integral from 0 to 4 of te^{-0.5t}

Lösung

\int_{0}^{4} t e^{- 0.5 t} d t

- 4 (- 1 + \frac{3}{e ^{2}})

Dezimale

2.37597 \dots

Schritte zur Lösung

\int_{0}^{4} t e^{- 0.5 t} d t

Wende U-Substitution an

= \int_{0}^{- 2} \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_{- 2}^{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_{- 2}^{0} e^{u} u d u

Vereinfache

= - 4 \cdot \int_{- 2}^{0} e^{u} u d u

Wende die partielle Integration an

= - 4 [e^{u} u - \int e^{u} d u]_{- 2}^{0}

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

= - 4 [e^{u} u - e^{u}]_{- 2}^{0}

Berechne die Grenzen:

- 1 + \frac{3}{e ^{2}}

= - 4 (- 1 + \frac{3}{e ^{2}})