What is the integral from 0 to 1 of te^{-st} ?

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

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 te^{-st}

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

\frac{- s e ^{- s} - e ^{- s} + 1}{s ^{2}}

Solution steps

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

Apply u-substitution

: \int_{0}^{- s} \frac{e ^{u} u}{s ^{2}} d u

= \int_{0}^{- s} \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

= \frac{1}{s ^{2}} \cdot \int_{0}^{- s} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{- s}

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

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

= \frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- s}

Simplify

\frac{1}{s ^{2}} [e^{u} u - e^{u}]_{0}^{- s} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- s}}{s ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- s}}{s ^{2}}

Compute the boundaries:

- s e^{- s} - e^{- s} + 1

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

\frac{d}{d x} (60 x^{4} + 180 x^{3} - 240 x^{2})

f (x) = (- 2 x^{2} - 4)^{3}

\frac{1}{1 + s}

\frac{d}{d x} (\frac{arcsin ( x )}{x})

\frac{\partial}{\partial x} (x^{2} y + 4 y)

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