integral from 0 to pi/5 of t^2sin(5t)

Lösung

\int_{0}^{\frac{π}{5}} t^{2} sin (5 t) d t

\frac{π ^{2} - 4}{125}

Dezimale

0.04695 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{5}} t^{2} sin (5 t) d t

Wende die partielle Integration an

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

\int - \frac{2}{5} t cos (5 t) d t = - \frac{2}{125} (5 t sin (5 t) + cos (5 t))

= [\frac{1}{5} (- t^{2} cos (5 t) - 5 (- \frac{2}{125} (5 t sin (5 t) + cos (5 t))))]_{0}^{\frac{π}{5}}

Vereinfache

[\frac{1}{5} (- t^{2} cos (5 t) - 5 (- \frac{2}{125} (5 t sin (5 t) + cos (5 t))))]_{0}^{\frac{π}{5}} : [\frac{1}{5} (- t^{2} cos (5 t) + \frac{2}{25} (5 t sin (5 t) + cos (5 t)))]_{0}^{\frac{π}{5}}

= [\frac{1}{5} (- t^{2} cos (5 t) + \frac{2}{25} (5 t sin (5 t) + cos (5 t)))]_{0}^{\frac{π}{5}}

Berechne die Grenzen:

\frac{π ^{2} - 2}{125} - \frac{2}{125}

= \frac{π ^{2} - 2}{125} - \frac{2}{125}

Vereinfache

= \frac{π ^{2} - 4}{125}

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