integral from 0 to sqrt(pi of)xsin^2(x)

Lösung

\int_{0}^{π} x sin^{2} (x) d x

\frac{1}{8} (2 π - 2 π sin (2 π) - cos (2 π)) + \frac{1}{8}

Dezimale

1.19927 \dots

Schritte zur Lösung

\int_{0}^{π} x sin^{2} (x) d x

Wende die partielle Integration an

= [\frac{1}{4} (x (2 x - sin (2 x)) - 4 \cdot \int \frac{1}{2} (x - \frac{1}{2} sin (2 x)) d x)]_{0}^{π}

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

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

Vereinfache

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

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

Berechne die Grenzen:

\frac{1}{8} (2 π - 2 π sin (2 π) - cos (2 π)) + \frac{1}{8}

= \frac{1}{8} (2 π - 2 π sin (2 π) - cos (2 π)) + \frac{1}{8}

\int_{0}^{5} \frac{1}{125 - x ^{3}} d x

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\int_{0}^{3} 6 d x

\int_{- 2}^{6} x d x

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