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^{-Ns+3N}-1)/(-s+3)

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^{3t}

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

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

Solution steps

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

Simplify

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

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

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

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

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

\int \frac{2}{3} x^{2} d x

\int \frac{x ^{3} + 3}{x ^{2}} d x

y^{^{'}} + 4 y = 3 sin (e^{4} x)

\int x^{m} d x

\int \frac{1}{x ^{4} + 4 x ^{2}} d x

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^{-Ns+3N}-1)/(-s+3)