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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int_{0}^{- s} \frac{e ^{u} u}{s ^{2}} d u

= \int_{0}^{- s} \frac{e ^{u} u}{s ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{s ^{2}} \cdot \int_{0}^{- s} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{- s}

= \frac{1}{s ^{2}} [e^{u} u - \int e^{u} d u]_{0}^{- s}

\int e^{u} d u = e^{u}

= \frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- s}

Simplify

\frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- s} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- s}}{s ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- s}}{s ^{2}}

Compute the boundaries:

- s e^{- s} - e^{- s} + 1

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

\int e^{- 4 t} d t

\frac{d y}{d t} = 6 y + cos (3 t)

\int (\frac{t ^{2} + t + 1}{( t + 3 ) ^{3}}) d t

- y + 10 e^{- t} = y^{^{'}}

\int ln (6 x) d x

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