integral from 1/2 to 1 of e^{sqrt(2x-1)}

Lösung

\int_{\frac{1}{2}}^{1} e^{2 x - 1} d x

1

Schritte zur Lösung

\int_{\frac{1}{2}}^{1} e^{2 x - 1} d x

Wende U-Substitution an

= \int_{0}^{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_{0}^{1} e^{u} d u

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende die partielle Integration an

= \frac{1}{2} \cdot 2 [e^{v} v - \int e^{v} d v]_{0}^{1}

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

= \frac{1}{2} \cdot 2 [e^{v} v - e^{v}]_{0}^{1}

Vereinfache

\frac{1}{2} \cdot 2 [e^{v} v - e^{v}]_{0}^{1} : [e^{v} v - e^{v}]_{0}^{1}

= [e^{v} v - e^{v}]_{0}^{1}

Berechne die Grenzen:

1

= 1

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