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

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

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

\int_{0}^{π} t e^{- s t} d t

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

Solution steps

\int_{0}^{π} t e^{- s t} d t

Apply u-substitution

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

= \int_{0}^{- s π} \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 π} e^{u} u d u

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

- π s e^{- π s} - e^{- π s} + 1

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

\int_{2}^{7} (2 - x) (x - 7) d x

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\int_{4}^{16} \frac{1}{z} d z

\int_{1}^{2} 1 + 9 x d x

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