integral from 0 to 3 of sqrt((9-x^2)^3)

Lösung

\int_{0}^{3} (9 - x^{2})^{3} d x

\frac{243 π}{16}

Dezimale

47.71293 \dots

Schritte zur Lösung

\int_{0}^{3} (9 - x^{2})^{3} d x

(9 - x^{2})^{3} = (9 - x^{2})^{\frac{3}{2}},

angenommen

(9 - x^{2}) \geq 0

= \int_{0}^{3} (9 - x^{2})^{\frac{3}{2}} d x

Wende die partielle Integration an

= [x (9 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} 9 - x^{2} d x]_{0}^{3}

\int - 3 x^{2} 9 - x^{2} d x = - \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x)))

= [x (9 - x^{2})^{\frac{3}{2}} - (- \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x))))]_{0}^{3}

Vereinfache

= [x (9 - x^{2})^{\frac{3}{2}} + \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x)))]_{0}^{3}

Berechne die Grenzen:

\frac{243 π}{16}

= \frac{243 π}{16}

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