{s ^{2}})]_{0}^{1}

Simplify

= [- e^{- s t} \frac{1}{s} (1 - t) + e^{- s t} \frac{1}{s ^{2}}]_{0}^{1}

Compute the boundaries:

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

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

Simplify

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

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