What is the integral from 0 to N of e^{-st}e^{2t} ?

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

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integral from 0 to N of e^{-st}e^{2t}

\int_{0}^{N} e^{- s t} e^{2 t} d t

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

Solution steps

\int_{0}^{N} e^{- s t} e^{2 t} d t

Simplify

e^{- s t} e^{2 t} : e^{- s t + 2 t}

= \int_{0}^{N} e^{- s t + 2 t} d t

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{- N s + 2 N} - 1

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

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