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^{-Ns+5N}-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 N} - 1}{- s + 5}

Solution steps

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

Simplify

e^{- s t} e^{5 t} : e^{- s t + 5 t}

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

Apply u-substitution

: \int_{0}^{- s N + 5 N} \frac{e ^{u}}{- s + 5} d u

= \int_{0}^{- s N + 5 N} \frac{e ^{u}}{- s + 5} d u

Take the constant out:

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

= \frac{1}{- s + 5} \cdot \int_{0}^{- s N + 5 N} e^{u} d u

Use the common integral:

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

= \frac{1}{- s + 5} [e^{u}]_{0}^{- s N + 5 N}

Simplify

\frac{1}{- s + 5} [e^{u}]_{0}^{- s N + 5 N} : \frac{[ e ^{u} ] _{0}^{- N s + 5 N}}{- s + 5}

= \frac{[ e ^{u} ] _{0}^{- N s + 5 N}}{- s + 5}

Compute the boundaries:

e^{- N s + 5 N} - 1

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

x \to 2 lim (- x + 6)

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\int e^{r} r + e^{r} d r

\int cos^{3} (x) \cdot sin (x) d x

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 e^{2 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^{-Ns+5N}-1)/(-s+5)