What is the integral from 0 to b of e^{-s*t}*t ?

The integral from 0 to b of e^{-s*t}*t is (-bse^{-bs}-e^{-bs}+1)/(s^2)

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integral from 0 to b of e^{-st}t

\int_{0}^{b} e^{- s \cdot t} \cdot t d t

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

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

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

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

\int_{e}^{4} \frac{1}{4 x} d x

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\int_{- 1}^{1} x^{3} sin (x^{2} + 1) d x

\int_{- 2 π}^{- π} sin (x) d x

What is the integral from 0 to b of e^{-s*t}*t ?
The integral from 0 to b of e^{-s*t}*t is (-bse^{-bs}-e^{-bs}+1)/(s^2)