What is the integral from 0 to 1 of e^{-sx}x ?

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

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integral from 0 to 1 of e^{-sx}x

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

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

Solution steps

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

Apply u-substitution

= \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

= \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}}

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