integral of 1/3 x^3sqrt(9-x^2)

Lösung

\int \frac{1}{3} x^{3} 9 - x^{2} d x

\frac{1}{15} (9 - x^{2})^{\frac{5}{2}} - (9 - x^{2})^{\frac{3}{2}} + C

Schritte zur Lösung

\int \frac{1}{3} x^{3} 9 - x^{2} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int x^{3} 9 - x^{2} d x

Wende U-Substitution an

= \frac{1}{3} \cdot \int - (- u^{2} + 9) u^{2} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} (- \int (- u^{2} + 9) u^{2} d u)

Multipliziere aus

(- u^{2} + 9) u^{2} : - u^{4} + 9 u^{2}

= \frac{1}{3} (- \int - u^{4} + 9 u^{2} d u)

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (- (- \int u^{4} d u + \int 9 u^{2} d u))

\int u^{4} d u = \frac{u ^{5}}{5}

\int 9 u^{2} d u = 3 u^{3}

= \frac{1}{3} (- (- \frac{u ^{5}}{5} + 3 u^{3}))

Setze in

u = 9 - x^{2}

ein

= \frac{1}{3} (- (- \frac{( 9 - x ^{2} ) ^{5}}{5} + 3 (9 - x^{2})^{3}))

Vereinfache

\frac{1}{3} (- (- \frac{( 9 - x ^{2} ) ^{5}}{5} + 3 (9 - x^{2})^{3})) : \frac{1}{15} (9 - x^{2})^{\frac{5}{2}} - (9 - x^{2})^{\frac{3}{2}}

= \frac{1}{15} (9 - x^{2})^{\frac{5}{2}} - (9 - x^{2})^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{15} (9 - x^{2})^{\frac{5}{2}} - (9 - x^{2})^{\frac{3}{2}} + C

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