What is the integral from 0 to N of e^{-st}e^{-5t} ?

The integral from 0 to N of e^{-st}e^{-5t} is -(e^{-N(s+5)}-1)/(s+5)

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integral from 0 to N of e^{-st}e^{-5t}

\int_{0}^{N} e^{- s t} e^{- 5 t} d t

- \frac{e ^{- N (s + 5)} - 1}{s + 5}

Solution steps

\int_{0}^{N} e^{- s t} e^{- 5 t} d t

Apply u-substitution

: \int_{1}^{e^{- 5 N}} - \frac{u ^{\frac{s}{5}}}{5} d u

= \int_{1}^{e^{- 5 N}} - \frac{u ^{\frac{s}{5}}}{5} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int_{1}^{e^{- 5 N}} u^{\frac{s}{5}} d u

Apply the Power Rule

: [\frac{5 u ^{\frac{s + 5}{5}}}{s + 5}]_{1}^{e^{- 5 N}}

= - \frac{1}{5} [\frac{5 u ^{\frac{s + 5}{5}}}{s + 5}]_{1}^{e^{- 5 N}}

Compute the boundaries:

\frac{5 e ^{- N (s + 5)}}{s + 5} - \frac{5}{s + 5}

= - \frac{1}{5} (\frac{5 e ^{- N (s + 5)}}{s + 5} - \frac{5}{s + 5})

Simplify

= - \frac{e ^{- N (s + 5)} - 1}{s + 5}

\frac{d}{d x} (\frac{- sin ( x )}{sec ( x ) tan ( x )})

\frac{y d y + x d x}{2 1 + x y} = 0

\frac{\partial}{\partial x} (4 x^{3} y^{\frac{1}{2}})

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 5)^{6})

x \to - 1 lim (\frac{x ^{2}}{x ^{3} - x})

What is the integral from 0 to N of e^{-st}e^{-5t} ?
The integral from 0 to N of e^{-st}e^{-5t} is -(e^{-N(s+5)}-1)/(s+5)