integral from 0 to 1 of r^2sqrt(1-r^2)

Lösung

\int_{0}^{1} r^{2} 1 - r^{2} d r

\frac{π}{16}

Dezimale

0.19634 \dots

Schritte zur Lösung

\int_{0}^{1} r^{2} 1 - r^{2} d r

Trigonometrische Substitution anwenden

= \int_{0}^{\frac{π}{2}} sin^{2} (u) cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 4 u )}{8} d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (4 u) 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}{8} (\int_{0}^{\frac{π}{2}} 1 d u - \int_{0}^{\frac{π}{2}} cos (4 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (4 u) d u = 0

= \frac{1}{8} (\frac{π}{2} - 0)

Vereinfache

\frac{1}{8} (\frac{π}{2} - 0) : \frac{π}{16}

= \frac{π}{16}

\int_{\frac{π}{3}}^{\frac{π}{2}} sin^{3} (x) d x

\int_{1}^{4} \frac{u - 9}{u} d u

\int_{0}^{1} \frac{4 sin ( x ^{2} )}{x} d x

\int_{1}^{4} \frac{2 + 5 x}{x} d x

\int_{0}^{1} \frac{4 x e ^{x}}{3} d x