integral from 0 to t of xe^{-4x}

Lösung

\int_{0}^{t} x e^{- 4 x} d x

\frac{1}{16} (- 4 t e^{- 4 t} - e^{- 4 t} + 1)

Schritte zur Lösung

\int_{0}^{t} x e^{- 4 x} d x

Wende U-Substitution an

= \int_{0}^{- 4 t} \frac{e ^{u} u}{16} d u

Entferne die Konstante:

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

= \frac{1}{16} \cdot \int_{0}^{- 4 t} e^{u} u d u

Wende die partielle Integration an

= \frac{1}{16} [e^{u} u - \int e^{u} d u]_{0}^{- 4 t}

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

= \frac{1}{16} [e^{u} u - e^{u}]_{0}^{- 4 t}

Berechne die Grenzen:

- 4 t e^{- 4 t} - e^{- 4 t} + 1

= \frac{1}{16} (- 4 t e^{- 4 t} - e^{- 4 t} + 1)

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