integral from 0 to 1 of te^{-t^2}

Lösung

\int_{0}^{1} t e^{- t^{2}} d t

\frac{e - 1}{2 e}

Dezimale

0.31606 \dots

Schritte zur Lösung

\int_{0}^{1} t e^{- t^{2}} d t

Wende U-Substitution an

= \int_{0}^{- 1} - \frac{e ^{u}}{2} d u

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

= - \int_{- 1}^{0} - \frac{e ^{u}}{2} d u

Entferne die Konstante:

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

= - (- \frac{1}{2} \cdot \int_{- 1}^{0} e^{u} d u)

Nutze das gemeinsame Integral :

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

= - (- \frac{1}{2} [e^{u}]_{- 1}^{0})

Vereinfache

= \frac{1}{2} [e^{u}]_{- 1}^{0}

Berechne die Grenzen:

1 - \frac{1}{e}

= \frac{1}{2} (1 - \frac{1}{e})

Vereinfache

= \frac{e - 1}{2 e}

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