integral from-2 to 0 of 2e^{4x-1}

Lösung

\int_{- 2}^{0} 2 e^{4 x - 1} d x

\frac{e ^{8} - 1}{2 e ^{9}}

Dezimale

0.18387 \dots

Schritte zur Lösung

\int_{- 2}^{0} 2 e^{4 x - 1} d x

Entferne die Konstante:

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

= 2 \cdot \int_{- 2}^{0} e^{4 x - 1} d x

Wende U-Substitution an

= 2 \cdot \int_{- 9}^{- 1} e^{u} \frac{1}{4} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{4} \cdot \int_{- 9}^{- 1} e^{u} d u

Nutze das gemeinsame Integral :

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

= 2 \cdot \frac{1}{4} [e^{u}]_{- 9}^{- 1}

Vereinfache

2 \cdot \frac{1}{4} [e^{u}]_{- 9}^{- 1} : \frac{1}{2} [e^{u}]_{- 9}^{- 1}

= \frac{1}{2} [e^{u}]_{- 9}^{- 1}

Berechne die Grenzen:

\frac{1}{e} - \frac{1}{e ^{9}}

= \frac{1}{2} (\frac{1}{e} - \frac{1}{e ^{9}})

Vereinfache

= \frac{e ^{8} - 1}{2 e ^{9}}

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