integral of t^2e^{-18t}

Lösung

\int t^{2} e^{- 18 t} d t

- \frac{1}{2916} e^{- 18 t} (162 t^{2} + 18 t + 1) + C

Schritte zur Lösung

\int t^{2} e^{- 18 t} d t

Wende U-Substitution an

= \int - \frac{e ^{u} u ^{2}}{5832} d u

Entferne die Konstante:

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

= - \frac{1}{5832} \cdot \int e^{u} u^{2} d u

Wende die partielle Integration an

= - \frac{1}{5832} (u^{2} e^{u} - \int 2 u e^{u} d u)

\int 2 u e^{u} d u = 2 (e^{u} u - e^{u})

= - \frac{1}{5832} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Setze in

u = - 18 t

ein

= - \frac{1}{5832} ((- 18 t)^{2} e^{- 18 t} - 2 (e^{- 18 t} (- 18 t) - e^{- 18 t}))

Vereinfache

- \frac{1}{5832} ((- 18 t)^{2} e^{- 18 t} - 2 (e^{- 18 t} (- 18 t) - e^{- 18 t})) : - \frac{1}{2916} e^{- 18 t} (162 t^{2} + 18 t + 1)

= - \frac{1}{2916} e^{- 18 t} (162 t^{2} + 18 t + 1)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2916} e^{- 18 t} (162 t^{2} + 18 t + 1) + C

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