What is the integral of e^{-(s-a)t} ?

The integral of e^{-(s-a)t} is 1/(a-s)e^{-(-a+s)t}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{-(s-a)t}

\int e^{- (s - a) t} d t

\frac{1}{a - s} e^{- (- a + s) t} + C

Solution steps

\int e^{- (s - a) t} d t

Apply u-substitution

: \int \frac{e ^{u}}{a - s} d u

= \int \frac{e ^{u}}{a - s} d u

Take the constant out:

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

= \frac{1}{a - s} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{a - s} e^{u}

Substitute back

u = - (- a + s) t

= \frac{1}{a - s} e^{- (- a + s) t}

Add a constant to the solution

= \frac{1}{a - s} e^{- (- a + s) t} + C

(3 + x^{5})^{\frac{2}{3}}

\frac{d}{d x} (π \cdot x)

\frac{d}{d t} (cos (t^{2}))

\frac{d}{d x} (\frac{π}{x})

\frac{d}{d x} (- 3 x^{4} - \frac{1}{x} + x x)

What is the integral of e^{-(s-a)t} ?
The integral of e^{-(s-a)t} is 1/(a-s)e^{-(-a+s)t}+C