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^{-Ns+4N}-1)/(-s+4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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 N} - 1}{- s + 4}

Solution steps

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

Simplify

e^{- s t} e^{4 t} : e^{- s t + 4 t}

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

Apply u-substitution

: \int_{0}^{- s N + 4 N} \frac{e ^{u}}{- s + 4} d u

= \int_{0}^{- s N + 4 N} \frac{e ^{u}}{- s + 4} d u

Take the constant out:

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

= \frac{1}{- s + 4} \cdot \int_{0}^{- s N + 4 N} e^{u} d u

Use the common integral:

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

= \frac{1}{- s + 4} [e^{u}]_{0}^{- s N + 4 N}

Simplify

\frac{1}{- s + 4} [e^{u}]_{0}^{- s N + 4 N} : \frac{[ e ^{u} ] _{0}^{- N s + 4 N}}{- s + 4}

= \frac{[ e ^{u} ] _{0}^{- N s + 4 N}}{- s + 4}

Compute the boundaries:

e^{- N s + 4 N} - 1

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

\frac{d}{d x} (\frac{- 9}{2 x ^{5}})

\frac{d y}{d x} = 1 - y^{2}

f (x) = 12 x, at x = 1

\frac{\partial}{\partial y} (\frac{x y z ^{2}}{x y + 1})

f (x) = 6 x, (6, 6)

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^{-Ns+4N}-1)/(-s+4)