What is the integral of e^{-st}f(t) ?

The integral of e^{-st}f(t) is f 1/(s^2)(-se^{-st}t-e^{-st})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{-st}f(t)

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

f \frac{1}{s ^{2}} (- s e^{- s t} t - e^{- s t}) + C

Solution steps

\int e^{- s t} f t d t

Take the constant out:

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

= f \cdot \int e^{- s t} t d t

Apply u-substitution

: \int \frac{e ^{u} u}{s ^{2}} d u

= f \cdot \int \frac{e ^{u} u}{s ^{2}} d u

Take the constant out:

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

= f \frac{1}{s ^{2}} \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= f \frac{1}{s ^{2}} (e^{u} u - \int e^{u} d u)

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

= f \frac{1}{s ^{2}} (e^{u} u - e^{u})

Substitute back

u = - s t

= f \frac{1}{s ^{2}} (e^{- s t} (- s t) - e^{- s t})

Simplify

= f \frac{1}{s ^{2}} (- s e^{- s t} t - e^{- s t})

Add a constant to the solution

= f \frac{1}{s ^{2}} (- s e^{- s t} t - e^{- s t}) + C

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

\int \frac{e ^{7 x}}{e ^{7 x} + 3} d x

\int sin^{6} (\frac{x}{3}) cos (\frac{x}{3}) d x

\int e^{x} \cdot e^{e^{x}} d x

\int \frac{- 2}{( x - 2 ) ^{2}} d x

What is the integral of e^{-st}f(t) ?
The integral of e^{-st}f(t) is f 1/(s^2)(-se^{-st}t-e^{-st})+C