What is the integral from 0 to N of e^{-st}e^{-4t} ?

The integral from 0 to N of e^{-st}e^{-4t} is -(e^{-N(s+4)}-1)/(s+4)

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integral from 0 to N of e^{-st}e^{-4t}

\int_{0}^{N} e^{- s t} e^{- 4 t} d t

- \frac{e ^{- N (s + 4)} - 1}{s + 4}

Solution steps

\int_{0}^{N} e^{- s t} e^{- 4 t} d t

Apply u-substitution

: \int_{1}^{e^{- 4 N}} - \frac{u ^{\frac{s}{4}}}{4} d u

= \int_{1}^{e^{- 4 N}} - \frac{u ^{\frac{s}{4}}}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int_{1}^{e^{- 4 N}} u^{\frac{s}{4}} d u

Apply the Power Rule

: [\frac{4 u ^{\frac{s + 4}{4}}}{s + 4}]_{1}^{e^{- 4 N}}

= - \frac{1}{4} [\frac{4 u ^{\frac{s + 4}{4}}}{s + 4}]_{1}^{e^{- 4 N}}

Compute the boundaries:

\frac{4 e ^{- N (s + 4)}}{s + 4} - \frac{4}{s + 4}

= - \frac{1}{4} (\frac{4 e ^{- N (s + 4)}}{s + 4} - \frac{4}{s + 4})

Simplify

= - \frac{e ^{- N (s + 4)} - 1}{s + 4}

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What is the integral from 0 to N of e^{-st}e^{-4t} ?
The integral from 0 to N of e^{-st}e^{-4t} is -(e^{-N(s+4)}-1)/(s+4)