integral from 1 to e^2 of 2piyln(y)

Lösung

\int_{1}^{e^{2}} 2 π y ln (y) d y

\frac{π ( 3 e ^{4} + 1 )}{2}

Dezimale

258.85851 \dots

Schritte zur Lösung

\int_{1}^{e^{2}} 2 π y ln (y) d y

Entferne die Konstante:

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

= 2 π \cdot \int_{1}^{e^{2}} y ln (y) d y

Wende die partielle Integration an

= 2 π [\frac{1}{2} (y^{2} ln (y) - 2 \cdot \int \frac{y}{2} d y)]_{1}^{e^{2}}

\int \frac{y}{2} d y = \frac{y ^{2}}{4}

= 2 π [\frac{1}{2} (y^{2} ln (y) - 2 \cdot \frac{y ^{2}}{4})]_{1}^{e^{2}}

2 \cdot \frac{y ^{2}}{4} = \frac{y ^{2}}{2}

= 2 π [\frac{1}{2} (y^{2} ln (y) - \frac{y ^{2}}{2})]_{1}^{e^{2}}

Berechne die Grenzen:

\frac{3 e ^{4}}{4} + \frac{1}{4}

= 2 π (\frac{3 e ^{4}}{4} + \frac{1}{4})

Vereinfache

= \frac{π ( 3 e ^{4} + 1 )}{2}

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