What is the integral from 0 to 1 of e^{-st}(t^2) ?

The integral from 0 to 1 of e^{-st}(t^2) is -(e^{-s})/s+(2(-se^{-s}-e^{-s})+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 1 of e^{-st}(t^2)

\int_{0}^{1} e^{- s t} (t^{2}) d t

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

Solution steps

\int_{0}^{1} e^{- s t} t^{2} d t

Apply Integration By Parts

: [- e^{- s t} \frac{1}{s} t^{2} - \int - e^{- s t} \frac{2}{s} t d t]_{0}^{1}

= [- e^{- s t} \frac{1}{s} t^{2} - \int - e^{- s t} \frac{2}{s} t d t]_{0}^{1}

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

= [- e^{- s t} \frac{1}{s} t^{2} - (- \frac{2}{s ^{3}} (- s e^{- s t} t - e^{- s t}))]_{0}^{1}

Simplify

= [- e^{- s t} \frac{1}{s} t^{2} + \frac{2}{s ^{3}} (- s e^{- s t} t - e^{- s t})]_{0}^{1}

Compute the boundaries:

- e^{- s} \frac{1}{s} + \frac{2}{s ^{3}} (- s e^{- s} - e^{- s}) + \frac{2}{s ^{3}}

= - e^{- s} \frac{1}{s} + \frac{2}{s ^{3}} (- s e^{- s} - e^{- s}) + \frac{2}{s ^{3}}

Simplify

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

\int_{0}^{2 π} (1 - cos (t))^{2} d t

\int_{1}^{x} t^{3} d t

\int_{- π}^{π} cos (n x) sin (m x) d x

\int_{2}^{6} \frac{x}{5 x ^{2} + 1} d x

\int_{0}^{8} y^{3} d y

What is the integral from 0 to 1 of e^{-st}(t^2) ?
The integral from 0 to 1 of e^{-st}(t^2) is -(e^{-s})/s+(2(-se^{-s}-e^{-s})+2)/(s^3)