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

Lösung

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

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

Dezimale

0.86466 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende U-Substitution an

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

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

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

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

Vereinfache

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

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

Berechne die Grenzen:

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

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

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