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

Lösung

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

\frac{4 e ( π erfi ( ln ( 2 ) + \frac{1}{2} ) - π erfi ( \frac{1}{2} ) )}{2}

Schritte zur Lösung

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

Wende U-Substitution an

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

Multipliziere aus

u (u + 1) : u^{2} + u

= \int_{0}^{l n (2)} e^{u^{2} + u} d u

Vervollständige das Quadrat

u^{2} + u : (u + \frac{1}{2})^{2} - \frac{1}{4}

= \int_{0}^{l n (2)} e^{(u + \frac{1}{2})^{2} - \frac{1}{4}} d u

Wende U-Substitution an

= \int_{\frac{1}{2}}^{l n (2) + \frac{1}{2}} e^{v^{2} - \frac{1}{4}} d v

Vereinfache

e^{v^{2} - \frac{1}{4}} : e^{v^{2}} 4 e

= \int_{\frac{1}{2}}^{l n (2) + \frac{1}{2}} e^{v^{2}} 4 e d v

Entferne die Konstante:

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

= 4 e \cdot \int_{\frac{1}{2}}^{l n (2) + \frac{1}{2}} e^{v^{2}} d v

Das ist ein nicht elementares Integral :

\int e^{v^{2}} d v = \frac{π}{2} erfi (v)

= 4 e [\frac{π}{2} erfi (v)]_{\frac{1}{2}}^{l n (2) + \frac{1}{2}}

Berechne die Grenzen:

\frac{π erfi ( ln ( 2 ) + \frac{1}{2} ) - π erfi ( \frac{1}{2} )}{2}

= 4 e \frac{π erfi ( ln ( 2 ) + \frac{1}{2} ) - π erfi ( \frac{1}{2} )}{2}

Vereinfache

= \frac{4 e ( π erfi ( ln ( 2 ) + \frac{1}{2} ) - π erfi ( \frac{1}{2} ) )}{2}

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