What is the integral from 0 to 3 of e^{-st}t ?

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

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integral from 0 to 3 of e^{-st}t

\int_{0}^{3} e^{- s t} t d t

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

Solution steps

\int_{0}^{3} e^{- s t} t d t

Apply u-substitution

: \int_{0}^{- s \cdot 3} \frac{e ^{u} u}{s ^{2}} d u

= \int_{0}^{- s \cdot 3} \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 \cdot 3} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{- 3 s}

= \frac{1}{s ^{2}} [e^{u} u - \int e^{u} d u]_{0}^{- 3 s}

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

= \frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- 3 s}

Simplify

\frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- 3 s} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- 3 s}}{s ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- 3 s}}{s ^{2}}

Compute the boundaries:

- 3 s e^{- 3 s} - e^{- 3 s} + 1

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

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