What is the integral from pi to 2pi of e^{-st} ?

The integral from pi to 2pi of e^{-st} is -(e^{-2pis}-e^{-pis})/s

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from pi to 2pi of e^{-st}

\int_{π}^{2 π} e^{- s t} d t

- \frac{e ^{- 2 π s} - e ^{- π s}}{s}

Solution steps

\int_{π}^{2 π} e^{- s t} d t

Apply u-substitution

: \int_{- s π}^{- s \cdot 2 π} - \frac{e ^{u}}{s} d u

= \int_{- s π}^{- s \cdot 2 π} - \frac{e ^{u}}{s} d u

Take the constant out:

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

= - \frac{1}{s} \cdot \int_{- s π}^{- s \cdot 2 π} e^{u} d u

Use the common integral:

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

= - \frac{1}{s} [e^{u}]_{- s π}^{- s \cdot 2 π}

Simplify

- \frac{1}{s} [e^{u}]_{- s π}^{- s \cdot 2 π} : - \frac{[ e ^{u} ] _{- π s}^{- 2 π s}}{s}

= - \frac{[ e ^{u} ] _{- π s}^{- 2 π s}}{s}

Compute the boundaries:

e^{- 2 π s} - e^{- π s}

= - \frac{e ^{- 2 π s} - e ^{- π s}}{s}

\int_{2}^{3} 95 + 972 t^{2} - 18 t d t

\int_{- 1}^{1} x^{2} (1 - ∣ x ∣) d x

\int_{- 1}^{1} (x^{2} + 4) (x - 1)^{6} d x

\int_{1}^{e} \frac{4}{x ( 10 + 9 ln ( x ) )} d x

\int_{0}^{2 π} \frac{32}{3} d θ

What is the integral from pi to 2pi of e^{-st} ?
The integral from pi to 2pi of e^{-st} is -(e^{-2pis}-e^{-pis})/s