What is the integral from 0 to pi of t^2e^{-st} ?

The integral from 0 to pi of t^2e^{-st} is (-pi^2s^2e^{-pis}+2(-pise^{-pis}-e^{-pis})+2)/(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 pi of t^2e^{-st}

\int_{0}^{π} t^{2} e^{- s t} d t

\frac{- π ^{2} s ^{2} e ^{- π s} + 2 ( - π s e ^{- π s} - e ^{- π s} ) + 2}{s ^{3}}

Solution steps

\int_{0}^{π} t^{2} e^{- s t} d t

Apply Integration By Parts

: [\frac{- e ^{- s t} t ^{2} - s \cdot \int - e ^{- s t} \frac{2}{s} t d t}{s}]_{0}^{π}

= [\frac{- e ^{- s t} t ^{2} - s \cdot \int - e ^{- s t} \frac{2}{s} t d t}{s}]_{0}^{π}

\int - e^{- s t} \frac{2}{s} t d t = - \frac{2}{s ^{3}} (- s e^{- s t} t - e^{- s t})

= [\frac{- e ^{- s t} t ^{2} - s ( - \frac{2}{s ^{3}} ( - s e ^{- s t} t - e ^{- s t} ) )}{s}]_{0}^{π}

Simplify

[\frac{- e ^{- s t} t ^{2} - s ( - \frac{2}{s ^{3}} ( - s e ^{- s t} t - e ^{- s t} ) )}{s}]_{0}^{π} : [\frac{- s ^{2} e ^{- s t} t ^{2} + 2 ( - s e ^{- s t} t - e ^{- s t} )}{s ^{3}}]_{0}^{π}

= [\frac{- s ^{2} e ^{- s t} t ^{2} + 2 ( - s e ^{- s t} t - e ^{- s t} )}{s ^{3}}]_{0}^{π}

Compute the boundaries:

\frac{- π ^{2} s ^{2} e ^{- π s} + 2 ( - π s e ^{- π s} - e ^{- π s} )}{s ^{3}} + \frac{2}{s ^{3}}

= \frac{- π ^{2} s ^{2} e ^{- π s} + 2 ( - π s e ^{- π s} - e ^{- π s} )}{s ^{3}} + \frac{2}{s ^{3}}

Simplify

= \frac{- π ^{2} s ^{2} e ^{- π s} + 2 ( - π s e ^{- π s} - e ^{- π s} ) + 2}{s ^{3}}

y = e^{x}, y = e^{- x}, x = 1

x e^{- 5 x}

\int \frac{x ^{3} + 3 x ^{2} + 6}{x ^{2} + 2} d x

\int \frac{x}{( x ^{2} + 4 )} d x

\int 4 + 3 x d x

What is the integral from 0 to pi of t^2e^{-st} ?
The integral from 0 to pi of t^2e^{-st} is (-pi^2s^2e^{-pis}+2(-pise^{-pis}-e^{-pis})+2)/(s^3)