integral from 0 to r of (ar^3+br)2pir

Lösung

\int_{0}^{r} (a r^{3} + b r) 2 π r d r

2 π (\frac{a r ^{5}}{5} + \frac{b r ^{3}}{3})

Schritte zur Lösung

\int_{0}^{r} (a r^{3} + b r) \cdot 2 π r d r

Entferne die Konstante:

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

= 2 π \cdot \int_{0}^{r} (a r^{3} + b r) r d r

Multipliziere aus

(a r^{3} + b r) r : a r^{4} + b r^{2}

= 2 π \cdot \int_{0}^{r} a r^{4} + b r^{2} d r

Wende die Summenregel an:

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

= 2 π (\int_{0}^{r} a r^{4} d r + \int_{0}^{r} b r^{2} d r)

\int_{0}^{r} a r^{4} d r = \frac{a r ^{5}}{5}

\int_{0}^{r} b r^{2} d r = \frac{b r ^{3}}{3}

= 2 π (\frac{a r ^{5}}{5} + \frac{b r ^{3}}{3})

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