integral from 0 to 4 of te^{-0.6t}

Lösung

\int_{0}^{4} t e^{- 0.6 t} d t

1.92099 \dots

Schritte zur Lösung

\int_{0}^{4} t e^{- 0.6 t} d t

Wende U-Substitution an

= \int_{0}^{- 2.4} \frac{5 e ^{u} u}{1.8} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2.4}^{0} \frac{5 e ^{u} u}{1.8} d u

Entferne die Konstante:

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

= - \frac{5}{1.8} \cdot \int_{- 2.4}^{0} e^{u} u d u

Vereinfache

= - 2.77777 \dots \cdot \int_{- 2.4}^{0} e^{u} u d u

Wende die partielle Integration an

= - 2.77777 \dots [e^{u} u - \int e^{u} d u]_{- 2.4}^{0}

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

= - 2.77777 \dots [e^{u} u - e^{u}]_{- 2.4}^{0}

Berechne die Grenzen:

- 0.69155 \dots

= - 2.77777 \dots (- 0.69155 \dots)

Vereinfache

= 1.92099 \dots

\int_{0}^{4} (x^{2} + x - 6) d x

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