integral from 0 to 2 of 64e^{2x}

Lösung

\int_{0}^{2} 64 e^{2 x} d x

32 (e^{4} - 1)

Dezimale

1715.14080 \dots

Schritte zur Lösung

\int_{0}^{2} 64 e^{2 x} d x

Entferne die Konstante:

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

= 64 \cdot \int_{0}^{2} e^{2 x} d x

Wende U-Substitution an

= 64 \cdot \int_{0}^{4} e^{u} \frac{1}{2} d u

Entferne die Konstante:

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

= 64 \cdot \frac{1}{2} \cdot \int_{0}^{4} e^{u} d u

Nutze das gemeinsame Integral :

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

= 64 \cdot \frac{1}{2} [e^{u}]_{0}^{4}

Vereinfache

64 \cdot \frac{1}{2} [e^{u}]_{0}^{4} : 32 [e^{u}]_{0}^{4}

= 32 [e^{u}]_{0}^{4}

Berechne die Grenzen:

e^{4} - 1

= 32 (e^{4} - 1)

\int_{0}^{3} x^{3} - 2 x^{2} - 3 x d x

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