integral from 0 to t of te^{-kt}

Lösung

\int_{0}^{t} t e^{- k t} d t

\frac{- k e ^{- k t} t - e ^{- k t} + 1}{k ^{2}}

Schritte zur Lösung

\int_{0}^{t} t e^{- k t} d t

Wende U-Substitution an

= \int_{0}^{- k t} \frac{e ^{u} u}{k ^{2}} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{k ^{2}} \cdot \int_{0}^{- k t} e^{u} u d u

Wende die partielle Integration an

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

- k e^{- k t} t - e^{- k t} + 1

= \frac{- k e ^{- k t} t - e ^{- k t} + 1}{k ^{2}}

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