integral from 0 to 3 of te^{-0.5t}

Lösung

\int_{0}^{3} t e^{- 0.5 t} d t

1.76869 \dots

Schritte zur Lösung

\int_{0}^{3} t e^{- 0.5 t} d t

Wende U-Substitution an

= \int_{0}^{- 1.5} \frac{2}{0.5} e^{u} u d u

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

= - \int_{- 1.5}^{0} \frac{2}{0.5} e^{u} u d u

Entferne die Konstante:

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

= - \frac{2}{0.5} \cdot \int_{- 1.5}^{0} e^{u} u d u

Vereinfache

= - 4 \cdot \int_{- 1.5}^{0} e^{u} u d u

Wende die partielle Integration an

= - 4 [e^{u} u - \int e^{u} d u]_{- 1.5}^{0}

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

= - 4 [e^{u} u - e^{u}]_{- 1.5}^{0}

Berechne die Grenzen:

- 0.44217 \dots

= - 4 (- 0.44217 \dots)

Vereinfache

= 1.76869 \dots

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