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

The integral from 0 to 3 of e^{-st}e^{3t} is (e^{-3s+9}-1)/(-s+3)

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

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

\frac{e ^{- 3 s + 9} - 1}{- s + 3}

Solution steps

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

Simplify

e^{- s t} e^{3 t} : e^{- s t + 3 t}

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

Apply u-substitution

: \int_{0}^{- 3 s + 9} \frac{e ^{u}}{- s + 3} d u

= \int_{0}^{- 3 s + 9} \frac{e ^{u}}{- s + 3} d u

Take the constant out:

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

= \frac{1}{- s + 3} \cdot \int_{0}^{- 3 s + 9} e^{u} d u

Use the common integral:

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

= \frac{1}{- s + 3} [e^{u}]_{0}^{- 3 s + 9}

Simplify

\frac{1}{- s + 3} [e^{u}]_{0}^{- 3 s + 9} : \frac{[ e ^{u} ] _{0}^{- 3 s + 9}}{- s + 3}

= \frac{[ e ^{u} ] _{0}^{- 3 s + 9}}{- s + 3}

Compute the boundaries:

e^{- 3 s + 9} - 1

= \frac{e ^{- 3 s + 9} - 1}{- s + 3}

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What is the integral from 0 to 3 of e^{-st}e^{3t} ?
The integral from 0 to 3 of e^{-st}e^{3t} is (e^{-3s+9}-1)/(-s+3)