integral from 0 to t of e^{-3u}

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

= - \frac{1}{3} \cdot \int_{0}^{- 3 t} e^{v} d v

Nutze das gemeinsame Integral :

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

= - \frac{1}{3} [e^{v}]_{0}^{- 3 t}

Berechne die Grenzen:

e^{- 3 t} - 1

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

\int_{a}^{b} [3 f (t) - 2 g (t)] d t

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