integral from 0 to ln(2) of e^{-t}

Lösung

\int_{0}^{l n (2)} e^{- t} d t

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

\int_{0}^{l n (2)} e^{- t} d t

Wende U-Substitution an

= \int_{0}^{- l n (2)} - e^{u} d u

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

= - \int_{- l n (2)}^{0} - e^{u} d u

Entferne die Konstante:

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

= - (- \int_{- l n (2)}^{0} e^{u} d u)

Nutze das gemeinsame Integral :

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

= - (- [e^{u}]_{- l n (2)}^{0})

Vereinfache

= [e^{u}]_{- l n (2)}^{0}

Berechne die Grenzen:

\frac{1}{2}

= \frac{1}{2}

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