integral from 0 to (5pi)/6 of x^3sin(2x)

Lösung

\int_{0}^{\frac{5 π}{6}} x^{3} sin (2 x) d x

\frac{27 ( - 25 π ^{2} + 10 3 π + 18 ) - 125 3 π ^{3}}{864 3}

Dezimale

- 7.63108 \dots

Schritte zur Lösung

\int_{0}^{\frac{5 π}{6}} x^{3} sin (2 x) d x

Wende die partielle Integration an

= [\frac{1}{2} (- x^{3} cos (2 x) - 2 \cdot \int - \frac{3}{2} x^{2} cos (2 x) d x)]_{0}^{\frac{5 π}{6}}

\int - \frac{3}{2} x^{2} cos (2 x) d x = - \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))

= [\frac{1}{2} (- x^{3} cos (2 x) - 2 (- \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))]_{0}^{\frac{5 π}{6}}

Vereinfache

[\frac{1}{2} (- x^{3} cos (2 x) - 2 (- \frac{3}{8} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x))))]_{0}^{\frac{5 π}{6}} : [\frac{1}{2} (- x^{3} cos (2 x) + \frac{3}{4} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)))]_{0}^{\frac{5 π}{6}}

= [\frac{1}{2} (- x^{3} cos (2 x) + \frac{3}{4} (2 x^{2} sin (2 x) + 2 x cos (2 x) - sin (2 x)))]_{0}^{\frac{5 π}{6}}

Berechne die Grenzen:

\frac{27 ( - 25 π ^{2} + 10 3 π + 18 ) - 125 3 π ^{3}}{864 3}

= \frac{27 ( - 25 π ^{2} + 10 3 π + 18 ) - 125 3 π ^{3}}{864 3}

\int_{0}^{1} (25 - 4 x) d x

\int_{0}^{π} sin (x - k x) d x

\int_{0}^{1} π x^{4} - x d x

\int_{0}^{1} (11 ln (x)) d x

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