integral of r^3sqrt(4-r^2)

Lösung

\int r^{3} 4 - r^{2} d r

\frac{1}{5} (4 - r^{2})^{\frac{5}{2}} - \frac{4}{3} (4 - r^{2})^{\frac{3}{2}} + C

Schritte zur Lösung

\int r^{3} 4 - r^{2} d r

Wende U-Substitution an

= \int - (- u^{2} + 4) u^{2} d u

Entferne die Konstante:

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

= - \int (- u^{2} + 4) u^{2} d u

Multipliziere aus

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

= - \int - u^{4} + 4 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

= - (- \int u^{4} d u + \int 4 u^{2} d u)

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

\int 4 u^{2} d u = \frac{4 u ^{3}}{3}

= - (- \frac{u ^{5}}{5} + \frac{4 u ^{3}}{3})

Setze in

u = 4 - r^{2}

ein

= - (- \frac{( 4 - r ^{2} ) ^{5}}{5} + \frac{4 ( 4 - r ^{2} ) ^{3}}{3})

Vereinfache

- (- \frac{( 4 - r ^{2} ) ^{5}}{5} + \frac{4 ( 4 - r ^{2} ) ^{3}}{3}) : \frac{1}{5} (4 - r^{2})^{\frac{5}{2}} - \frac{4}{3} (4 - r^{2})^{\frac{3}{2}}

= \frac{1}{5} (4 - r^{2})^{\frac{5}{2}} - \frac{4}{3} (4 - r^{2})^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} (4 - r^{2})^{\frac{5}{2}} - \frac{4}{3} (4 - r^{2})^{\frac{3}{2}} + C

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