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

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

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

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

\frac{e ^{- N (s + 3)} - 1}{- s - 3}

Solution steps

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

Factor

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

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

Apply u-substitution

: \int_{0}^{- N (s + 3)} \frac{e ^{u}}{- s - 3} d u

= \int_{0}^{- N (s + 3)} \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}^{- N (s + 3)} e^{u} d u

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{- N (s + 3)} - 1

= \frac{e ^{- N (s + 3)} - 1}{- s - 3}

y^{^{'}} - (4 x - y + 1)^{2} = 0

\frac{d}{d x} (\frac{1}{0})

\int sin (2) d x

\frac{d y}{d x} + \frac{3 y}{x} + 2 = 3 x

12 x^{2} y + y^{^{'}} = 5 x^{2}

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