What is the integral from 0 to 2 of te^{-st} ?

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

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integral from 0 to 2 of te^{-st}

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

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

Solution steps

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

Apply u-substitution

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

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

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

\int_{0}^{3} \frac{2}{x + 1} d x

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\int_{0}^{x^{2}} 3 sin (x) d x

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