integral from 0 to ln(2) of e^{2x+1}

Lösung

\int_{0}^{l n (2)} e^{2 x + 1} d x

\frac{3 e}{2}

Dezimale

4.07742 \dots

Schritte zur Lösung

\int_{0}^{l n (2)} e^{2 x + 1} d x

Wende U-Substitution an

= \int_{1}^{2 l n (2) + 1} e^{u} \frac{1}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{1}^{2 l n (2) + 1} e^{u} d u

Nutze das gemeinsame Integral :

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

= \frac{1}{2} [e^{u}]_{1}^{2 l n (2) + 1}

Berechne die Grenzen:

3 e

= \frac{1}{2} \cdot 3 e

Vereinfache

= \frac{3 e}{2}

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