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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of (2-t)e^{-st}

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int e^{- s t} \frac{1}{s} d t = - e^{- s t} \frac{1}{s ^{2}}

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

\int_{1}^{e} \frac{t ^{5} + 1}{t} d t

\int_{0}^{300} 2 x + 500 d x

\int_{0}^{4} x^{2} - 4 d x

\int_{1}^{2} \frac{v ^{4} + 4 v ^{8}}{v ^{5}} d v

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

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