integral from 0 to 6 of e^{-1/3 t}

Lösung

\int_{0}^{6} e^{- \frac{1}{3} t} d t

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

Dezimale

2.59399 \dots

Schritte zur Lösung

\int_{0}^{6} e^{- \frac{1}{3} t} d t

Vereinfache

- \frac{1}{3} t : - \frac{t}{3}

= \int_{0}^{6} e^{- \frac{t}{3}} d t

Wende U-Substitution an

= \int_{0}^{- 2} - 3 e^{u} d u

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

= - \int_{- 2}^{0} - 3 e^{u} d u

Entferne die Konstante:

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

= - (- 3 \cdot \int_{- 2}^{0} e^{u} d u)

Nutze das gemeinsame Integral :

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

= - (- 3 [e^{u}]_{- 2}^{0})

Vereinfache

= 3 [e^{u}]_{- 2}^{0}

Berechne die Grenzen:

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

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

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