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

The integral from 0 to 1 of (2t+1)e^{-st} is (-3e^{-s}+1)/s+(-2e^{-s}+2)/(s^2)

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

\int_{0}^{1} (2 t + 1) e^{- s t} d t

\frac{- 3 e ^{- s} + 1}{s} + \frac{- 2 e ^{- s} + 2}{s ^{2}}

Solution steps

\int_{0}^{1} (2 t + 1) e^{- s t} d t

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

- e^{- s} \frac{3}{s} - e^{- s} \frac{2}{s ^{2}} + \frac{1}{s} + \frac{2}{s ^{2}}

= - e^{- s} \frac{3}{s} - e^{- s} \frac{2}{s ^{2}} + \frac{1}{s} + \frac{2}{s ^{2}}

Simplify

= \frac{- 3 e ^{- s} + 1}{s} + \frac{- 2 e ^{- s} + 2}{s ^{2}}

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What is the integral from 0 to 1 of (2t+1)e^{-st} ?
The integral from 0 to 1 of (2t+1)e^{-st} is (-3e^{-s}+1)/s+(-2e^{-s}+2)/(s^2)