integral from 0 to pi/6 of xsin(2x)

Lösung

\int_{0}^{\frac{π}{6}} x sin (2 x) d x

\frac{- π + 3 3}{24}

Dezimale

0.08560 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{6}} x sin (2 x) d x

Wende die partielle Integration an

= [\frac{1}{2} (- x cos (2 x) - 2 \cdot \int - \frac{1}{2} cos (2 x) d x)]_{0}^{\frac{π}{6}}

\int - \frac{1}{2} cos (2 x) d x = - \frac{1}{4} sin (2 x)

= [\frac{1}{2} (- x cos (2 x) - 2 (- \frac{1}{4} sin (2 x)))]_{0}^{\frac{π}{6}}

Vereinfache

[\frac{1}{2} (- x cos (2 x) - 2 (- \frac{1}{4} sin (2 x)))]_{0}^{\frac{π}{6}} : [\frac{1}{2} (- x cos (2 x) + \frac{1}{2} sin (2 x))]_{0}^{\frac{π}{6}}

= [\frac{1}{2} (- x cos (2 x) + \frac{1}{2} sin (2 x))]_{0}^{\frac{π}{6}}

Berechne die Grenzen:

\frac{- π + 3 3}{24}

= \frac{- π + 3 3}{24}

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