integral from 1 to e of 8t^2ln(t)

Lösung

\int_{1}^{e} 8 t^{2} ln (t) d t

\frac{8 ( 2 e ^{3} + 1 )}{9}

Dezimale

36.59651 \dots

Schritte zur Lösung

\int_{1}^{e} 8 t^{2} ln (t) d t

Entferne die Konstante:

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

= 8 \cdot \int_{1}^{e} t^{2} ln (t) d t

Wende die partielle Integration an

= 8 [\frac{1}{3} (t^{3} ln (t) - 3 \cdot \int \frac{t ^{2}}{3} d t)]_{1}^{e}

\int \frac{t ^{2}}{3} d t = \frac{t ^{3}}{9}

= 8 [\frac{1}{3} (t^{3} ln (t) - 3 \cdot \frac{t ^{3}}{9})]_{1}^{e}

3 \cdot \frac{t ^{3}}{9} = \frac{t ^{3}}{3}

= 8 [\frac{1}{3} (t^{3} ln (t) - \frac{t ^{3}}{3})]_{1}^{e}

Berechne die Grenzen:

\frac{2 e ^{3}}{9} + \frac{1}{9}

= 8 (\frac{2 e ^{3}}{9} + \frac{1}{9})

Vereinfache

= \frac{8 ( 2 e ^{3} + 1 )}{9}

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