integral of rsqrt(2-r^2)

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

= \int - \frac{u}{2} d u

Entferne die Konstante:

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

= - \frac{1}{2} \cdot \int u d u

Wende die Potenzregel an

= - \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Setze in

u = 2 - r^{2}

ein

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

Vereinfache

- \frac{1}{2} \cdot \frac{2}{3} (2 - r^{2})^{\frac{3}{2}} : - \frac{1}{3} (2 - r^{2})^{\frac{3}{2}}

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{1}{( 4 x + 1 ) 16 x ^{2} + 4} d x

t an g e n t f (x) = - 1 - 8 x^{2}, at x = - 2

\int_{0}^{1} (x) - (x^{2}) d x

\int (x - 7)^{2} d x

n = 1 \sum \infty \frac{1}{2 ^{n - 1}}