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

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

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

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

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

Solution steps

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

\frac{d}{d θ} (4 sin (3 θ))

(2 x + e^{y}) d x + x e^{y} d y = 0

\int t e^{3 t} + e^{t} d t

\int \frac{11 x ^{3}}{2 + x ^{2}} d x

\frac{d}{d x} ((1 + \frac{7}{x})^{x})

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